https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Anellipticcurveoverq&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-25T03:32:16Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_8&diff=149472 2021 AIME I Problems/Problem 8 2021-03-14T17:03:17Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation &lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;y = |x|.&lt;/math&gt; Then the equation becomes &lt;math&gt;\left|\left|20y-y^2\right|-c\right| = 21&lt;/math&gt;, or &lt;math&gt;\left|y^2-20y\right| = c \pm 21&lt;/math&gt;. Note that since &lt;math&gt;y = |x|&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; is nonnegative, so we only care about nonnegative solutions in &lt;math&gt;y&lt;/math&gt;. Notice that each positive solution in &lt;math&gt;y&lt;/math&gt; gives two solutions in &lt;math&gt;x&lt;/math&gt; (&lt;math&gt;x = \pm y&lt;/math&gt;), whereas if &lt;math&gt;y = 0&lt;/math&gt; is a solution, this only gives one solution in &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;x = 0&lt;/math&gt;. Since the total number of solutions in &lt;math&gt;x&lt;/math&gt; is even, &lt;math&gt;y = 0&lt;/math&gt; must not be a solution. Hence, we require that &lt;math&gt;\left|y^2-20y\right| = c \pm 21&lt;/math&gt; has exactly &lt;math&gt;6&lt;/math&gt; positive solutions and is not solved by &lt;math&gt;y = 0.&lt;/math&gt;<br /> <br /> If &lt;math&gt;c &lt; 21&lt;/math&gt;, then &lt;math&gt;c - 21&lt;/math&gt; is negative, and therefore cannot be the absolute value of &lt;math&gt;y^2 - 20y&lt;/math&gt;. This means the equation's only solutions are in &lt;math&gt;\left|y^2-20y\right| = c + 21&lt;/math&gt;. There is no way for this equation to have &lt;math&gt;6&lt;/math&gt; solutions, since the quadratic &lt;math&gt;y^2-20y&lt;/math&gt; can only take on each of the two values &lt;math&gt;\pm(c + 21)&lt;/math&gt; at most twice, yielding at most &lt;math&gt;4&lt;/math&gt; solutions. Hence, &lt;math&gt;c \ge 21&lt;/math&gt;. &lt;math&gt;c&lt;/math&gt; also can't equal &lt;math&gt;21&lt;/math&gt;, since this would mean &lt;math&gt;y = 0&lt;/math&gt; would solve the equation. Hence, &lt;math&gt;c &gt; 21.&lt;/math&gt;<br /> <br /> At this point, the equation &lt;math&gt;y^2-20y = c \pm 21&lt;/math&gt; will always have exactly &lt;math&gt;2&lt;/math&gt; positive solutions, since &lt;math&gt;y^2-20y&lt;/math&gt; takes on each positive value exactly once when &lt;math&gt;y&lt;/math&gt; is restricted to positive values (graph it to see this), and &lt;math&gt;c \pm 21&lt;/math&gt; are both positive. Therefore, we just need &lt;math&gt;y^2-20y = -(c \pm 21)&lt;/math&gt; to have the remaining &lt;math&gt;4&lt;/math&gt; solutions exactly. This means the horizontal lines at &lt;math&gt;-(c \pm 21)&lt;/math&gt; each intersect the parabola &lt;math&gt;y^2 - 20y&lt;/math&gt; in two places. This occurs when the two lines are above the parabola's vertex &lt;math&gt;(10,-100)&lt;/math&gt;. Hence we have:<br /> &lt;cmath&gt;-(c + 21) &gt; -100&lt;/cmath&gt;<br /> &lt;cmath&gt;c + 21 &lt; 100&lt;/cmath&gt;<br /> &lt;cmath&gt;c &lt; 79&lt;/cmath&gt;<br /> <br /> Hence, the integers &lt;math&gt;c&lt;/math&gt; satisfying the conditions are those satisfying &lt;math&gt;21 &lt; c &lt; 79.&lt;/math&gt; There are &lt;math&gt;\boxed{057}&lt;/math&gt; such integers.<br /> Note: Be careful of counting at the end, you may mess up and get 59.<br /> <br /> ==Solution 2 (also graphing)==<br /> <br /> Graph &lt;math&gt;y=|20|x|-x^2|&lt;/math&gt; (If you are having trouble, look at the description in the next two lines and/or the diagram in solution 3). Notice that we want this to be equal to &lt;math&gt;c-21&lt;/math&gt; and &lt;math&gt;c+21&lt;/math&gt;.<br /> <br /> We see that from left to right, the graph first dips from very positive to &lt;math&gt;0&lt;/math&gt; at &lt;math&gt;x=-20&lt;/math&gt;, then rebounds up to &lt;math&gt;100&lt;/math&gt; at &lt;math&gt;x=-10&lt;/math&gt;, then falls back down to &lt;math&gt;0&lt;/math&gt; at &lt;math&gt;x=0&lt;/math&gt;.<br /> <br /> The positive &lt;math&gt;x&lt;/math&gt; are symmetric, so the graph re-ascends to &lt;math&gt;100&lt;/math&gt; at &lt;math&gt;x=10&lt;/math&gt;, falls back to &lt;math&gt;0&lt;/math&gt; at &lt;math&gt;x=10&lt;/math&gt;, and rises to arbitrarily large values afterwards.<br /> <br /> Now we analyze the &lt;math&gt;y&lt;/math&gt; (varied by &lt;math&gt;c&lt;/math&gt;) values. At &lt;math&gt;y=k&lt;0&lt;/math&gt;, we will have no solutions, as the line &lt;math&gt;y=k&lt;/math&gt; will have no intersections with our graph.<br /> <br /> At &lt;math&gt;y=0&lt;/math&gt;, we will have exactly &lt;math&gt;3&lt;/math&gt; solutions for the three zeroes.<br /> <br /> At &lt;math&gt;y=n&lt;/math&gt; for any &lt;math&gt;n&lt;/math&gt; strictly between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt;, we will have exactly &lt;math&gt;6&lt;/math&gt; solutions.<br /> <br /> At &lt;math&gt;y=100&lt;/math&gt;, we will have &lt;math&gt;4&lt;/math&gt; solutions, because local maxima are reached at &lt;math&gt;x= \pm 10&lt;/math&gt;.<br /> <br /> At &lt;math&gt;y=m&gt;100&lt;/math&gt;, we will have exactly &lt;math&gt;2&lt;/math&gt; solutions.<br /> <br /> To get &lt;math&gt;12&lt;/math&gt; distinct solutions for &lt;math&gt;y=|20|x|-x^2|=c \pm 21&lt;/math&gt;, both &lt;math&gt;c +21&lt;/math&gt; and &lt;math&gt;c-21&lt;/math&gt; must produce &lt;math&gt;6&lt;/math&gt; solutions.<br /> <br /> Thus &lt;math&gt;0&lt;c-21&lt;/math&gt; and &lt;math&gt;c+21&lt;100&lt;/math&gt;, so &lt;math&gt;c \in \{ 22, 23, \dots , 77, 78 \}&lt;/math&gt; is required.<br /> <br /> It is easy to verify that all of these choices of &lt;math&gt;c&lt;/math&gt; produce &lt;math&gt;12&lt;/math&gt; distinct solutions (none overlap), so our answer is &lt;math&gt;\boxed{057}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Piecewise Functions: Analyses and Graphs)==<br /> We take cases for the outermost absolute value, then rearrange: &lt;cmath&gt;\left|20|x|-x^2\right|-c=\pm21.&lt;/cmath&gt;<br /> Let &lt;math&gt;f(x)=\left|20|x|-x^2\right|.&lt;/math&gt; We will rewrite &lt;math&gt;f(x)&lt;/math&gt; as a piecewise function without using any absolute value:<br /> &lt;cmath&gt;f(x) = \begin{cases}<br /> \left|-20x-x^2\right| &amp; \text{if} \ x \le 0 <br /> \begin{cases}<br /> 20x+x^2 &amp; \text{if} \ x\le-20 \\<br /> -20x-x^2 &amp; \text{if} \ -20&lt;x\leq0<br /> \end{cases} \\<br /> \left|20x-x^2\right| &amp; \text{if} \ x &gt; 0<br /> \begin{cases}<br /> 20x-x^2 &amp; \text{if} \ 0&lt;x\leq20 \\<br /> -20x+x^2 &amp; \text{if} \ x&gt;20<br /> \end{cases}<br /> \end{cases}.&lt;/cmath&gt;<br /> We graph &lt;math&gt;f(x)&lt;/math&gt; as shown below, with some key points labeled. The fact that &lt;math&gt;f(x)&lt;/math&gt; is an even function (&lt;math&gt;f(x)=f(-x)&lt;/math&gt; holds for all real numbers &lt;math&gt;x,&lt;/math&gt; from which the graph of &lt;math&gt;f(x)&lt;/math&gt; is symmetric about the &lt;math&gt;y\text{-}&lt;/math&gt;axis) should facilitate the process of graphing.<br /> <br /> [[File:2021 AIME I Problem 8.png|center]]<br /> Graph in Desmos: https://www.desmos.com/calculator/fwvhtltxjr<br /> <br /> Since &lt;math&gt;f(x)-c=\pm21&lt;/math&gt; has &lt;math&gt;12&lt;/math&gt; distinct real solutions, it is clear that each case has &lt;math&gt;6&lt;/math&gt; distinct real solutions geometrically. We need to shift the graph of &lt;math&gt;f(x)&lt;/math&gt; down by &lt;math&gt;c&lt;/math&gt; units:<br /> # For &lt;math&gt;f(x)-c=21&lt;/math&gt; to have &lt;math&gt;6&lt;/math&gt; distinct real solutions, we get &lt;math&gt;0&lt;c&lt;79.&lt;/math&gt;<br /> # For &lt;math&gt;f(x)-c=-21&lt;/math&gt; to have &lt;math&gt;6&lt;/math&gt; distinct real solutions, we get &lt;math&gt;21&lt;c&lt;121.&lt;/math&gt;<br /> Taking the intersection of these two cases gives &lt;math&gt;21&lt;c&lt;79,&lt;/math&gt; from which there are &lt;math&gt;79-21-1=\boxed{057}&lt;/math&gt; such integers &lt;math&gt;c.&lt;/math&gt;<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 4==<br /> <br /> Removing the absolute value bars from the equation successively, we get<br /> &lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;<br /> &lt;cmath&gt;|20|x|-x^2|= c \pm21&lt;/cmath&gt;<br /> &lt;cmath&gt;20|x|-x^2 = \pm c \pm 21&lt;/cmath&gt;<br /> &lt;cmath&gt;x^2 \pm 20x \pm c \pm21 = 0&lt;/cmath&gt;<br /> <br /> The discriminant of this equation is<br /> &lt;cmath&gt;\sqrt{400-4(\pm c \pm 21)}&lt;/cmath&gt;<br /> <br /> Equating the discriminant to &lt;math&gt;0&lt;/math&gt;, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval &lt;math&gt;-79 &lt; c &lt; 79&lt;/math&gt;. However, the number of zeros the equation &lt;math&gt;ax^2+b|x|+k&lt;/math&gt; has is determined by where &lt;math&gt;ax^2+bx+k&lt;/math&gt; and &lt;math&gt;ax^2-bx+k&lt;/math&gt; intersect, namely at &lt;math&gt;(0,k)&lt;/math&gt;. When &lt;math&gt;k&lt;0&lt;/math&gt;, &lt;math&gt;a&gt;0&lt;/math&gt;, &lt;math&gt;ax^2+b|x|+k&lt;/math&gt; will have only &lt;math&gt;2&lt;/math&gt; solutions, and when &lt;math&gt;k&gt;0&lt;/math&gt;, &lt;math&gt;a&gt;0&lt;/math&gt;, then there will be &lt;math&gt;4&lt;/math&gt; real solutions, if they exist at all.<br /> In order to have &lt;math&gt;12&lt;/math&gt; solutions here, we thus need to ensure &lt;math&gt;-c+21&lt;0&lt;/math&gt;, so that exactly &lt;math&gt;2&lt;/math&gt; out of the &lt;math&gt;4&lt;/math&gt; possible equations of the form &lt;math&gt;ax^2+b|x|+k&lt;/math&gt; given above have y-intercepts below &lt;math&gt;0&lt;/math&gt; and only &lt;math&gt;2&lt;/math&gt; real solutions, while the remaining &lt;math&gt;2&lt;/math&gt; equations have &lt;math&gt;4&lt;/math&gt; solutions. This occurs when &lt;math&gt;c&gt;21&lt;/math&gt;, so our final bounds are &lt;math&gt;21&lt;c&lt;79&lt;/math&gt;, giving us &lt;math&gt;\boxed{057}&lt;/math&gt; valid values of &lt;math&gt;c&lt;/math&gt;.<br /> <br /> ==Remark==<br /> The graph of &lt;math&gt;F(x)=\left||20|x|-x^2|-c\right|&lt;/math&gt; is shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp<br /> <br /> Move the slider around for &lt;math&gt;21&lt;c&lt;79&lt;/math&gt; to observe how the graph of &lt;math&gt;F(x)&lt;/math&gt; intersects the line &lt;math&gt;y=21&lt;/math&gt; for &lt;math&gt;12&lt;/math&gt; times.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=131908 2005 AIME II Problems/Problem 11 2020-08-16T15:28:06Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=JfxNr7lv7iQ<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_13&diff=130204 2000 AIME II Problems/Problem 13 2020-08-02T01:42:59Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The [[equation]] &lt;math&gt;2000x^6+100x^5+10x^3+x-2=0&lt;/math&gt; has exactly two real roots, one of which is &lt;math&gt;\frac{m+\sqrt{n}}r&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are integers, &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are relatively prime, and &lt;math&gt;r&gt;0&lt;/math&gt;. Find &lt;math&gt;m+n+r&lt;/math&gt;.<br /> <br /> == Solution ==<br /> We may factor the equation as:{{ref|1}}<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 2000x^6+100x^5+10x^3+x-2&amp;=0\\<br /> 2(1000x^6-1) + x(100x^4+10x^2+1)&amp;=0\\<br /> 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&amp;=0\\<br /> 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&amp;=0\\<br /> (20x^2+x-2)(100x^4+10x^2+1)&amp;=0\\<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Now &lt;math&gt;100x^4+10x^2+1\ge 1&gt;0&lt;/math&gt; for real &lt;math&gt;x&lt;/math&gt;. Thus the real roots must be the roots of the equation &lt;math&gt;20x^2+x-2=0&lt;/math&gt;. By the [[quadratic formula]] the roots of this are:<br /> <br /> &lt;cmath&gt;x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;r=\frac{-1+\sqrt{161}}{40}&lt;/math&gt;, and so the final answer is &lt;math&gt;-1+161+40 = \boxed{200}&lt;/math&gt;.<br /> <br /> &lt;br /&gt;<br /> {{note|1}} A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of &lt;math&gt;x&lt;/math&gt; with half of the polynomial's degree (in this case, divide through by &lt;math&gt;x^3&lt;/math&gt;), and then to use one of the substitutions &lt;math&gt;t = x \pm \frac{1}{x}&lt;/math&gt;. In this case, the substitution &lt;math&gt;t = x\sqrt{10} - \frac{1}{x\sqrt{10}}&lt;/math&gt; gives &lt;math&gt;t^2 + 2 = 10x^2 + \frac 1{10x^2}&lt;/math&gt; and &lt;math&gt;2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}&lt;/math&gt;, which reduces the polynomial to just &lt;math&gt;(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0&lt;/math&gt;. Then one can backwards solve for &lt;math&gt;x&lt;/math&gt;.<br /> <br /> == Solution 2 (Complex Bash)==<br /> It would be really nice if the coefficients were symmetrical. What if we make the substitution, &lt;math&gt;x = -\frac{i}{\sqrt{10}}y&lt;/math&gt;. The the polynomial becomes <br /> <br /> &lt;math&gt;-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2&lt;/math&gt;<br /> <br /> It's symmetric! Dividing by &lt;math&gt;y^3&lt;/math&gt; and rearranging, we get <br /> <br /> &lt;math&gt;-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})&lt;/math&gt;<br /> <br /> Now, if we let &lt;math&gt;z = y + \frac{1}{y}&lt;/math&gt;, we can get the equations <br /> <br /> &lt;math&gt;z = y + \frac{1}{y}&lt;/math&gt; <br /> <br /> &lt;math&gt;z^2 - 2 = y^2 + \frac{1}{y^2}&lt;/math&gt;<br /> <br /> and<br /> <br /> &lt;math&gt;z^3 - 3z = y^3 + \frac{1}{y^3}&lt;/math&gt;<br /> <br /> (These come from squaring &lt;math&gt;z&lt;/math&gt; and subtracting &lt;math&gt;2&lt;/math&gt;, then multiplying that result by &lt;math&gt;z&lt;/math&gt; and subtracting &lt;math&gt;z&lt;/math&gt;) <br /> Plugging this into our polynomial, expanding, and rearranging, we get <br /> <br /> &lt;math&gt;-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})&lt;/math&gt;<br /> <br /> Now, we see that the two &lt;math&gt;i&lt;/math&gt; terms must cancel in order to get this polynomial equal to &lt;math&gt;0&lt;/math&gt;, so what squared equals 3? Plugging in &lt;math&gt;z = \sqrt{3}&lt;/math&gt; into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying &lt;math&gt;z = -\sqrt{3}&lt;/math&gt;, we see that it also works! Great, we use long division on the polynomial by <br /> <br /> &lt;math&gt;(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)&lt;/math&gt; and we get <br /> <br /> &lt;math&gt;2z -(\frac{i}{\sqrt{10}}) = 0&lt;/math&gt;. <br /> <br /> We know that the other two solutions for z wouldn't result in real solutions for &lt;math&gt;x&lt;/math&gt; since we have to solve a quadratic with a negative discriminant, then multiply by &lt;math&gt;-(\frac{i}{\sqrt{10}})&lt;/math&gt;. We get that &lt;math&gt;z = (\frac{i}{-2\sqrt{10}})&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt; (using &lt;math&gt;y + \frac{1}{y} = z&lt;/math&gt;) we get that &lt;math&gt;y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}&lt;/math&gt;, and multiplying this by &lt;math&gt;-(\frac{i}{\sqrt{10}})&lt;/math&gt; (because &lt;math&gt;x = -(\frac{i}{\sqrt{10}})y&lt;/math&gt;) we get that &lt;math&gt;x = \frac{-1 \pm \sqrt{161}}{40}&lt;/math&gt; for a final answer of &lt;math&gt;-1 + 161 + 40 = \boxed{200}&lt;/math&gt;<br /> <br /> -GrizzyProblemSolver79c<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=mAXDdKX52TM<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&diff=130199 2014 AIME II Problems/Problem 14 2020-08-02T00:53:07Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> In &lt;math&gt;\triangle{ABC}, AB=10, \angle{A}=30^\circ&lt;/math&gt; , and &lt;math&gt;\angle{C=45^\circ}&lt;/math&gt;. Let &lt;math&gt;H, D,&lt;/math&gt; and &lt;math&gt;M&lt;/math&gt; be points on the line &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;AH\perp{BC}&lt;/math&gt;, &lt;math&gt;\angle{BAD}=\angle{CAD}&lt;/math&gt;, and &lt;math&gt;BM=CM&lt;/math&gt;. Point &lt;math&gt;N&lt;/math&gt; is the midpoint of the segment &lt;math&gt;HM&lt;/math&gt;, and point &lt;math&gt;P&lt;/math&gt; is on ray &lt;math&gt;AD&lt;/math&gt; such that &lt;math&gt;PN\perp{BC}&lt;/math&gt;. Then &lt;math&gt;AP^2=\dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> unitsize(20);<br /> pair A = MP(&quot;A&quot;,(-5sqrt(3),0)), B = MP(&quot;B&quot;,(0,5),N), C = MP(&quot;C&quot;,(5,0)), M = D(MP(&quot;M&quot;,0.5(B+C),NE)), D = MP(&quot;D&quot;,IP(L(A,incenter(A,B,C),0,2),B--C),N), H = MP(&quot;H&quot;,foot(A,B,C),N), N = MP(&quot;N&quot;,0.5(H+M),NE), P = MP(&quot;P&quot;,IP(A--D,L(N,N-(1,1),0,10)));<br /> D(A--B--C--cycle);<br /> D(B--H--A,blue+dashed);<br /> D(A--D);<br /> D(P--N);<br /> markscalefactor = 0.05;<br /> D(rightanglemark(A,H,B));<br /> D(rightanglemark(P,N,D));<br /> MP(&quot;10&quot;,0.5(A+B)-(-0.1,0.1),NW);<br /> &lt;/asy&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let us just drop the perpendicular from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; and label the point of intersection &lt;math&gt;O&lt;/math&gt;. We will use this point later in the problem.<br /> As we can see, <br /> <br /> <br /> &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;HM&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;AHC&lt;/math&gt; is a &lt;math&gt;45-45-90&lt;/math&gt; triangle, so &lt;math&gt;\angle{HAB}=15^\circ&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;AHD&lt;/math&gt; is &lt;math&gt;30-60-90&lt;/math&gt; triangle.<br /> <br /> <br /> &lt;math&gt;AH&lt;/math&gt; and &lt;math&gt;PN&lt;/math&gt; are parallel lines so &lt;math&gt;PND&lt;/math&gt; is &lt;math&gt;30-60-90&lt;/math&gt; triangle also.<br /> <br /> <br /> Then if we use those informations we get &lt;math&gt;AD=2HD&lt;/math&gt; and <br /> <br /> <br /> &lt;math&gt;PD=2ND&lt;/math&gt; and &lt;math&gt;AP=AD-PD=2HD-2ND=2HN&lt;/math&gt; or &lt;math&gt;AP=2HN=HM&lt;/math&gt;<br /> <br /> <br /> Now we know that &lt;math&gt;HM=AP&lt;/math&gt;, we can find for &lt;math&gt;HM&lt;/math&gt; which is simpler to find.<br /> <br /> <br /> We can use point &lt;math&gt;B&lt;/math&gt; to split it up as &lt;math&gt;HM=HB+BM&lt;/math&gt;,<br /> <br /> <br /> We can chase those lengths and we would get<br /> <br /> <br /> &lt;math&gt;AB=10&lt;/math&gt;, so &lt;math&gt;OB=5&lt;/math&gt;, so &lt;math&gt;BC=5\sqrt{2}&lt;/math&gt;, so &lt;math&gt;BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}&lt;/math&gt;<br /> <br /> <br /> We can also use Law of Sines:<br /> <br /> &lt;cmath&gt;\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}&lt;/cmath&gt;<br /> <br /> Then using right triangle &lt;math&gt;AHB&lt;/math&gt;, we have &lt;math&gt;HB=10 \sin 15^\circ&lt;/math&gt;<br /> <br /> <br /> So &lt;math&gt;HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{2}&lt;/math&gt;.<br /> <br /> <br /> And we know that &lt;math&gt;AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}&lt;/math&gt;.<br /> <br /> <br /> Finally if we calculate &lt;math&gt;(AP)^2&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}&lt;/math&gt;. So our final answer is &lt;math&gt;75+2=77&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;m+n=\boxed{077}&lt;/math&gt;<br /> <br /> <br /> Thank you.<br /> <br /> <br /> -Gamjawon<br /> -edited by srisainandan6 to clarify and correct a small mistake<br /> <br /> ==Solution 2==<br /> Here's a solution that doesn't need &lt;math&gt;\sin 15^\circ&lt;/math&gt;.<br /> <br /> As above, get to &lt;math&gt;AP=HM&lt;/math&gt;. As in the figure, let &lt;math&gt;O&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;BCO&lt;/math&gt; is a 45-45-90 triangle, and &lt;math&gt;ABO&lt;/math&gt; is a 30-60-90 triangle. So &lt;math&gt;BO=5&lt;/math&gt; and &lt;math&gt;AO=5\sqrt{3}&lt;/math&gt;; also, &lt;math&gt;CO=5&lt;/math&gt;, &lt;math&gt;BC=5\sqrt2&lt;/math&gt;, and &lt;math&gt;MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}&lt;/math&gt;. But &lt;math&gt;MO&lt;/math&gt; and &lt;math&gt;AH&lt;/math&gt; are parallel, both being orthogonal to &lt;math&gt;BC&lt;/math&gt;. Therefore &lt;math&gt;MH:AO=MC:CO&lt;/math&gt;, or &lt;math&gt;MH=\dfrac{5\sqrt3}{\sqrt2}&lt;/math&gt;, and we're done.<br /> <br /> ==Solution 3==<br /> Break our diagram into 2 special right triangle by dropping an altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; we then get that &lt;cmath&gt;AC=5+5\sqrt{3}, BC=5\sqrt{2}.&lt;/cmath&gt;<br /> Since &lt;math&gt;\triangle{HCA}&lt;/math&gt; is a 45-45-90,<br /> <br /> &lt;cmath&gt;HC=\frac{5\sqrt2+5\sqrt6}{2}&lt;/cmath&gt; <br /> &lt;math&gt;MC=\frac{BM}{2},&lt;/math&gt; <br /> &lt;cmath&gt;HM=\frac{5\sqrt6}{2}&lt;/cmath&gt; <br /> &lt;cmath&gt;HN=\frac{5\sqrt6}{4}&lt;/cmath&gt; <br /> We know that &lt;math&gt;\triangle{AHD}\simeq \triangle{PND}&lt;/math&gt; and are 30-60-90. <br /> Thus, &lt;cmath&gt;AP=2 \cdot HN=\frac{5\sqrt6}{2}.&lt;/cmath&gt;<br /> <br /> &lt;math&gt;(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}&lt;/math&gt;. So our final answer is &lt;math&gt;75+2=\boxed{077}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> &lt;asy&gt;<br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); <br /> <br /> draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); <br /> /* draw figures */<br /> draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); <br /> draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */<br /> draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); <br /> draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */<br /> draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); <br /> draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); <br /> draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); <br /> draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */<br /> draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); <br /> draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); <br /> draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); <br /> draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); <br /> draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); <br /> draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); <br /> draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); <br /> /* dots and labels */<br /> dot((-1.4934334172297545,2.6953043701763835),dotstyle); <br /> label(&quot;$A$&quot;, (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); <br /> dot((1.1286284157632023,-6.954814372303504),dotstyle); <br /> label(&quot;$B$&quot;, (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); <br /> dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); <br /> label(&quot;$C$&quot;, (8.292420110475998,-6.741880204396438), NE * labelscalefactor); <br /> dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); <br /> label(&quot;$H$&quot;, (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); <br /> dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); <br /> label(&quot;$E$&quot;, (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); <br /> dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); <br /> label(&quot;$L$&quot;, (4.705181710331174,6.441792132441429), NE * labelscalefactor); <br /> dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); <br /> label(&quot;$O$&quot;, (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); <br /> dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); <br /> label(&quot;$D$&quot;, (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); <br /> dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); <br /> label(&quot;$F$&quot;, (4.750879014791618,-7.701523598065745), NE * labelscalefactor); <br /> dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); <br /> label(&quot;$G$&quot;, (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); <br /> dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); <br /> label(&quot;$K$&quot;, (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); <br /> dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> /* end of picture */<br /> &lt;/asy&gt;<br /> Draw the &lt;math&gt;45-45-90 \triangle AHC&lt;/math&gt;. Now, take the perpendicular bisector of &lt;math&gt;BC&lt;/math&gt; to intersect the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; at &lt;math&gt;F, L, G&lt;/math&gt; as shown, and denote &lt;math&gt;O&lt;/math&gt; to be the circumcenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. It is not difficult to see by angle chasing that &lt;math&gt;AHBGO&lt;/math&gt; is cyclic, namely with diameter &lt;math&gt;AB&lt;/math&gt;. Then, by symmetry, &lt;math&gt;EH = HB&lt;/math&gt; and as &lt;math&gt;HB, OG&lt;/math&gt; are both subtended by equal arcs they are equal. Hence, &lt;math&gt;EH = GO&lt;/math&gt;. Now, draw line &lt;math&gt;HL&lt;/math&gt; and intersect it at &lt;math&gt;AC&lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt; in the diagram. It is not hard to use angle chase to arrive at &lt;math&gt;AEOL&lt;/math&gt; a parallelogram, and from our length condition derived earlier, &lt;math&gt;AL \parallel HG&lt;/math&gt;. From here, it is clear that &lt;math&gt;AK = KG&lt;/math&gt;; that is, &lt;math&gt;P&lt;/math&gt; is just the intersection of the perpendicular from &lt;math&gt;K&lt;/math&gt; down to &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt;! After this point, note that &lt;math&gt;AP = PF&lt;/math&gt;. It is easily derived that the circumradius of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\frac{10}{\sqrt{2}}&lt;/math&gt;. Now, &lt;math&gt;APO&lt;/math&gt; is a &lt;math&gt;30-60-90&lt;/math&gt; triangle, and from here it is easy to arrive at the final answer of &lt;math&gt;\boxed{077}&lt;/math&gt;. ~awang11's sol<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=SvJ0wDJphdU<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&diff=130198 2000 AIME I Problems/Problem 9 2020-08-02T00:52:14Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;\log ab = \log a + \log b&lt;/math&gt;, we can reduce the equations to a more recognizable form:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &amp;=&amp; 3 - \log 2000\\<br /> -\log y \log z + \log y + \log z - 1 &amp;=&amp; - \log 2\\<br /> -\log x \log z + \log x + \log z - 1 &amp;=&amp; -1\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be &lt;math&gt;\log x, \log y, \log z&lt;/math&gt; respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a - 1)(b - 1) &amp;=&amp; \log 2 \\<br /> (b-1)(c-1) &amp;=&amp; \log 2 \\<br /> (a-1)(c-1) &amp;=&amp; 1 <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> From here, multiplying the three equations gives <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &amp;=&amp; (\log 2)^2\\<br /> (a-1)(b-1)(c-1) &amp;=&amp; \pm\log 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Dividing the third equation of (*) from this equation, &lt;math&gt;b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1&lt;/math&gt;. This gives &lt;math&gt;y_1 = 20, y_2 = 5&lt;/math&gt;, and the answer is &lt;math&gt;y_1 + y_2 = \boxed{025}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Subtracting the second equation from the first equation yields<br /> &lt;cmath&gt;\begin{align*}<br /> \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &amp;= 3 \\<br /> \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &amp;= 3 \\<br /> \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> \log\frac{x}{z}(1-\log y) &amp;= 0 \\<br /> \end{align*}&lt;/cmath&gt;<br /> If &lt;math&gt;1-\log y=0&lt;/math&gt; then &lt;math&gt;y=10&lt;/math&gt;. Substituting into the first equation yields &lt;math&gt;\log20000=4&lt;/math&gt; which is not possible.<br /> <br /> If &lt;math&gt;\log\frac{x}{z}=0&lt;/math&gt; then &lt;math&gt;\frac{x}{z}=1\Longrightarrow x=z&lt;/math&gt;. Substituting into the third equation gets<br /> &lt;cmath&gt;\begin{align*}<br /> \log x^2-(\log x)(\log x) &amp;= 0 \\<br /> \log x^2-\log x^x &amp;= 0 \\<br /> \log x^{2-x} &amp;= 0 \\<br /> x^{2-x} &amp;= 1 \\<br /> \end{align*}&lt;/cmath&gt;<br /> Thus either &lt;math&gt;x=1&lt;/math&gt; or &lt;math&gt;2-x=0\Longrightarrow x=2&lt;/math&gt;. (Note that here &lt;math&gt;x\neq-1&lt;/math&gt; since logarithm isn't defined for negative number.)<br /> <br /> Substituting &lt;math&gt;x=1&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; into the first equation will obtain &lt;math&gt;y=5&lt;/math&gt; and &lt;math&gt;y=20&lt;/math&gt;, respectively. Thus &lt;math&gt;y_1+y_2=\boxed{25}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> == Solution 3 == <br /> <br /> Let &lt;math&gt;a = \log x&lt;/math&gt;, &lt;math&gt;b = \log y&lt;/math&gt; and &lt;math&gt;c = \log z&lt;/math&gt;. Then the given equations become: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \log 2 + a + b - ab = 1 \\<br /> \log 2 + b + c - bc = 1 \\<br /> a+c = ac \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the first and second equations, solving, and factoring, we get &lt;math&gt;a(1-b) = c(1-b) \implies{a = c}&lt;/math&gt;. Plugging this result into the third equation, we get &lt;math&gt;c = 0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. Substituting each of these values of &lt;math&gt;c&lt;/math&gt; into the second equation, we get &lt;math&gt;b = 1 - \log 2&lt;/math&gt; and &lt;math&gt;b = 1 + \log 2&lt;/math&gt;. Substituting backwards from our original substitution, we get &lt;math&gt;y = 5&lt;/math&gt; and &lt;math&gt;y = 20&lt;/math&gt;, respectively, so our answer is &lt;math&gt;\boxed{025}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=sOyLnGJjVvc&amp;t<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_15&diff=130156 1995 AIME Problems/Problem 15 2020-08-01T20:50:33Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;p_{}&lt;/math&gt; be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encounter a run of &lt;math&gt;5&lt;/math&gt; heads before one encounters a run of &lt;math&gt;2&lt;/math&gt; tails. Given that &lt;math&gt;p_{}&lt;/math&gt; can be written in the form &lt;math&gt;m/n&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Think of the problem as a sequence of &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s. No two &lt;tt&gt;T&lt;/tt&gt;'s can occur in a row, so the sequence is blocks of &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s separated by &lt;tt&gt;T&lt;/tt&gt;'s and ending in &lt;math&gt;5&lt;/math&gt; &lt;tt&gt;H&lt;/tt&gt;'s. Since the first letter could be &lt;tt&gt;T&lt;/tt&gt; or the sequence could start with a block of &lt;tt&gt;H&lt;/tt&gt;'s, the total probability is that &lt;math&gt;3/2&lt;/math&gt; of it has to start with an &lt;tt&gt;H&lt;/tt&gt;. <br /> <br /> The answer to the problem is then the sum of all numbers of the form &lt;math&gt;\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5&lt;/math&gt;, where &lt;math&gt;a,b,c \ldots &lt;/math&gt; are all numbers &lt;math&gt;1-4&lt;/math&gt;, since the blocks of &lt;tt&gt;H&lt;/tt&gt;'s can range from &lt;math&gt;1-4&lt;/math&gt; in length. The sum of all numbers of the form &lt;math&gt;(1/2)^a&lt;/math&gt; is &lt;math&gt;1/2+1/4+1/8+1/16=15/16&lt;/math&gt;, so if there are n blocks of &lt;tt&gt;H&lt;/tt&gt;'s before the final five &lt;tt&gt;H&lt;/tt&gt;'s, the answer can be rewritten as the sum of all numbers of the form &lt;math&gt;\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; ranges from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\infty&lt;/math&gt;, since that's how many blocks of &lt;tt&gt;H&lt;/tt&gt;'s there can be before the final five. This is an infinite geometric series whose sum is &lt;math&gt;\frac{3/64}{1-(15/32)}=\frac{3}{34}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;p_H, p_T&lt;/math&gt; respectively denote the probabilities that a string beginning with &lt;tt&gt;H&lt;/tt&gt;'s and &lt;tt&gt;T&lt;/tt&gt;'s are successful. Thus, <br /> &lt;center&gt;&lt;math&gt;p_T = \frac 12p_H.&lt;/math&gt;&lt;/center&gt;<br /> <br /> A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with &lt;tt&gt;H&lt;/tt&gt; (as there cannot be &lt;math&gt;2&lt;/math&gt; &lt;tt&gt;T&lt;/tt&gt;'s in a row, or be the string HHHHH.<br /> <br /> There is a &lt;math&gt;\frac{1}{16}&lt;/math&gt; probability we roll &lt;math&gt;4&lt;/math&gt; consecutive &lt;tt&gt;H&lt;/tt&gt;'s, and there is a &lt;math&gt;\frac{15}{16}&lt;/math&gt; probability we roll a &lt;tt&gt;T&lt;/tt&gt;. Thus, <br /> <br /> &lt;center&gt;&lt;math&gt;p_H = \left(\frac{15}{16}\right) \cdot \left(\frac 12\right) p_H + \frac{1}{32} \Longrightarrow p_H = \frac{1}{17}.&lt;/math&gt;&lt;/center&gt;<br /> <br /> The answer is &lt;math&gt;p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}&lt;/math&gt;, and &lt;math&gt;m+n = \boxed{037}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> For simplicity, let's compute the complement, namely the probability of getting to &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads.<br /> <br /> Let &lt;math&gt;h_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive heads. Similarly, let &lt;math&gt;t_{i}&lt;/math&gt; denote the probability that we get &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads, given that we have &lt;math&gt;i&lt;/math&gt; consecutive tails. Specifically, &lt;math&gt;h_{5} = 0&lt;/math&gt; and &lt;math&gt;t_{2} = 1&lt;/math&gt;. If we can solve for &lt;math&gt;h_{1}&lt;/math&gt; and &lt;math&gt;t_{1}&lt;/math&gt;, we are done; the answer is simply &lt;math&gt;1/2 * (h_{1} + t_{1})&lt;/math&gt;, since on our first roll, we have equal chances of getting a string with &quot;1 consecutive head&quot; or &quot;1 consecutive tail.&quot;<br /> <br /> Consider solving for &lt;math&gt;t_{1}&lt;/math&gt;. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} h_{1}&lt;/math&gt;<br /> &lt;/center&gt;<br /> <br /> Applying similar logic, we get the equations:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> h_{1} &amp;= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\<br /> h_{2} &amp;= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\<br /> h_{3} &amp;= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\<br /> h_{4} &amp;= \frac{1}{2} h_{5} + \frac{1}{2} t_{1}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;h_{5} = 0&lt;/math&gt;, we get &lt;math&gt;h_{4} = \frac{1}{2} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{3} = \frac{3}{4} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{2} = \frac{7}{8} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} t_{1}&lt;/math&gt; &lt;math&gt;\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}&lt;/math&gt; &lt;math&gt;\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}&lt;/math&gt;.<br /> <br /> So, the probability of reaching &lt;math&gt;2&lt;/math&gt; tails before &lt;math&gt;5&lt;/math&gt; heads is &lt;math&gt;\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}&lt;/math&gt;; we want the complement, &lt;math&gt;\frac{3}{34}&lt;/math&gt;, yielding an answer of &lt;math&gt;3 + 34 = \boxed{037}&lt;/math&gt;.<br /> <br /> Note: The same approach still works if we tried solving the original problem rather than the complement; we would have simply used &lt;math&gt;h_{5} = 1&lt;/math&gt; and &lt;math&gt;t_{2} = 0&lt;/math&gt; instead. The repeated back-substitution is cleaner because we used the complement.<br /> <br /> ==Solution 4(fast)==<br /> Consider what happens in the &quot;endgame&quot; or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are &lt;math&gt;\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}&lt;/math&gt; respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are &lt;math&gt; \frac{3}{64}&lt;/math&gt;. The sum of all these endgame outcomes are &lt;math&gt;\frac{34}{64}&lt;/math&gt;, hence the desired probability is &lt;math&gt;\frac{3}{34}&lt;/math&gt;, and in this case &lt;math&gt;m=3,n=34&lt;/math&gt; so we have &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;<br /> -vsamc<br /> <br /> <br /> ==solution 5 (has lots of variables and steps, but it is very easy to think up) ==<br /> <br /> so the states are<br /> (h=heads t=tails)<br /> <br /> &lt;math&gt;h_1&lt;/math&gt; which is the probability of having exactly 1 H in a row.<br /> &lt;math&gt;h_2&lt;/math&gt; which is the probability of having exactly 2 Hs in a row.<br /> &lt;math&gt;h_3&lt;/math&gt; which is the probability of having exactly 3 Hs in a row.<br /> &lt;math&gt;h_4&lt;/math&gt; which is the probability of having exactly 4 Hs in a row.<br /> &lt;math&gt;h_5&lt;/math&gt; which is the probability of having exactly 5 Hs in a row.<br /> &lt;math&gt;t_1&lt;/math&gt; which is the probability of having exactly 1 T in a row.<br /> &lt;math&gt;t_2&lt;/math&gt; which is the probability of having exactly 2 Ts in a row.<br /> <br /> <br /> (1)it is easy to see that(if you dont get the skipped steps, just ask) &lt;math&gt;h_2 = \frac{h_1}{2}&lt;/math&gt;, &lt;math&gt;h_3 = \frac{h_1}{4}&lt;/math&gt;, &lt;math&gt;h_4 = \frac{h_1}{8}&lt;/math&gt; and &lt;math&gt;h_5 = \frac{h_1}{16}&lt;/math&gt;<br /> <br /> (2)we can also see &lt;math&gt;h_1 = \frac{1}{2} + \frac{t_1}{2}&lt;/math&gt; because the first move has probability &lt;math&gt;\frac{1}{2}&lt;/math&gt; of landing heads, and any time there is tails, there is a &lt;math&gt;\frac{1}{2}&lt;/math&gt; chance of getting a head.<br /> <br /> (3)obviously &lt;math&gt;t_2 = \frac{t_1}{2}&lt;/math&gt; <br /> <br /> (4)we also have &lt;math&gt;t_1 = \frac{1}{2} + \frac{h_1}{2} +\frac{h_2}{2} +\frac{h_3}{2} + \frac{h_4}{2}&lt;/math&gt; and by then by using (1), &lt;math&gt;t_1 = \frac{1}{2} + \frac{15h_1}{16}&lt;/math&gt;<br /> <br /> (5)so using (2) and (4) we can solve for &lt;math&gt;t_1&lt;/math&gt; and &lt;math&gt;h_1&lt;/math&gt;. so we will get &lt;math&gt;h_1 = \frac{24}{17}&lt;/math&gt;, and &lt;math&gt;t_1 = \frac{31}{17}&lt;/math&gt;. <br /> <br /> so since &lt;math&gt;h_5 = \frac{h_1}{16}&lt;/math&gt;, we now know &lt;math&gt;h_5 = \frac{3}{34}&lt;/math&gt;, and since &lt;math&gt;t_2 = \frac{t_1}{2}&lt;/math&gt;, &lt;math&gt;t_2 = \frac{31}{34}&lt;/math&gt;.<br /> <br /> what we are looking for is &lt;math&gt;\frac{h_5}{h_5+t_2}&lt;/math&gt; which gives the answer &lt;math&gt;\frac{3}{34}&lt;/math&gt;<br /> <br /> <br /> <br /> intelligence_20<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=Vo-4w5Cor9w&amp;t<br /> <br /> == See also ==<br /> {{AIME box|year=1995|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_10&diff=129240 2000 AIME I Problems/Problem 10 2020-07-25T19:42:53Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> A [[sequence]] of numbers &lt;math&gt;x_{1},x_{2},x_{3},\ldots,x_{100}&lt;/math&gt; has the property that, for every [[integer]] &lt;math&gt;k&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100,&lt;/math&gt; inclusive, the number &lt;math&gt;x_{k}&lt;/math&gt; is &lt;math&gt;k&lt;/math&gt; less than the sum of the other &lt;math&gt;99&lt;/math&gt; numbers. Given that &lt;math&gt;x_{50} = m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Let the sum of all of the terms in the sequence be &lt;math&gt;\mathbb{S}&lt;/math&gt;. Then for each integer &lt;math&gt;k&lt;/math&gt;, &lt;math&gt;x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k&lt;/math&gt;. Summing this up for all &lt;math&gt;k&lt;/math&gt; from &lt;math&gt;1, 2, \ldots, 100&lt;/math&gt;,<br /> <br /> &lt;cmath&gt;\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &amp;= 1 + 2 + \cdots + 100\\<br /> 100\mathbb{S} - 2\mathbb{S} &amp;= \frac{100 \cdot 101}{2} = 5050\\<br /> \mathbb{S}&amp;=\frac{2525}{49}\end{align*}&lt;/cmath&gt;<br /> <br /> Now, substituting for &lt;math&gt;x_{50}&lt;/math&gt;, we get &lt;math&gt;2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}&lt;/math&gt;, and the answer is &lt;math&gt;75+98=\boxed{173}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Consider &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;x_{k+1}&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the sum of the rest 98 terms. Then &lt;math&gt;x_k+k=S+x_{k+1}&lt;/math&gt; and &lt;math&gt;x_{k+1}+(k+1)=S+x_k.&lt;/math&gt; Eliminating &lt;math&gt;S&lt;/math&gt; we have &lt;math&gt;x_{k+1}-x_k=-\dfrac{1}{2}.&lt;/math&gt; So the sequence is arithmetic with common difference &lt;math&gt;-\dfrac{1}{2}.&lt;/math&gt; <br /> <br /> In terms of &lt;math&gt;x_{50},&lt;/math&gt; the sequence is &lt;math&gt;x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.&lt;/math&gt; Therefore &lt;math&gt;x_{50}+50=99x_{50}-\dfrac{50}{2}&lt;/math&gt;. We are done by solving for &lt;math&gt;x_{50}&lt;/math&gt;.<br /> <br /> -JZ<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=TdvxgrSZTQw<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_15&diff=128826 2006 AIME II Problems/Problem 15 2020-07-21T22:03:44Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Given that &lt;math&gt; x, y, &lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are [[real number]]s that satisfy:<br /> <br /> &lt;center&gt;&lt;math&gt; x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} &lt;/math&gt; &lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt; y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} &lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt; z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}&lt;/math&gt;&lt;/center&gt;<br /> <br /> and that &lt;math&gt; x+y+z = \frac{m}{\sqrt{n}}, &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are positive integers and &lt;math&gt; n &lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt; m+n.&lt;/math&gt;<br /> <br /> == Solution ==<br /> Let &lt;math&gt;\triangle XYZ&lt;/math&gt; be a triangle with sides of length &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;, and suppose this triangle is acute (so all [[altitude]]s are on the interior of the triangle).<br /> Let the altitude to the side of length &lt;math&gt;x&lt;/math&gt; be of length &lt;math&gt;h_x&lt;/math&gt;, and similarly for &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. Then we have by two applications of the [[Pythagorean Theorem]] that &lt;math&gt;x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}&lt;/math&gt;. As a [[function]] of &lt;math&gt;h_x&lt;/math&gt;, the [[RHS]] of this equation is strictly decreasing, so it takes each value in its [[range]] exactly once. Thus we must have that &lt;math&gt;h_x^2 = \frac1{16}&lt;/math&gt; and so &lt;math&gt;h_x = \frac{1}4&lt;/math&gt; and similarly &lt;math&gt;h_y = \frac15&lt;/math&gt; and &lt;math&gt;h_z = \frac16&lt;/math&gt;.<br /> <br /> Since the area of the triangle must be the same no matter how we measure, &lt;math&gt;x\cdot h_x = y\cdot h_y = z \cdot h_z&lt;/math&gt; and so &lt;math&gt;\frac x4 = \frac y5 = \frac z6 = 2A&lt;/math&gt; and &lt;math&gt;x = 8A, y = 10A&lt;/math&gt; and &lt;math&gt;z = 12A&lt;/math&gt;. The [[semiperimeter]] of the triangle is &lt;math&gt;s = \frac{8A + 10A + 12A}{2} = 15A&lt;/math&gt; so by [[Heron's formula]] we have &lt;math&gt;A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}&lt;/math&gt;. Thus &lt;math&gt;A = \frac{1}{15\sqrt{7}}&lt;/math&gt; and &lt;math&gt;x + y + z = 30A = \frac2{\sqrt{7}}&lt;/math&gt; and the answer is &lt;math&gt;2 + 7 = \boxed{9}&lt;/math&gt;.<br /> <br /> <br /> Justification that there is an acute triangle with sides of length &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;:<br /> <br /> Note that &lt;math&gt;x, y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; are each the sum of two positive [[square root]]s of real numbers, so &lt;math&gt;x, y, z \geq 0&lt;/math&gt;. (Recall that, by [[AIME]] [[mathematical convention| convention]], all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, &lt;math&gt;\sqrt{y^2-\frac{1}{16}} &lt; \sqrt{y^2} = y&lt;/math&gt;, so we have &lt;math&gt;x &lt; y + z&lt;/math&gt;, &lt;math&gt;y &lt; z + x&lt;/math&gt; and &lt;math&gt;z &lt; x + y&lt;/math&gt;. But these conditions are exactly those of the [[triangle inequality]], so there does exist such a triangle.<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=M6sC26dzb_I<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_5&diff=128825 2010 AIME II Problems/Problem 5 2020-07-21T21:56:13Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Positive numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; satisfy &lt;math&gt;xyz = 10^{81}&lt;/math&gt; and &lt;math&gt;(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468&lt;/math&gt;. Find &lt;math&gt;\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}&lt;/math&gt;.<br /> <br /> == Solution == <br /> Using the properties of logarithms, &lt;math&gt;\log_{10}xyz = 81&lt;/math&gt; by taking the log base 10 of both sides, and &lt;math&gt;(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468&lt;/math&gt; by using the fact that &lt;math&gt;\log_{10}ab = \log_{10}a + \log_{10}b&lt;/math&gt;. <br /> <br /> Through further simplification, we find that &lt;math&gt;\log_{10}x+\log_{10}y+\log_{10}z = 81&lt;/math&gt;. It can be seen that there is enough information to use the formula &lt;math&gt;\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc&lt;/math&gt;, as we have both &lt;math&gt;\ a+b+c&lt;/math&gt; and &lt;math&gt;\ 2ab+2ac+2bc&lt;/math&gt;, and we want to find &lt;math&gt;\sqrt {a^2 + b^2 + c^2}&lt;/math&gt;. <br /> <br /> After plugging in the values into the equation, we find that &lt;math&gt;\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2&lt;/math&gt; is equal to &lt;math&gt;\ 6561 - 936 = 5625&lt;/math&gt;. <br /> <br /> However, we want to find &lt;math&gt;\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}&lt;/math&gt;, so we take the square root of &lt;math&gt;\ 5625&lt;/math&gt;, or &lt;math&gt;\boxed{075}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;a = \log{x}&lt;/math&gt;<br /> <br /> &lt;math&gt;b = \log{y}&lt;/math&gt;<br /> <br /> &lt;math&gt;c = \log{z}&lt;/math&gt;<br /> <br /> &lt;math&gt;xyz = 10^{81}&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{xyz} = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{x} + \log{y} + \log{z} = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;a + b + c = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468&lt;/math&gt;<br /> <br /> &lt;math&gt;a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561&lt;/math&gt;<br /> <br /> &lt;math&gt;a^2 + b^2 + c^2 = 5625 = 75^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}&lt;/math&gt;<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=Ix6czB_A_Js&amp;t<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=4|num-a=6|n=II}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_5&diff=128822 2010 AIME II Problems/Problem 5 2020-07-21T21:55:24Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Positive numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; satisfy &lt;math&gt;xyz = 10^{81}&lt;/math&gt; and &lt;math&gt;(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468&lt;/math&gt;. Find &lt;math&gt;\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}&lt;/math&gt;.<br /> <br /> == Solution == <br /> Using the properties of logarithms, &lt;math&gt;\log_{10}xyz = 81&lt;/math&gt; by taking the log base 10 of both sides, and &lt;math&gt;(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468&lt;/math&gt; by using the fact that &lt;math&gt;\log_{10}ab = \log_{10}a + \log_{10}b&lt;/math&gt;. <br /> <br /> Through further simplification, we find that &lt;math&gt;\log_{10}x+\log_{10}y+\log_{10}z = 81&lt;/math&gt;. It can be seen that there is enough information to use the formula &lt;math&gt;\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc&lt;/math&gt;, as we have both &lt;math&gt;\ a+b+c&lt;/math&gt; and &lt;math&gt;\ 2ab+2ac+2bc&lt;/math&gt;, and we want to find &lt;math&gt;\sqrt {a^2 + b^2 + c^2}&lt;/math&gt;. <br /> <br /> After plugging in the values into the equation, we find that &lt;math&gt;\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2&lt;/math&gt; is equal to &lt;math&gt;\ 6561 - 936 = 5625&lt;/math&gt;. <br /> <br /> However, we want to find &lt;math&gt;\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}&lt;/math&gt;, so we take the square root of &lt;math&gt;\ 5625&lt;/math&gt;, or &lt;math&gt;\boxed{075}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;a = \log{x}&lt;/math&gt;<br /> <br /> &lt;math&gt;b = \log{y}&lt;/math&gt;<br /> <br /> &lt;math&gt;c = \log{z}&lt;/math&gt;<br /> <br /> &lt;math&gt;xyz = 10^{81}&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{xyz} = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{x} + \log{y} + \log{z} = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;a + b + c = 81&lt;/math&gt;<br /> <br /> &lt;math&gt;\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468&lt;/math&gt;<br /> <br /> &lt;math&gt;a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561&lt;/math&gt;<br /> <br /> &lt;math&gt;a^2 + b^2 + c^2 = 5625 = 75^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}&lt;/math&gt;<br /> <br /> ==Video solution==<br /> <br /> https://www.youtube.com/watch?v=Ix6czB_A_Js&amp;t=2s<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=4|num-a=6|n=II}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&diff=127976 2000 AIME I Problems/Problem 9 2020-07-10T12:41:21Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;\log ab = \log a + \log b&lt;/math&gt;, we can reduce the equations to a more recognizable form:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &amp;=&amp; 3 - \log 2000\\<br /> -\log y \log z + \log y + \log z - 1 &amp;=&amp; - \log 2\\<br /> -\log x \log z + \log x + \log z - 1 &amp;=&amp; -1\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be &lt;math&gt;\log x, \log y, \log z&lt;/math&gt; respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a - 1)(b - 1) &amp;=&amp; \log 2 \\<br /> (b-1)(c-1) &amp;=&amp; \log 2 \\<br /> (a-1)(c-1) &amp;=&amp; 1 <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> From here, multiplying the three equations gives <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &amp;=&amp; (\log 2)^2\\<br /> (a-1)(b-1)(c-1) &amp;=&amp; \pm\log 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Dividing the third equation of (*) from this equation, &lt;math&gt;b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1&lt;/math&gt;. This gives &lt;math&gt;y_1 = 20, y_2 = 5&lt;/math&gt;, and the answer is &lt;math&gt;y_1 + y_2 = \boxed{025}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Subtracting the second equation from the first equation yields<br /> &lt;cmath&gt;\begin{align*}<br /> \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &amp;= 3 \\<br /> \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &amp;= 3 \\<br /> \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> \log\frac{x}{z}(1-\log y) &amp;= 0 \\<br /> \end{align*}&lt;/cmath&gt;<br /> If &lt;math&gt;1-\log y=0&lt;/math&gt; then &lt;math&gt;y=10&lt;/math&gt;. Substituting into the first equation yields &lt;math&gt;\log20000=4&lt;/math&gt; which is not possible.<br /> <br /> If &lt;math&gt;\log\frac{x}{z}=0&lt;/math&gt; then &lt;math&gt;\frac{x}{z}=1\Longrightarrow x=z&lt;/math&gt;. Substituting into the third equation gets<br /> &lt;cmath&gt;\begin{align*}<br /> \log x^2-(\log x)(\log x) &amp;= 0 \\<br /> \log x^2-\log x^x &amp;= 0 \\<br /> \log x^{2-x} &amp;= 0 \\<br /> x^{2-x} &amp;= 1 \\<br /> \end{align*}&lt;/cmath&gt;<br /> Thus either &lt;math&gt;x=1&lt;/math&gt; or &lt;math&gt;2-x=0\Longrightarrow x=2&lt;/math&gt;. (Note that here &lt;math&gt;x\neq-1&lt;/math&gt; since logarithm isn't defined for negative number.)<br /> <br /> Substituting &lt;math&gt;x=1&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; into the first equation will obtain &lt;math&gt;y=5&lt;/math&gt; and &lt;math&gt;y=20&lt;/math&gt;, respectively. Thus &lt;math&gt;y_1+y_2=\boxed{25}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> == Solution 3 == <br /> <br /> Let &lt;math&gt;a = \log x&lt;/math&gt;, &lt;math&gt;b = \log y&lt;/math&gt; and &lt;math&gt;c = \log z&lt;/math&gt;. Then the given equations become: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \log 2 + a + b - ab = 1 \\<br /> \log 2 + b + c - bc = 1 \\<br /> a+c = ac \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the first and second equations, solving, and factoring, we get &lt;math&gt;a(1-b) = c(1-b) \implies{a = c}&lt;/math&gt;. Plugging this result into the third equation, we get &lt;math&gt;c = 0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. Substituting each of these values of &lt;math&gt;c&lt;/math&gt; into the second equation, we get &lt;math&gt;b = 1 - \log 2&lt;/math&gt; and &lt;math&gt;b = 1 + \log 2&lt;/math&gt;. Substituting backwards from our original substitution, we get &lt;math&gt;y = 5&lt;/math&gt; and &lt;math&gt;y = 20&lt;/math&gt;, respectively, so our answer is &lt;math&gt;\boxed{025}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&diff=127925 2000 AIME I Problems/Problem 9 2020-07-09T21:20:18Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;\log ab = \log a + \log b&lt;/math&gt;, we can reduce the equations to a more recognizable form:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &amp;=&amp; 3 - \log 2000\\<br /> -\log y \log z + \log y + \log z - 1 &amp;=&amp; - \log 2\\<br /> -\log x \log z + \log x + \log z - 1 &amp;=&amp; -1\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be &lt;math&gt;\log x, \log y, \log z&lt;/math&gt; respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a - 1)(b - 1) &amp;=&amp; \log 2 \\<br /> (b-1)(c-1) &amp;=&amp; \log 2 \\<br /> (a-1)(c-1) &amp;=&amp; 1 <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> From here, multiplying the three equations gives <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &amp;=&amp; (\log 2)^2\\<br /> (a-1)(b-1)(c-1) &amp;=&amp; \pm\log 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Dividing the third equation of (*) from this equation, &lt;math&gt;b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1&lt;/math&gt;. This gives &lt;math&gt;y_1 = 20, y_2 = 5&lt;/math&gt;, and the answer is &lt;math&gt;y_1 + y_2 = \boxed{025}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Subtracting the second equation from the first equation yields<br /> &lt;cmath&gt;\begin{align*}<br /> \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &amp;= 3 \\<br /> \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &amp;= 3 \\<br /> \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> \log\frac{x}{z}(1-\log y) &amp;= 0 \\<br /> \end{align*}&lt;/cmath&gt;<br /> If &lt;math&gt;1-\log y=0&lt;/math&gt; then &lt;math&gt;y=10&lt;/math&gt;. Substituting into the first equation yields &lt;math&gt;\log20000=4&lt;/math&gt; which is not possible.<br /> <br /> If &lt;math&gt;\log\frac{x}{z}=0&lt;/math&gt; then &lt;math&gt;\frac{x}{z}=1\Longrightarrow x=z&lt;/math&gt;. Substituting into the third equation gets<br /> &lt;cmath&gt;\begin{align*}<br /> \log x^2-(\log x)(\log x) &amp;= 0 \\<br /> \log x^2-\log x^x &amp;= 0 \\<br /> \log x^{2-x} &amp;= 0 \\<br /> x^{2-x} &amp;= 1 \\<br /> \end{align*}&lt;/cmath&gt;<br /> Thus either &lt;math&gt;x=1&lt;/math&gt; or &lt;math&gt;2-x=0\Longrightarrow x=2&lt;/math&gt;. (Note that here &lt;math&gt;x\neq-1&lt;/math&gt; since logarithm isn't defined for negative number.)<br /> <br /> Substituting &lt;math&gt;x=1&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; into the first equation will obtain &lt;math&gt;y=5&lt;/math&gt; and &lt;math&gt;y=20&lt;/math&gt;, respectively. Thus &lt;math&gt;y_1+y_2=\boxed{25}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> === Solution 3 === <br /> <br /> Let &lt;math&gt;a = \log x&lt;/math&gt;, &lt;math&gt;b = \log y&lt;/math&gt; and &lt;math&gt;c = \log z&lt;/math&gt;. Then the given equations become: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \log 2 + a + b - ab = 1 \\<br /> \log 2 + b + c - bc = 1 \\<br /> a+c = ac \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the first and second equations, solving, and factoring, we get &lt;math&gt;a(1-b) = c(1-b) \implies{a = c}&lt;/math&gt;. Plugging this result into the third equation, we get &lt;math&gt;c = 0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. Substituting each of these values of &lt;math&gt;c&lt;/math&gt; into the second equation, we get &lt;math&gt;b = 1 - \log 2&lt;/math&gt; and &lt;math&gt;b = 1 + \log 2&lt;/math&gt;. Substituting backwards from our original substitution, we get &lt;math&gt;y = 5&lt;/math&gt; and &lt;math&gt;y = 20&lt;/math&gt;, respectively, so our answer is &lt;math&gt;\boxed{025}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&diff=127923 2000 AIME I Problems/Problem 9 2020-07-09T21:19:21Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;\log ab = \log a + \log b&lt;/math&gt;, we can reduce the equations to a more recognizable form:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &amp;=&amp; 3 - \log 2000\\<br /> -\log y \log z + \log y + \log z - 1 &amp;=&amp; - \log 2\\<br /> -\log x \log z + \log x + \log z - 1 &amp;=&amp; -1\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be &lt;math&gt;\log x, \log y, \log z&lt;/math&gt; respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a - 1)(b - 1) &amp;=&amp; \log 2 \\<br /> (b-1)(c-1) &amp;=&amp; \log 2 \\<br /> (a-1)(c-1) &amp;=&amp; 1 <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> From here, multiplying the three equations gives <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &amp;=&amp; (\log 2)^2\\<br /> (a-1)(b-1)(c-1) &amp;=&amp; \pm\log 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Dividing the third equation of (*) from this equation, &lt;math&gt;b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1&lt;/math&gt;. This gives &lt;math&gt;y_1 = 20, y_2 = 5&lt;/math&gt;, and the answer is &lt;math&gt;y_1 + y_2 = \boxed{025}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Subtracting the second equation from the first equation yields<br /> &lt;cmath&gt;\begin{align*}<br /> \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &amp;= 3 \\<br /> \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &amp;= 3 \\<br /> \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> \log\frac{x}{z}(1-\log y) &amp;= 0 \\<br /> \end{align*}&lt;/cmath&gt;<br /> If &lt;math&gt;1-\log y=0&lt;/math&gt; then &lt;math&gt;y=10&lt;/math&gt;. Substituting into the first equation yields &lt;math&gt;\log20000=4&lt;/math&gt; which is not possible.<br /> <br /> If &lt;math&gt;\log\frac{x}{z}=0&lt;/math&gt; then &lt;math&gt;\frac{x}{z}=1\Longrightarrow x=z&lt;/math&gt;. Substituting into the third equation gets<br /> &lt;cmath&gt;\begin{align*}<br /> \log x^2-(\log x)(\log x) &amp;= 0 \\<br /> \log x^2-\log x^x &amp;= 0 \\<br /> \log x^{2-x} &amp;= 0 \\<br /> x^{2-x} &amp;= 1 \\<br /> \end{align*}&lt;/cmath&gt;<br /> Thus either &lt;math&gt;x=1&lt;/math&gt; or &lt;math&gt;2-x=0\Longrightarrow x=2&lt;/math&gt;. (Note that here &lt;math&gt;x\neq-1&lt;/math&gt; since logarithm isn't defined for negative number.)<br /> <br /> Substituting &lt;math&gt;x=1&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; into the first equation will obtain &lt;math&gt;y=5&lt;/math&gt; and &lt;math&gt;y=20&lt;/math&gt;, respectively. Thus &lt;math&gt;y_1+y_2=\boxed{25}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> === Solution 3 === <br /> <br /> Let &lt;math&gt;a = \log x&lt;/math&gt;, &lt;math&gt;b = \log y&lt;/math&gt; and &lt;math&gt;c = \log z&lt;/math&gt;. Then the given equations become: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \log 2 + a + b - ab = 1 \\<br /> \log 2 + b + c - bc = 1 \\<br /> a+c = ac \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the first and second equations, solving, and factoring, we get &lt;math&gt;a(1-b) = c(1-b) \implies{a = c}&lt;/math&gt;. Plugging this result into the third equation, we get &lt;math&gt;c = 0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. Substituting each of these values of &lt;math&gt;c&lt;/math&gt; into the second equation, we get &lt;math&gt;b = 1 - \log_{10}2&lt;/math&gt; and &lt;math&gt;b = 1 + \log_{10}2&lt;/math&gt;. Substituting backwards from our original substitution, we get &lt;math&gt;y = 5&lt;/math&gt; and &lt;math&gt;y = 20&lt;/math&gt;, respectively, so our answer is &lt;math&gt;\boxed{025}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_9&diff=127922 2000 AIME I Problems/Problem 9 2020-07-09T21:16:51Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The system of equations<br /> &lt;cmath&gt;\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) &amp; = &amp; 4 \\<br /> \log_{10}(2yz) - (\log_{10}y)(\log_{10}z) &amp; = &amp; 1 \\<br /> \log_{10}(zx) - (\log_{10}z)(\log_{10}x) &amp; = &amp; 0 \\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> has two solutions &lt;math&gt;(x_{1},y_{1},z_{1})&lt;/math&gt; and &lt;math&gt;(x_{2},y_{2},z_{2})&lt;/math&gt;. Find &lt;math&gt;y_{1} + y_{2}&lt;/math&gt;.<br /> <br /> == Solution ==<br /> Since &lt;math&gt;\log ab = \log a + \log b&lt;/math&gt;, we can reduce the equations to a more recognizable form:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &amp;=&amp; 3 - \log 2000\\<br /> -\log y \log z + \log y + \log z - 1 &amp;=&amp; - \log 2\\<br /> -\log x \log z + \log x + \log z - 1 &amp;=&amp; -1\\<br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a,b,c&lt;/math&gt; be &lt;math&gt;\log x, \log y, \log z&lt;/math&gt; respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*)<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a - 1)(b - 1) &amp;=&amp; \log 2 \\<br /> (b-1)(c-1) &amp;=&amp; \log 2 \\<br /> (a-1)(c-1) &amp;=&amp; 1 <br /> \end{eqnarray*}&lt;/cmath&gt;<br /> <br /> From here, multiplying the three equations gives <br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &amp;=&amp; (\log 2)^2\\<br /> (a-1)(b-1)(c-1) &amp;=&amp; \pm\log 2\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Dividing the third equation of (*) from this equation, &lt;math&gt;b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1&lt;/math&gt;. This gives &lt;math&gt;y_1 = 20, y_2 = 5&lt;/math&gt;, and the answer is &lt;math&gt;y_1 + y_2 = \boxed{025}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Subtracting the second equation from the first equation yields<br /> &lt;cmath&gt;\begin{align*}<br /> \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &amp;= 3 \\<br /> \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &amp;= 3 \\<br /> \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &amp;= 3 \\<br /> \log\frac{x}{z}(1-\log y) &amp;= 0 \\<br /> \end{align*}&lt;/cmath&gt;<br /> If &lt;math&gt;1-\log y=0&lt;/math&gt; then &lt;math&gt;y=10&lt;/math&gt;. Substituting into the first equation yields &lt;math&gt;\log20000=4&lt;/math&gt; which is not possible.<br /> <br /> If &lt;math&gt;\log\frac{x}{z}=0&lt;/math&gt; then &lt;math&gt;\frac{x}{z}=1\Longrightarrow x=z&lt;/math&gt;. Substituting into the third equation gets<br /> &lt;cmath&gt;\begin{align*}<br /> \log x^2-(\log x)(\log x) &amp;= 0 \\<br /> \log x^2-\log x^x &amp;= 0 \\<br /> \log x^{2-x} &amp;= 0 \\<br /> x^{2-x} &amp;= 1 \\<br /> \end{align*}&lt;/cmath&gt;<br /> Thus either &lt;math&gt;x=1&lt;/math&gt; or &lt;math&gt;2-x=0\Longrightarrow x=2&lt;/math&gt;. (Note that here &lt;math&gt;x\neq-1&lt;/math&gt; since logarithm isn't defined for negative number.)<br /> <br /> Substituting &lt;math&gt;x=1&lt;/math&gt; and &lt;math&gt;x=2&lt;/math&gt; into the first equation will obtain &lt;math&gt;y=5&lt;/math&gt; and &lt;math&gt;y=20&lt;/math&gt;, respectively. Thus &lt;math&gt;y_1+y_2=\boxed{25}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> <br /> === Solution 3 === <br /> <br /> Let &lt;math&gt;a = \log x&lt;/math&gt;, &lt;math&gt;b = \log y&lt;/math&gt; and &lt;math&gt;c = \log z&lt;/math&gt;. Then the given equations become: <br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \log 2 + a + b - ab = 1 \\<br /> \log 2 + b + c - bc = 1 \\<br /> a+c = ac \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the first and second equations, solving, and factoring, we get &lt;math&gt;a(1-b) = c(1-b) \implies{a = c}&lt;/math&gt;. Plugging this result into the third equation, we get &lt;math&gt;c = 0&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt;. Substituting each of these values for &lt;math&gt;c&lt;/math&gt; into the second equation, we get &lt;math&gt;b = 1 - \log_{10}2&lt;/math&gt; and &lt;math&gt;b = 1 + \log_{10}2&lt;/math&gt;. Substituting backwards from our original substitution, we get &lt;math&gt;y = 5&lt;/math&gt; and &lt;math&gt;y = 20&lt;/math&gt;, respectively, so our answer is &lt;math&gt;\boxed{025}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2000|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_8&diff=127454 2007 AIME I Problems/Problem 8 2020-07-04T17:09:54Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> The [[polynomial]] &lt;math&gt;P(x)&lt;/math&gt; is [[cubic polynomial | cubic]]. What is the largest value of &lt;math&gt;k&lt;/math&gt; for which the polynomials &lt;math&gt;Q_1(x) = x^2 + (k-29)x - k&lt;/math&gt; and &lt;math&gt;Q_2(x) = 2x^2+ (2k-43)x + k&lt;/math&gt; are both [[factor]]s of &lt;math&gt;P(x)&lt;/math&gt;?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> We can see that &lt;math&gt;Q_1&lt;/math&gt; and &lt;math&gt;Q_2&lt;/math&gt; must have a [[root]] in common for them to both be [[factor]]s of the same cubic.<br /> <br /> Let this root be &lt;math&gt;a&lt;/math&gt;.<br /> <br /> We then know that &lt;math&gt;a&lt;/math&gt; is a root of<br /> &lt;math&gt;<br /> Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0<br /> &lt;/math&gt;<br /> , so &lt;math&gt;x = \frac{-k}{5}&lt;/math&gt;.<br /> <br /> We then know that &lt;math&gt;\frac{-k}{5}&lt;/math&gt; is a root of &lt;math&gt;Q_{1}&lt;/math&gt; so we get:<br /> &lt;math&gt;<br /> \frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k<br /> &lt;/math&gt;<br /> or &lt;math&gt;k^{2}=30k&lt;/math&gt;, so &lt;math&gt;k=30&lt;/math&gt; is the highest.<br /> <br /> We can trivially check into the original equations to find that &lt;math&gt;k=30&lt;/math&gt; produces a root in common, so the answer is &lt;math&gt;030&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Again, let the common root be &lt;math&gt;a&lt;/math&gt;; let the other two roots be &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. We can write that &lt;math&gt;(x - a)(x - m) = x^2 + (k - 29)x - k&lt;/math&gt; and that &lt;math&gt;2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)&lt;/math&gt;.<br /> <br /> Therefore, we can write four equations (and we have four [[variable]]s), &lt;math&gt;a + m = 29 - k&lt;/math&gt;, &lt;math&gt;a + n = \frac{43}{2} - k&lt;/math&gt;, &lt;math&gt;am = -k&lt;/math&gt;, and &lt;math&gt;an = \frac{k}{2}&lt;/math&gt;. <br /> <br /> The first two equations show that &lt;math&gt;m - n = 29 - \frac{43}{2} = \frac{15}{2}&lt;/math&gt;. The last two equations show that &lt;math&gt;\frac{m}{n} = -2&lt;/math&gt;. Solving these show that &lt;math&gt;m = 5&lt;/math&gt; and that &lt;math&gt;n = -\frac{5}{2}&lt;/math&gt;. Substituting back into the equations, we eventually find that &lt;math&gt;k = 30&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Since &lt;math&gt;Q_1(x)&lt;/math&gt; and &lt;math&gt;Q_2(x)&lt;/math&gt; are both factors of &lt;math&gt;P(x)&lt;/math&gt;, which is cubic, we know the other factors associated with each of &lt;math&gt;Q_1(x)&lt;/math&gt; and &lt;math&gt;Q_2(x)&lt;/math&gt; must be linear. Let &lt;math&gt;Q_1(x)R(x) = Q_2(x)S(x) = P(x)&lt;/math&gt;, where &lt;math&gt;R(x) = ax + b&lt;/math&gt; and &lt;math&gt;S(x) = cx + d&lt;/math&gt;. Then we have that &lt;math&gt;((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)&lt;/math&gt;. Equating coefficients, we get the following system of equations: <br /> <br /> &lt;cmath&gt;\begin{align} <br /> a = 2c \\<br /> b = -d \\<br /> 2c(k - 29) - d = c(2k - 43) + 2d \\<br /> -d(k - 29) - 2ck = d(2k - 43) + ck <br /> \end{align}&lt;/cmath&gt;<br /> <br /> Using equations &lt;math&gt;(1)&lt;/math&gt; and &lt;math&gt;(2)&lt;/math&gt; to make substitutions into equation &lt;math&gt;(3)&lt;/math&gt;, we see that the &lt;math&gt;k&lt;/math&gt;'s drop out and we're left with &lt;math&gt;d = -5c&lt;/math&gt;. Substituting this expression for &lt;math&gt;d&lt;/math&gt; into equation &lt;math&gt;(4)&lt;/math&gt; and solving, we see that &lt;math&gt;k&lt;/math&gt; must be &lt;math&gt;\boxed {30}&lt;/math&gt;. <br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=127252 2010 AIME I Problems/Problem 6 2020-07-02T02:00:21Z <p>Anellipticcurveoverq: </p> <hr /> <div>__TOC__<br /> == Problem ==<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a [[quadratic]] polynomial with real coefficients satisfying &lt;math&gt;x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt;, and suppose &lt;math&gt;P(11) = 181&lt;/math&gt;. Find &lt;math&gt;P(16)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br /> <br /> real P(real x) { return 8*(x-1)^2/5+1; }<br /> real Q(real x) { return (x-1)^2+1; }<br /> real R(real x) { return 2*(x-1)^2+1; }<br /> draw(graph(P,min,max),dark);<br /> draw(graph(Q,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> draw(graph(R,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> dot((1,1));<br /> label(&quot;$P(x)$&quot;,(max,P(max)),E,fontsize(10));<br /> label(&quot;$Q(x)$&quot;,(max,Q(max)),E,fontsize(10));<br /> label(&quot;$R(x)$&quot;,(max,R(max)),E,fontsize(10));<br /> <br /> /* axes */<br /> Label f; f.p=fontsize(8);<br /> xaxis(-2, 3, Ticks(f, 5, 1));<br /> yaxis(-1, 5, Ticks(f, 6, 1)); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;Q(x) = x^2 - 2x + 2&lt;/math&gt;, &lt;math&gt;R(x) = 2x^2 - 4x + 3&lt;/math&gt;. [[Completing the square]], we have &lt;math&gt;Q(x) = (x-1)^2 + 1&lt;/math&gt;, and &lt;math&gt;R(x) = 2(x-1)^2 + 1&lt;/math&gt;, so it follows that &lt;math&gt;P(x) \ge Q(x) \ge 1&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; (by the [[Trivial Inequality]]). <br /> <br /> Also, &lt;math&gt;1 = Q(1) \le P(1) \le R(1) = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; obtains its minimum at the point &lt;math&gt;(1,1)&lt;/math&gt;. Then &lt;math&gt;P(x)&lt;/math&gt; must be of the form &lt;math&gt;c(x-1)^2 + 1&lt;/math&gt; for some constant &lt;math&gt;c&lt;/math&gt;; substituting &lt;math&gt;P(11) = 181&lt;/math&gt; yields &lt;math&gt;c = \frac 95&lt;/math&gt;. Finally, &lt;math&gt;P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> It can be seen that the function &lt;math&gt;P(x)&lt;/math&gt; must be in the form &lt;math&gt;P(x) = ax^2 - 2ax + c&lt;/math&gt; for some real &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. This is because the [[derivative]] of &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;2ax - 2a&lt;/math&gt;, and a global minimum occurs only at &lt;math&gt;x = 1&lt;/math&gt; (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at &lt;math&gt;\frac{-b}{2a}&lt;/math&gt;). Substituting &lt;math&gt;(1,1)&lt;/math&gt; and &lt;math&gt;(11, 181)&lt;/math&gt; we obtain two equations:<br /> <br /> &lt;center&gt;&lt;math&gt;P(11) = 99a + c = 181&lt;/math&gt;, and &lt;math&gt;P(1) = -a + c = 1&lt;/math&gt;.&lt;/center&gt;<br /> <br /> Solving, we get &lt;math&gt;a = \frac{9}{5}&lt;/math&gt; and &lt;math&gt;c = \frac{14}{5}&lt;/math&gt;, so &lt;math&gt;P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}&lt;/math&gt;. Therefore, &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;y = x^2 - 2x + 2&lt;/math&gt;; note that &lt;math&gt;2y - 1 = 2x^2 - 4x + 3&lt;/math&gt;. Setting &lt;math&gt;y = 2y - 1&lt;/math&gt;, we find that equality holds when &lt;math&gt;y = 1&lt;/math&gt; and therefore when &lt;math&gt;x^2 - 2x + 2 = 1&lt;/math&gt;; this is true iff &lt;math&gt;x = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - x&lt;/math&gt;; clearly &lt;math&gt;Q(1) = 0&lt;/math&gt;, so we can write &lt;math&gt;Q(x) = (x - 1)Q'(x)&lt;/math&gt;, where &lt;math&gt;Q'(x)&lt;/math&gt; is some linear function. Plug &lt;math&gt;Q(x)&lt;/math&gt; into the given inequality:<br /> <br /> &lt;math&gt;x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3&lt;/math&gt;<br /> <br /> &lt;math&gt;(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)&lt;/math&gt;, and thus<br /> <br /> &lt;math&gt;x - 2 \le Q'(x) \le 2x - 3&lt;/math&gt;<br /> <br /> For all &lt;math&gt;x &gt; 1&lt;/math&gt;; note that the inequality signs are flipped if &lt;math&gt;x &lt; 1&lt;/math&gt;, and that the division is invalid for &lt;math&gt;x = 1&lt;/math&gt;. However,<br /> <br /> &lt;math&gt;\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1&lt;/math&gt;,<br /> <br /> and thus by the [[sandwich theorem]] &lt;math&gt;\lim_{x \to 1} Q'(x) = -1&lt;/math&gt;; by the definition of a continuous function, &lt;math&gt;Q'(1) = -1&lt;/math&gt;. Also, &lt;math&gt;Q(11) = 170&lt;/math&gt;, so &lt;math&gt;Q'(11) = 170/(11-1) = 17&lt;/math&gt;; plugging in and solving, &lt;math&gt;Q'(x) = (9/5)(x - 1) - 1&lt;/math&gt;. Thus &lt;math&gt;Q(16) = 390&lt;/math&gt;, and so &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - (x^2-2x+2)&lt;/math&gt;, then &lt;math&gt;0\le Q(x) \le (x-1)^2&lt;/math&gt; (note this is derived from the given inequality chain). Therefore, &lt;math&gt;0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2&lt;/math&gt; for some real value A.<br /> <br /> &lt;math&gt;Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> <br /> Let &lt;math&gt;P(x) = ax^2 + bx + c&lt;/math&gt;. Plugging in &lt;math&gt;x = 1&lt;/math&gt; to the expressions on both sides of the inequality, we see that &lt;math&gt;a + b + c = 1&lt;/math&gt;. We see from the problem statement that &lt;math&gt;121a + 11b + c = 181&lt;/math&gt;. Since we know the vertex of &lt;math&gt;P(x)&lt;/math&gt; lies at &lt;math&gt;x = 1&lt;/math&gt;, by symmetry we get &lt;math&gt;81a -9b + c = 181&lt;/math&gt; as well. Since we now have three equations, we can solve this trivial system and get our answer of &lt;math&gt;\boxed{406}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=127250 2010 AIME I Problems/Problem 6 2020-07-02T01:59:48Z <p>Anellipticcurveoverq: </p> <hr /> <div>__TOC__<br /> == Problem ==<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a [[quadratic]] polynomial with real coefficients satisfying &lt;math&gt;x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt;, and suppose &lt;math&gt;P(11) = 181&lt;/math&gt;. Find &lt;math&gt;P(16)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br /> <br /> real P(real x) { return 8*(x-1)^2/5+1; }<br /> real Q(real x) { return (x-1)^2+1; }<br /> real R(real x) { return 2*(x-1)^2+1; }<br /> draw(graph(P,min,max),dark);<br /> draw(graph(Q,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> draw(graph(R,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> dot((1,1));<br /> label(&quot;$P(x)$&quot;,(max,P(max)),E,fontsize(10));<br /> label(&quot;$Q(x)$&quot;,(max,Q(max)),E,fontsize(10));<br /> label(&quot;$R(x)$&quot;,(max,R(max)),E,fontsize(10));<br /> <br /> /* axes */<br /> Label f; f.p=fontsize(8);<br /> xaxis(-2, 3, Ticks(f, 5, 1));<br /> yaxis(-1, 5, Ticks(f, 6, 1)); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;Q(x) = x^2 - 2x + 2&lt;/math&gt;, &lt;math&gt;R(x) = 2x^2 - 4x + 3&lt;/math&gt;. [[Completing the square]], we have &lt;math&gt;Q(x) = (x-1)^2 + 1&lt;/math&gt;, and &lt;math&gt;R(x) = 2(x-1)^2 + 1&lt;/math&gt;, so it follows that &lt;math&gt;P(x) \ge Q(x) \ge 1&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; (by the [[Trivial Inequality]]). <br /> <br /> Also, &lt;math&gt;1 = Q(1) \le P(1) \le R(1) = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; obtains its minimum at the point &lt;math&gt;(1,1)&lt;/math&gt;. Then &lt;math&gt;P(x)&lt;/math&gt; must be of the form &lt;math&gt;c(x-1)^2 + 1&lt;/math&gt; for some constant &lt;math&gt;c&lt;/math&gt;; substituting &lt;math&gt;P(11) = 181&lt;/math&gt; yields &lt;math&gt;c = \frac 95&lt;/math&gt;. Finally, &lt;math&gt;P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> It can be seen that the function &lt;math&gt;P(x)&lt;/math&gt; must be in the form &lt;math&gt;P(x) = ax^2 - 2ax + c&lt;/math&gt; for some real &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. This is because the [[derivative]] of &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;2ax - 2a&lt;/math&gt;, and a global minimum occurs only at &lt;math&gt;x = 1&lt;/math&gt; (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at &lt;math&gt;\frac{-b}{2a}&lt;/math&gt;). Substituting &lt;math&gt;(1,1)&lt;/math&gt; and &lt;math&gt;(11, 181)&lt;/math&gt; we obtain two equations:<br /> <br /> &lt;center&gt;&lt;math&gt;P(11) = 99a + c = 181&lt;/math&gt;, and &lt;math&gt;P(1) = -a + c = 1&lt;/math&gt;.&lt;/center&gt;<br /> <br /> Solving, we get &lt;math&gt;a = \frac{9}{5}&lt;/math&gt; and &lt;math&gt;c = \frac{14}{5}&lt;/math&gt;, so &lt;math&gt;P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}&lt;/math&gt;. Therefore, &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;y = x^2 - 2x + 2&lt;/math&gt;; note that &lt;math&gt;2y - 1 = 2x^2 - 4x + 3&lt;/math&gt;. Setting &lt;math&gt;y = 2y - 1&lt;/math&gt;, we find that equality holds when &lt;math&gt;y = 1&lt;/math&gt; and therefore when &lt;math&gt;x^2 - 2x + 2 = 1&lt;/math&gt;; this is true iff &lt;math&gt;x = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - x&lt;/math&gt;; clearly &lt;math&gt;Q(1) = 0&lt;/math&gt;, so we can write &lt;math&gt;Q(x) = (x - 1)Q'(x)&lt;/math&gt;, where &lt;math&gt;Q'(x)&lt;/math&gt; is some linear function. Plug &lt;math&gt;Q(x)&lt;/math&gt; into the given inequality:<br /> <br /> &lt;math&gt;x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3&lt;/math&gt;<br /> <br /> &lt;math&gt;(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)&lt;/math&gt;, and thus<br /> <br /> &lt;math&gt;x - 2 \le Q'(x) \le 2x - 3&lt;/math&gt;<br /> <br /> For all &lt;math&gt;x &gt; 1&lt;/math&gt;; note that the inequality signs are flipped if &lt;math&gt;x &lt; 1&lt;/math&gt;, and that the division is invalid for &lt;math&gt;x = 1&lt;/math&gt;. However,<br /> <br /> &lt;math&gt;\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1&lt;/math&gt;,<br /> <br /> and thus by the [[sandwich theorem]] &lt;math&gt;\lim_{x \to 1} Q'(x) = -1&lt;/math&gt;; by the definition of a continuous function, &lt;math&gt;Q'(1) = -1&lt;/math&gt;. Also, &lt;math&gt;Q(11) = 170&lt;/math&gt;, so &lt;math&gt;Q'(11) = 170/(11-1) = 17&lt;/math&gt;; plugging in and solving, &lt;math&gt;Q'(x) = (9/5)(x - 1) - 1&lt;/math&gt;. Thus &lt;math&gt;Q(16) = 390&lt;/math&gt;, and so &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - (x^2-2x+2)&lt;/math&gt;, then &lt;math&gt;0\le Q(x) \le (x-1)^2&lt;/math&gt; (note this is derived from the given inequality chain). Therefore, &lt;math&gt;0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2&lt;/math&gt; for some real value A.<br /> <br /> &lt;math&gt;Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> <br /> Let &lt;math&gt;P(x) = ax^2 + bx + c&lt;/math&gt;. Plugging in &lt;math&gt;x = 1&lt;/math&gt; to the expressions on both sides of the inequality, we see that &lt;math&gt;a + b + c = 1&lt;/math&gt;. We see from the problem statement that &lt;math&gt;121a + 11b + c = 181&lt;/math&gt;. Since we know the vertex of &lt;math&gt;P(x)&lt;/math&gt; lies at &lt;math&gt;x = 1&lt;/math&gt;, by symmetry we get &lt;math&gt;81a -9b + c&lt;/math&gt;. Since we now have three equations, we can solve this trivial system and get our answer of &lt;math&gt;\boxed{406}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_6&diff=127249 2010 AIME I Problems/Problem 6 2020-07-02T01:59:32Z <p>Anellipticcurveoverq: </p> <hr /> <div>__TOC__<br /> == Problem ==<br /> Let &lt;math&gt;P(x)&lt;/math&gt; be a [[quadratic]] polynomial with real coefficients satisfying &lt;math&gt;x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt;, and suppose &lt;math&gt;P(11) = 181&lt;/math&gt;. Find &lt;math&gt;P(16)&lt;/math&gt;.<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);<br /> <br /> real P(real x) { return 8*(x-1)^2/5+1; }<br /> real Q(real x) { return (x-1)^2+1; }<br /> real R(real x) { return 2*(x-1)^2+1; }<br /> draw(graph(P,min,max),dark);<br /> draw(graph(Q,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> draw(graph(R,min,max),linetype(&quot;6 2&quot;)+linewidth(0.7));<br /> dot((1,1));<br /> label(&quot;$P(x)$&quot;,(max,P(max)),E,fontsize(10));<br /> label(&quot;$Q(x)$&quot;,(max,Q(max)),E,fontsize(10));<br /> label(&quot;$R(x)$&quot;,(max,R(max)),E,fontsize(10));<br /> <br /> /* axes */<br /> Label f; f.p=fontsize(8);<br /> xaxis(-2, 3, Ticks(f, 5, 1));<br /> yaxis(-1, 5, Ticks(f, 6, 1)); <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Let &lt;math&gt;Q(x) = x^2 - 2x + 2&lt;/math&gt;, &lt;math&gt;R(x) = 2x^2 - 4x + 3&lt;/math&gt;. [[Completing the square]], we have &lt;math&gt;Q(x) = (x-1)^2 + 1&lt;/math&gt;, and &lt;math&gt;R(x) = 2(x-1)^2 + 1&lt;/math&gt;, so it follows that &lt;math&gt;P(x) \ge Q(x) \ge 1&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; (by the [[Trivial Inequality]]). <br /> <br /> Also, &lt;math&gt;1 = Q(1) \le P(1) \le R(1) = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;, and &lt;math&gt;P&lt;/math&gt; obtains its minimum at the point &lt;math&gt;(1,1)&lt;/math&gt;. Then &lt;math&gt;P(x)&lt;/math&gt; must be of the form &lt;math&gt;c(x-1)^2 + 1&lt;/math&gt; for some constant &lt;math&gt;c&lt;/math&gt;; substituting &lt;math&gt;P(11) = 181&lt;/math&gt; yields &lt;math&gt;c = \frac 95&lt;/math&gt;. Finally, &lt;math&gt;P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> It can be seen that the function &lt;math&gt;P(x)&lt;/math&gt; must be in the form &lt;math&gt;P(x) = ax^2 - 2ax + c&lt;/math&gt; for some real &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. This is because the [[derivative]] of &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;2ax - 2a&lt;/math&gt;, and a global minimum occurs only at &lt;math&gt;x = 1&lt;/math&gt; (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at &lt;math&gt;\frac{-b}{2a}&lt;/math&gt;). Substituting &lt;math&gt;(1,1)&lt;/math&gt; and &lt;math&gt;(11, 181)&lt;/math&gt; we obtain two equations:<br /> <br /> &lt;center&gt;&lt;math&gt;P(11) = 99a + c = 181&lt;/math&gt;, and &lt;math&gt;P(1) = -a + c = 1&lt;/math&gt;.&lt;/center&gt;<br /> <br /> Solving, we get &lt;math&gt;a = \frac{9}{5}&lt;/math&gt; and &lt;math&gt;c = \frac{14}{5}&lt;/math&gt;, so &lt;math&gt;P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}&lt;/math&gt;. Therefore, &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> <br /> Let &lt;math&gt;y = x^2 - 2x + 2&lt;/math&gt;; note that &lt;math&gt;2y - 1 = 2x^2 - 4x + 3&lt;/math&gt;. Setting &lt;math&gt;y = 2y - 1&lt;/math&gt;, we find that equality holds when &lt;math&gt;y = 1&lt;/math&gt; and therefore when &lt;math&gt;x^2 - 2x + 2 = 1&lt;/math&gt;; this is true iff &lt;math&gt;x = 1&lt;/math&gt;, so &lt;math&gt;P(1) = 1&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - x&lt;/math&gt;; clearly &lt;math&gt;Q(1) = 0&lt;/math&gt;, so we can write &lt;math&gt;Q(x) = (x - 1)Q'(x)&lt;/math&gt;, where &lt;math&gt;Q'(x)&lt;/math&gt; is some linear function. Plug &lt;math&gt;Q(x)&lt;/math&gt; into the given inequality:<br /> <br /> &lt;math&gt;x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3&lt;/math&gt;<br /> <br /> &lt;math&gt;(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)&lt;/math&gt;, and thus<br /> <br /> &lt;math&gt;x - 2 \le Q'(x) \le 2x - 3&lt;/math&gt;<br /> <br /> For all &lt;math&gt;x &gt; 1&lt;/math&gt;; note that the inequality signs are flipped if &lt;math&gt;x &lt; 1&lt;/math&gt;, and that the division is invalid for &lt;math&gt;x = 1&lt;/math&gt;. However,<br /> <br /> &lt;math&gt;\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1&lt;/math&gt;,<br /> <br /> and thus by the [[sandwich theorem]] &lt;math&gt;\lim_{x \to 1} Q'(x) = -1&lt;/math&gt;; by the definition of a continuous function, &lt;math&gt;Q'(1) = -1&lt;/math&gt;. Also, &lt;math&gt;Q(11) = 170&lt;/math&gt;, so &lt;math&gt;Q'(11) = 170/(11-1) = 17&lt;/math&gt;; plugging in and solving, &lt;math&gt;Q'(x) = (9/5)(x - 1) - 1&lt;/math&gt;. Thus &lt;math&gt;Q(16) = 390&lt;/math&gt;, and so &lt;math&gt;P(16) = \boxed{406}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> <br /> Let &lt;math&gt;Q(x) = P(x) - (x^2-2x+2)&lt;/math&gt;, then &lt;math&gt;0\le Q(x) \le (x-1)^2&lt;/math&gt; (note this is derived from the given inequality chain). Therefore, &lt;math&gt;0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2&lt;/math&gt; for some real value A.<br /> <br /> &lt;math&gt;Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}&lt;/math&gt;.<br /> <br /> &lt;math&gt;Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}&lt;/math&gt;<br /> <br /> === Solution 5 ===<br /> <br /> Let &lt;math&gt;P(X) = ax^2 + bx + c&lt;/math&gt;. Plugging in &lt;math&gt;x = 1&lt;/math&gt; to the expressions on both sides of the inequality, we see that &lt;math&gt;a + b + c = 1&lt;/math&gt;. We see from the problem statement that &lt;math&gt;121a + 11b + c = 181&lt;/math&gt;. Since we know the vertex of &lt;math&gt;P(x)&lt;/math&gt; lies at &lt;math&gt;x = 1&lt;/math&gt;, by symmetry we get &lt;math&gt;81a -9b + c&lt;/math&gt;. Since we now have three equations, we can solve this trivial system and get our answer of &lt;math&gt;\boxed{406}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> {{AIME box|year=2010|num-b=5|num-a=7|n=I}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_9&diff=126286 1999 AIME Problems/Problem 9 2020-06-23T19:03:25Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> A function &lt;math&gt;f&lt;/math&gt; is defined on the [[complex number]]s by &lt;math&gt;f(z)=(a+bi)z,&lt;/math&gt; where &lt;math&gt;a_{}&lt;/math&gt; and &lt;math&gt;b_{}&lt;/math&gt; are positive numbers. This [[function]] has the property that the image of each point in the complex plane is [[equidistant]] from that point and the [[origin]]. Given that &lt;math&gt;|a+bi|=8&lt;/math&gt; and that &lt;math&gt;b^2=m/n,&lt;/math&gt; where &lt;math&gt;m_{}&lt;/math&gt; and &lt;math&gt;n_{}&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;m+n.&lt;/math&gt;<br /> <br /> __TOC__<br /> <br /> === Solution 1 ===<br /> Suppose we pick an arbitrary point on the [[complex plane]], say &lt;math&gt;(1,1)&lt;/math&gt;. According to the definition of &lt;math&gt;f(z)&lt;/math&gt;, &lt;cmath&gt;f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,&lt;/cmath&gt; this image must be equidistant to &lt;math&gt;(1,1)&lt;/math&gt; and &lt;math&gt;(0,0)&lt;/math&gt;. Thus the image must lie on the line with slope &lt;math&gt;-1&lt;/math&gt; and which passes through &lt;math&gt;\left(\frac 12, \frac12\right)&lt;/math&gt;, so its graph is &lt;math&gt;x + y = 1&lt;/math&gt;. Substituting &lt;math&gt;x = (a-b)&lt;/math&gt; and &lt;math&gt;y = (a+b)&lt;/math&gt;, we get &lt;math&gt;2a = 1 \Rightarrow a = \frac 12&lt;/math&gt;. <br /> <br /> By the [[Pythagorean Theorem]], we have &lt;math&gt;\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{259}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Plugging in &lt;math&gt;z=1&lt;/math&gt; yields &lt;math&gt;f(1) = a+bi&lt;/math&gt;. This implies that &lt;math&gt;a+bi&lt;/math&gt; must fall on the line &lt;math&gt;Re(z)=a=\frac{1}{2}&lt;/math&gt;, given the equidistant rule. By &lt;math&gt;|a+bi|=8&lt;/math&gt;, we get &lt;math&gt;a^2 + b^2 = 64&lt;/math&gt;, and plugging in &lt;math&gt;a=\frac{1}{2}&lt;/math&gt; yields &lt;math&gt;b^2=\frac{255}{4}&lt;/math&gt;. The answer is thus &lt;math&gt;\boxed{259}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We are given that &lt;math&gt;(a + bi)z&lt;/math&gt; is equidistant from the origin and &lt;math&gt;z.&lt;/math&gt; This translates to<br /> &lt;cmath&gt;<br /> \begin{eqnarray*} |(a + bi)z - z| &amp; = &amp; |(a + bi)z| \\<br /> |z(a - 1) + bzi| &amp; = &amp; |az + bzi| \\<br /> |z||(a - 1) + bi| &amp; = &amp; |z||a + bi| \\<br /> (a - 1)^2 + b^2 &amp; = &amp; a^2 + b^2 \\<br /> &amp; \Rightarrow &amp; a = \frac 12 \end{eqnarray*}<br /> &lt;/cmath&gt;<br /> Since &lt;math&gt;|a + bi| = 8,&lt;/math&gt; &lt;math&gt;a^2 + b^2 = 64.&lt;/math&gt; But &lt;math&gt;a = \frac 12,&lt;/math&gt; thus &lt;math&gt;b^2 = \frac {255}4.&lt;/math&gt; So the answer is &lt;math&gt;259&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the points in the complex plane represented by &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;(a+bi)z&lt;/math&gt;, respectively. &lt;math&gt;|a+bi| = 8&lt;/math&gt; implies &lt;math&gt;OQ = 8OP&lt;/math&gt;. Also, we are given &lt;math&gt;OQ = PQ&lt;/math&gt;, so &lt;math&gt;OPQ&lt;/math&gt; is isosceles with base &lt;math&gt;OP&lt;/math&gt;. Notice that the base angle of this isosceles triangle is equal to the argument &lt;math&gt;\theta&lt;/math&gt; of the complex number &lt;math&gt;a + bi&lt;/math&gt;, because &lt;math&gt;(a+bi)z&lt;/math&gt; forms an angle of &lt;math&gt;\theta&lt;/math&gt; with &lt;math&gt;z&lt;/math&gt;. Drop the altitude/median from &lt;math&gt;Q&lt;/math&gt; to base &lt;math&gt;OP&lt;/math&gt;, and you end up with a right triangle that shows &lt;math&gt;\cos \theta = \frac{\frac{1}{2}OP}{8OQ} = \frac{\frac{1}{2}|z|}{8|z|} = \frac{1}{16}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive, &lt;math&gt;z&lt;/math&gt; lies in the first quadrant and &lt;math&gt;\theta &lt; \pi/2&lt;/math&gt;; hence by right triangle trigonometry &lt;math&gt;\sin \theta = \frac{\sqrt{255}}{16}&lt;/math&gt;. Finally, &lt;math&gt;b = |a+bi|\sin\theta = 8\frac{\sqrt{255}}{16} = \frac{\sqrt{255}}{2}&lt;/math&gt;, and &lt;math&gt;b^2 = \frac{255}{4}&lt;/math&gt;, so the answer is &lt;math&gt;259&lt;/math&gt;.<br /> <br /> === Solution 5 === <br /> Similarly to in Solution 3, we see that &lt;math&gt;|(a + bi)z - z| = |(a + bi)z|&lt;/math&gt;. Letting the point &lt;math&gt;z = c + di&lt;/math&gt;, we have &lt;math&gt;\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \sqrt{(ac-bd)^2+(ad+bc)^2}&lt;/math&gt;. Expanding both sides of this equation (after squaring, of course) and canceling terms, we get &lt;math&gt;(d^2+c^2)(-2a+1) = 0&lt;/math&gt;. Of course, &lt;math&gt;(d^2+c^2)&lt;/math&gt; can't be zero because this property of the function holds for all complex &lt;math&gt;z&lt;/math&gt;. Therefore, &lt;math&gt;a = \frac{1}{2}&lt;/math&gt; and we proceed as above to get &lt;math&gt;\boxed{259}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1999|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=125791 2005 AIME II Problems/Problem 11 2020-06-18T15:49:15Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_11&diff=125790 2005 AIME II Problems/Problem 11 2020-06-18T15:48:14Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;m &lt;/math&gt; be a positive integer, and let &lt;math&gt; a_0, a_1,\ldots,a_m &lt;/math&gt; be a sequence of reals such that &lt;math&gt;a_0 = 37, a_1 = 72, a_m = 0, &lt;/math&gt; and &lt;math&gt; a_{k+1} = a_{k-1} - \frac 3{a_k} &lt;/math&gt; for &lt;math&gt; k = 1,2,\ldots, m-1. &lt;/math&gt; Find &lt;math&gt;m. &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;0 &lt; k &lt; m&lt;/math&gt;, we have<br /> <br /> &lt;center&gt;<br /> &lt;math&gt;a_{k}a_{k+1} = a_{k-1}a_{k} - 3 &lt;/math&gt;.<br /> &lt;/center&gt;<br /> <br /> Thus the product &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; is a [[monovariant]]: it decreases by 3 each time &lt;math&gt;k&lt;/math&gt; increases by 1. For &lt;math&gt;k = 0&lt;/math&gt; we have &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72&lt;/math&gt;, so when &lt;math&gt;k = \frac{37 \cdot 72}{3} = 888&lt;/math&gt;, &lt;math&gt;a_{k}a_{k+1}&lt;/math&gt; will be zero for the first time, which implies that &lt;math&gt;m = \boxed{889}&lt;/math&gt;, our answer.<br /> <br /> ==Solution 2==<br /> <br /> Plugging in &lt;math&gt;k = m-1&lt;/math&gt; to the given relation, we get &lt;math&gt;0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}&lt;/math&gt;. Inspecting the value of &lt;math&gt;a_{k}a{k+1}&lt;/math&gt; for small values of &lt;math&gt;k&lt;/math&gt;, we see that &lt;math&gt;a_{k}a_{k+1} = 37\cdot 72 - 3k&lt;/math&gt;. Setting the RHS of this equation equal to &lt;math&gt;3&lt;/math&gt;, we find that &lt;math&gt;m&lt;/math&gt; must be &lt;math&gt; \boxed{889}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=125684 2011 AIME II Problems/Problem 15 2020-06-17T03:07:48Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;P(x) = x^2 - 3x - 9&lt;/math&gt;. A real number &lt;math&gt;x&lt;/math&gt; is chosen at random from the interval &lt;math&gt;5 \le x \le 15&lt;/math&gt;. The probability that &lt;math&gt;\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; is equal to &lt;math&gt;\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}&lt;/math&gt; , where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b + c + d + e&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Table of values of &lt;math&gt;P(x)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> P(5) &amp;= 1 \\<br /> P(6) &amp;= 9 \\<br /> P(7) &amp;= 19 \\<br /> P(8) &amp;= 31 \\<br /> P(9) &amp;= 45 \\<br /> P(10) &amp;= 61 \\<br /> P(11) &amp;= 79 \\<br /> P(12) &amp;= 99 \\<br /> P(13) &amp;= 121 \\<br /> P(14) &amp;= 145 \\<br /> P(15) &amp;= 171 \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> In order for &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; to hold, &lt;math&gt;\sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; must be an integer and hence &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; must be a perfect square. This limits &lt;math&gt;x&lt;/math&gt; to &lt;math&gt;5 \le x &lt; 6&lt;/math&gt; or &lt;math&gt;6 \le x &lt; 7&lt;/math&gt; or &lt;math&gt;13 \le x &lt; 14&lt;/math&gt; since, from the table above, those are the only values of &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; is an perfect square. However, in order for &lt;math&gt;\sqrt{P(x)}&lt;/math&gt; to be rounded down to &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt;, &lt;math&gt;P(x)&lt;/math&gt; must be less than the next perfect square after &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; (for the said intervals). Now, we consider the three cases:<br /> <br /> <br /> Case &lt;math&gt;5 \le x &lt; 6&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;1&lt;/math&gt;, which is &lt;math&gt;4&lt;/math&gt;, ''i.e.'':<br /> <br /> &lt;math&gt;1 \le P(x) &lt; 4&lt;/math&gt; (because &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = 1&lt;/math&gt; implies &lt;math&gt;1 \le \sqrt{P(x)} &lt; 2&lt;/math&gt;)<br /> <br /> Since &lt;math&gt;P(x)&lt;/math&gt; is increasing for &lt;math&gt;x \ge 5&lt;/math&gt;, we just need to find the value &lt;math&gt;v \ge 5&lt;/math&gt; where &lt;math&gt;P(v) = 4&lt;/math&gt;, which will give us the working range &lt;math&gt;5 \le x &lt; v&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 4 \\<br /> v &amp;= \frac{3 + \sqrt{61}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;5 \le x &lt; \frac{3 + \sqrt{61}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;6 \le x &lt; 7&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;9&lt;/math&gt;, which is &lt;math&gt;16&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 16 \\<br /> v &amp;= \frac{3 + \sqrt{109}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;6 \le x &lt; \frac{3 + \sqrt{109}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;13 \le x &lt; 14&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;121&lt;/math&gt;, which is &lt;math&gt;144&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 144 \\<br /> v &amp;= \frac{3 + \sqrt{621}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;13 \le x &lt; \frac{3 + \sqrt{621}}{2}&lt;/math&gt;.<br /> <br /> Now, we find the length of the working intervals and divide it by the length of the total interval, &lt;math&gt;15 - 5 = 10&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\<br /> &amp;= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus, the answer is &lt;math&gt;61 + 109 + 621 + 39 + 20 = \fbox{850}&lt;/math&gt;.<br /> <br /> P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.<br /> <br /> == Solution 2 == <br /> <br /> Make the substitution &lt;math&gt;y=2x-3&lt;/math&gt;, so &lt;math&gt;P(x)=\frac{y^2-45}{4}.&lt;/math&gt; We're looking for solutions to &lt;cmath&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}&lt;/cmath&gt;with the new bounds &lt;math&gt;y\in{[7,27]}&lt;/math&gt;. Since the right side is an integer, it must be that &lt;math&gt;\frac{\lfloor{y\rfloor}^2-45}{4}&lt;/math&gt; is a perfect square. For simplicity, write &lt;math&gt;\lfloor{y\rfloor}=a&lt;/math&gt; and &lt;cmath&gt;a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.&lt;/cmath&gt;Since &lt;math&gt;a-2b&lt;a+2b&lt;/math&gt;, it must be that &lt;math&gt;(a-2b,a+2b)=(1,45),(3,15),(5,9)&lt;/math&gt;, which gives solutions &lt;math&gt;(23,11),(9,3),(7,1)&lt;/math&gt;, respectively. But this gives us three cases to check:<br /> <br /> Case 1: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;11\leq{\sqrt{\frac{y^2-45}{4}}}&lt;12\implies{y\in{[23,\sqrt{621})}}.&lt;/cmath&gt;<br /> Case 2: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;3\leq{\sqrt{\frac{y^2-45}{4}}}&lt;4\implies{y\in{[9,\sqrt{109})}}.&lt;/cmath&gt;<br /> Case 3: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1&lt;/math&gt;<br /> <br /> In this case, we have &lt;cmath&gt;1\leq{\sqrt{\frac{y^2-45}{4}}}&lt;2\implies{y\in{[7,\sqrt{61})}}.&lt;/cmath&gt;<br /> To finish, the total length of the interval from which we choose &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;27-7=20&lt;/math&gt;. The total length of the success intervals is &lt;cmath&gt;(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,&lt;/cmath&gt;which means the probability is &lt;cmath&gt;\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.&lt;/cmath&gt;AIMEifying the answer gives &lt;math&gt;\boxed{850}&lt;/math&gt;.<br /> <br /> == Solution 3 (Graphing) == <br /> <br /> It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let &lt;math&gt;A = \lfloor\sqrt{P(x)}\rfloor&lt;/math&gt; and &lt;math&gt;B = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt;. The graph of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will look like this, with &lt;math&gt;A&lt;/math&gt; having only integral y-values and &lt;math&gt;B&lt;/math&gt; having only integral x-values:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 1.png|400px]]<br /> <br /> As both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; consist of a bunch of line segments, the probability that &lt;math&gt;A = B&lt;/math&gt; is the &quot;length&quot; of the overlap between the segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; divided by the total length of the segments of &lt;math&gt;B&lt;/math&gt;. <br /> <br /> Looking at the graph, we see that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will overlap only when &lt;math&gt;B&lt;/math&gt; is an integer. Specifically, each region of overlap will begin when &lt;math&gt;\sqrt{P(x)}\ = k, 5 \le x \le 15&lt;/math&gt; has solutions for integral &lt;math&gt;k&lt;/math&gt; in the range of &lt;math&gt;A&lt;/math&gt;, which consists of the integers &lt;math&gt;1-13&lt;/math&gt;, and end when &lt;math&gt;A&lt;/math&gt; jumps up to its next y-value.<br /> <br /> Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of &lt;math&gt;\frac{3 + \sqrt{45 + 4k^2}}{2}&lt;/math&gt; for &lt;math&gt;k&lt;/math&gt; in the specified range, meaning &lt;math&gt;45 + 4k^2&lt;/math&gt; must be a perfect square. Plugging in all the possible values of &lt;math&gt;k&lt;/math&gt;, we get &lt;math&gt;k = 1, 3, 11&lt;/math&gt;, corresponding to start points of &lt;math&gt;x = 5, 6, 13&lt;/math&gt;. As already stated, the endpoints will occur when &lt;math&gt;A&lt;/math&gt; jumps up to the next integer &lt;math&gt;k+1&lt;/math&gt; at each of these segments, at which point the x-value will be &lt;math&gt;\frac{3 + \sqrt{45 + 4(k+1)^2}}{2}&lt;/math&gt;. On the graph, the overlapping segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; would be represented by the highlighted green segments below:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 2.png|400px]]<br /> <br /> <br /> Taking the difference between this second x-value and the start point for each of our start points &lt;math&gt;x = 5, 6, 13&lt;/math&gt; and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of &lt;math&gt;B&lt;/math&gt;) to give us the probability of overlap between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> Doing so gives us:<br /> <br /> &lt;math&gt;\frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies{61 + 109 + 621 + 39 + 20 = \fbox{850}}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Solution 4==<br /> Note that all the &quot;bounds&quot; have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is &lt;math&gt;\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}&lt;/math&gt;<br /> <br /> ~Lcz<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=14 | after=Last Problem}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_11&diff=125037 1988 AIME Problems/Problem 11 2020-06-11T21:29:22Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; be [[complex number]]s. A line &lt;math&gt;L&lt;/math&gt; in the [[complex plane]] is called a mean [[line]] for the [[point]]s &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; if &lt;math&gt;L&lt;/math&gt; contains points (complex numbers) &lt;math&gt;z_1, z_2, \dots, z_n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> \sum_{k = 1}^n (z_k - w_k) = 0.<br /> &lt;/cmath&gt;<br /> For the numbers &lt;math&gt;w_1 = 32 + 170i&lt;/math&gt;, &lt;math&gt;w_2 = - 7 + 64i&lt;/math&gt;, &lt;math&gt;w_3 = - 9 + 200i&lt;/math&gt;, &lt;math&gt;w_4 = 1 + 27i&lt;/math&gt;, and &lt;math&gt;w_5 = - 14 + 43i&lt;/math&gt;, there is a unique mean line with &lt;math&gt;y&lt;/math&gt;-intercept 3. Find the [[slope]] of this mean line.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> &lt;math&gt;\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> Each &lt;math&gt;z_k = x_k + y_ki&lt;/math&gt; lies on the complex line &lt;math&gt;y = mx + 3&lt;/math&gt;, so we can rewrite this as <br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)&lt;/math&gt;<br /> <br /> Matching the real parts and the imaginary parts, we get that &lt;math&gt;\sum_{k=1}^5 x_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 (mx_k + 3) = 504&lt;/math&gt;. Simplifying the second summation, we find that &lt;math&gt;m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489&lt;/math&gt;, and substituting, the answer is &lt;math&gt;m \cdot 3 = 489 \Longrightarrow m = 163&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 w_k = 3 + 504i&lt;/math&gt;<br /> <br /> And because the sum of the 5 &lt;math&gt;z&lt;/math&gt;'s must cancel this out,<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> We write the numbers in the form &lt;math&gt;a + bi&lt;/math&gt; and we know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 a_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 b_k = 504&lt;/math&gt;<br /> <br /> The line is of equation &lt;math&gt;y=mx+3&lt;/math&gt;. Substituting in the polar coordinates, we have &lt;math&gt;b_k = ma_k + 3&lt;/math&gt;.<br /> <br /> Summing all 5 of the equations given for each &lt;math&gt;k&lt;/math&gt;, we get <br /> <br /> &lt;math&gt;504 = 3m + 15&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;m&lt;/math&gt;, the slope, we get &lt;math&gt;\boxed{163}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> The mean line for &lt;math&gt;w_1, . . ., w_5&lt;/math&gt; must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is &lt;math&gt;(\frac{3}{5}, \frac{504i}{5})&lt;/math&gt;. Since we now have two points, namely that one and &lt;math&gt;(0, 3i)&lt;/math&gt;, we can simply find the slope between them, which is &lt;math&gt;\boxed{163}&lt;/math&gt; by the good ol' slope formula.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_11&diff=125036 1988 AIME Problems/Problem 11 2020-06-11T21:28:06Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; be [[complex number]]s. A line &lt;math&gt;L&lt;/math&gt; in the [[complex plane]] is called a mean [[line]] for the [[point]]s &lt;math&gt;w_1, w_2, \dots, w_n&lt;/math&gt; if &lt;math&gt;L&lt;/math&gt; contains points (complex numbers) &lt;math&gt;z_1, z_2, \dots, z_n&lt;/math&gt; such that<br /> &lt;cmath&gt;<br /> \sum_{k = 1}^n (z_k - w_k) = 0.<br /> &lt;/cmath&gt;<br /> For the numbers &lt;math&gt;w_1 = 32 + 170i&lt;/math&gt;, &lt;math&gt;w_2 = - 7 + 64i&lt;/math&gt;, &lt;math&gt;w_3 = - 9 + 200i&lt;/math&gt;, &lt;math&gt;w_4 = 1 + 27i&lt;/math&gt;, and &lt;math&gt;w_5 = - 14 + 43i&lt;/math&gt;, there is a unique mean line with &lt;math&gt;y&lt;/math&gt;-intercept 3. Find the [[slope]] of this mean line.<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> &lt;math&gt;\sum_{k=1}^5 z_k - \sum_{k=1}^5 w_k = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> Each &lt;math&gt;z_k = x_k + y_ki&lt;/math&gt; lies on the complex line &lt;math&gt;y = mx + 3&lt;/math&gt;, so we can rewrite this as <br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = \sum_{k=1}^5 x_k + \sum_{k=1}^n y_ki&lt;/math&gt;<br /> <br /> &lt;math&gt;3 + 504i = \sum_{k=1}^5 x_k + i \sum_{k=1}^5 (mx_k + 3)&lt;/math&gt;<br /> <br /> Matching the real parts and the imaginary parts, we get that &lt;math&gt;\sum_{k=1}^5 x_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 (mx_k + 3) = 504&lt;/math&gt;. Simplifying the second summation, we find that &lt;math&gt;m\sum_{k=1}^5 x_k = 504 - 3 \cdot 5 = 489&lt;/math&gt;, and substituting, the answer is &lt;math&gt;m \cdot 3 = 489 \Longrightarrow m = 163&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> We know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 w_k = 3 + 504i&lt;/math&gt;<br /> <br /> And because the sum of the 5 &lt;math&gt;z&lt;/math&gt;'s must cancel this out,<br /> <br /> &lt;math&gt;\sum_{k=1}^5 z_k = 3 + 504i&lt;/math&gt;<br /> <br /> We write the numbers in the form &lt;math&gt;a + bi&lt;/math&gt; and we know that <br /> <br /> &lt;math&gt;\sum_{k=1}^5 a_k = 3&lt;/math&gt; and &lt;math&gt;\sum_{k=1}^5 b_k = 504&lt;/math&gt;<br /> <br /> The line is of equation &lt;math&gt;y=mx+3&lt;/math&gt;. Substituting in the polar coordinates, we have &lt;math&gt;b_k = ma_k + 3&lt;/math&gt;.<br /> <br /> Summing all 5 of the equations given for each &lt;math&gt;k&lt;/math&gt;, we get <br /> <br /> &lt;math&gt;504 = 3m + 15&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;m&lt;/math&gt;, the slope, we get &lt;math&gt;\boxed{163}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> <br /> The mean line for &lt;math&gt;w_1, . . ., w_5&lt;/math&gt; must pass through the mean (the center of mass) of these points, which, if we graph these points on the complex plane, is &lt;math&gt;(\frac{3}{5}, \frac{504i}{5})&lt;/math&gt;. Since we now have two points, namely that one and &lt;math&gt;(0, 3i)&lt;/math&gt;, we can simply find the slope between them, which is &lt;math&gt;\boxed{163}&lt;/math&gt; by the good ol' slope formula.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123245 2018 AIME II Problems/Problem 10 2020-05-29T14:07:09Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123139 2018 AIME II Problems/Problem 10 2020-05-28T03:59:26Z <p>Anellipticcurveoverq: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 3==<br /> <br /> Note that there are &lt;math&gt;5^5&lt;/math&gt; possible functions &lt;math&gt;f(x)&lt;/math&gt;. <br /> <br /> Now consider the probability of picking a function from those &lt;math&gt;3125&lt;/math&gt; functions that satisfies, exclusively, one of the following criteria:<br /> <br /> &lt;math&gt;P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} = f(f(x)) = f(f(f(x)))) = \frac{1}{5^4}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x = f(x) \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5}&lt;/math&gt;<br /> <br /> Thus the number of functions &lt;math&gt;f(x)&lt;/math&gt; satisfying the condition stated in the problem is given by: <br /> <br /> &lt;math&gt;\frac{5^5}{5^5} + \frac{5^5}{5^4} + \frac{5^5}{5^2} + \frac{5^5}{5} = 1 + 5 + 125 + 625 = \boxed{756}&lt;/math&gt;<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123138 2018 AIME II Problems/Problem 10 2020-05-28T03:52:06Z <p>Anellipticcurveoverq: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 3==<br /> <br /> Note that there are &lt;math&gt;5^5&lt;/math&gt; possible functions &lt;math&gt;f(x)&lt;/math&gt;. <br /> <br /> Now consider the probability of picking a function from those &lt;math&gt;3125&lt;/math&gt; functions that satisfies, exclusively, one of the following criteria:<br /> <br /> &lt;math&gt;P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} = f(f(x)) = f(f(f(x)))) = \frac{1}{5^4}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x = f(x) \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5}&lt;/math&gt;<br /> <br /> Thus the number of functions &lt;math&gt;f(x)&lt;/math&gt; satisfying the given condition is given by: <br /> <br /> &lt;math&gt;\frac{5^5}{5^5} + \frac{5^5}{5^4} + \frac{5^5}{5^2} + \frac{5^5}{5} = 1 + 5 + 125 + 625 = \boxed{756}&lt;/math&gt;<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123137 2018 AIME II Problems/Problem 10 2020-05-28T03:51:32Z <p>Anellipticcurveoverq: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 3==<br /> <br /> Note that there are &lt;math&gt;5^5&lt;/math&gt; possible functions &lt;math&gt;f(x)&lt;/math&gt;. <br /> <br /> Now consider the probability of picking a function from those &lt;math&gt;3125&lt;/math&gt; functions that satisfies, exclusively, the following criteria:<br /> <br /> &lt;math&gt;P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} = f(f(x)) = f(f(f(x)))) = \frac{1}{5^4}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x = f(x) \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5}&lt;/math&gt;<br /> <br /> Thus the number of functions &lt;math&gt;f(x)&lt;/math&gt; satisfying the given condition is given by: <br /> <br /> &lt;math&gt;\frac{5^5}{5^5} + \frac{5^5}{5^4} + \frac{5^5}{5^2} + \frac{5^5}{5} = 1 + 5 + 125 + 625 = \boxed{756}&lt;/math&gt;<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123077 2018 AIME II Problems/Problem 10 2020-05-27T21:05:55Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 3==<br /> <br /> Note that there are &lt;math&gt;5^5&lt;/math&gt; possible functions &lt;math&gt;f(x)&lt;/math&gt;. <br /> <br /> Then:<br /> <br /> &lt;math&gt;P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} = f(f(x)) = f(f(f(x)))) = \frac{1}{5^4}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x = f(x) \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5^2}&lt;/math&gt;<br /> <br /> &lt;math&gt;P(x \neq{f(x)} \neq{f(f(x))} = f(f(f(x)))) = \frac{1}{5}&lt;/math&gt;<br /> <br /> Thus the number of functions &lt;math&gt;f(x)&lt;/math&gt; satisfying the given condition is given by: <br /> <br /> &lt;math&gt;\frac{5^5}{5^5} + \frac{5^5}{5^4} + \frac{5^5}{5^2} + \frac{5^5}{5} = 1 + 5 + 125 + 625 = \boxed{756}&lt;/math&gt;<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_10&diff=123075 2018 AIME II Problems/Problem 10 2020-05-27T20:46:11Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Find the number of functions &lt;math&gt;f(x)&lt;/math&gt; from &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; to &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt; that satisfy &lt;math&gt;f(f(x)) = f(f(f(x)))&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;\{1, 2, 3, 4, 5\}&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Just to visualize solution 1. If we list all possible &lt;math&gt;(x,f(x))&lt;/math&gt;, from &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; to &lt;math&gt;{1,2,3,4,5}&lt;/math&gt; in a specific order, we get &lt;math&gt;5*5 = 25&lt;/math&gt; different &lt;math&gt;(x,f(x))&lt;/math&gt; 's.<br /> Namely:<br /> <br /> &lt;math&gt;(1,1) (1,2) (1,3) (1,4) (1,5)<br /> (2,1) (2,2) (2,3) (2,4) (2,5) <br /> (3,1) (3,2) (3,3) (3,4) (3,5) <br /> (4,1) (4,2) (4,3) (4,4) (4,5)<br /> (5,1) (5,2) (5,3) (5,4) (5,5)&lt;/math&gt;<br /> <br /> To list them in this specific order makes it a lot easier to solve this problem. We notice, In order to solve this problem at least one pair of &lt;math&gt;(x,x)&lt;/math&gt; where &lt;math&gt;x\in{1,2,3,4,5}&lt;/math&gt; must exist.In this case I rather &quot;go backwards&quot;. First fixing &lt;math&gt;5&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt;, (the diagonal of our table) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{5!} = 1&lt;/math&gt; way. Then fixing &lt;math&gt;4&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;1&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in <br /> &lt;math&gt;4\cdot\frac{5!}{4!} = 20&lt;/math&gt; ways. Then fixing &lt;math&gt;3&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (The diagonal minus &lt;math&gt;2&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot3\cdot6\cdot3)}{3!2!} + \frac{(5\cdot4\cdot3\cdot6\cdot1)}{3!} = 150&lt;/math&gt; ways.<br /> Fixing &lt;math&gt;2&lt;/math&gt; pairs &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;3&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!3!} + \frac{(5\cdot4\cdot6\cdot4\cdot2)}{2!2!} + \frac{(5\cdot4\cdot6\cdot2\cdot1)}{2!2!} = 380&lt;/math&gt; ways.<br /> Lastely, fixing &lt;math&gt;1&lt;/math&gt; pair &lt;math&gt;(x,x)&lt;/math&gt; (the diagonal minus &lt;math&gt;4&lt;/math&gt;) and map them to the other fitting pairs &lt;math&gt;(x,f(x))&lt;/math&gt;. You can do this in &lt;math&gt;\frac{5!}{4!} + 4\cdot\frac{5!}{3!} + 5! = 205&lt;/math&gt;<br /> <br /> So &lt;math&gt;1 + 20 + 150 + 380 + 205 = \framebox{756}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> <br /> We can do some caseworks about the special points of functions &lt;math&gt;f&lt;/math&gt; for &lt;math&gt;x\in\{1,2,3,4,5\}&lt;/math&gt;. Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; be three different elements in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt;. There must be elements such like &lt;math&gt;k&lt;/math&gt; in set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; satisfies &lt;math&gt;f(k)=k&lt;/math&gt;, and we call the points such like &lt;math&gt;(k,k)&lt;/math&gt; on functions &lt;math&gt;f&lt;/math&gt; are &quot;Good Points&quot; (Actually its academic name is &quot;fixed-points&quot;). The only thing we need to consider is the &quot;steps&quot; to get &quot;Good Points&quot;. Notice that the &quot;steps&quot; must less than &lt;math&gt;3&lt;/math&gt; because the highest iterations of function &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt;. Now we can classify &lt;math&gt;3&lt;/math&gt; cases of “Good points” of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; One &quot;step&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=x&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(x))=f(x)=x&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt; and &lt;math&gt;f(y)=y&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=y&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(y)=y&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; Three &quot;steps&quot; to &quot;Good Points&quot;: Assume that &lt;math&gt;f(x)=y&lt;/math&gt;, &lt;math&gt;f(y)=z&lt;/math&gt; and &lt;math&gt;f(z)=z&lt;/math&gt;, then we get &lt;math&gt;f(f(x))=f(y)=z&lt;/math&gt;, and &lt;math&gt;f(f(f(x)))=f(f(y))=f(z)=z&lt;/math&gt;, so &lt;math&gt;f(f(f(x)))=f(f(x))&lt;/math&gt;.<br /> <br /> Divide set &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; into three parts which satisfy these three cases, respectively. Let the first part has &lt;math&gt;a&lt;/math&gt; elements, the second part has &lt;math&gt;b&lt;/math&gt; elements and the third part has &lt;math&gt;c&lt;/math&gt; elements, it is easy to see that &lt;math&gt;a+b+c=5&lt;/math&gt;. First, there are &lt;math&gt;\binom{5}{a}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 1. Second, we have &lt;math&gt;\binom{5-a}{b}&lt;/math&gt; ways to select &lt;math&gt;x&lt;/math&gt; for Case 2. After that we map all elements that satisfy Case 2 to Case 1, and the total number of ways of this operation is &lt;math&gt;a^b&lt;/math&gt;. Finally, we map all the elements that satisfy Case 3 to Case 2, and the total number of ways of this operation is &lt;math&gt;b^c&lt;/math&gt;. As a result, the number of such functions &lt;math&gt;f&lt;/math&gt; can be represented in an algebraic expression contains &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;\boxed{\binom{5}{a}\cdot \binom{5-a}{b}\cdot a^b\cdot b^c}&lt;/math&gt;<br /> <br /> Now it's time to consider about the different values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; and the total number of functions &lt;math&gt;f&lt;/math&gt; satisfy these values of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;:<br /> <br /> For &lt;math&gt;a=5&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{5}=1&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{4}\cdot \binom{1}{1}\cdot 4^1\cdot 1^0=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{1}\cdot 3^1\cdot 1^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=3&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{3}\cdot \binom{2}{2}\cdot 3^2\cdot 2^0=90&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{1}\cdot 2^1\cdot 1^2=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{2}\cdot 2^2\cdot 2^1=240&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{2}\cdot \binom{3}{3}\cdot 2^3\cdot 3^0=80&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt; and &lt;math&gt;c=3&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{1}\cdot 1^1\cdot 1^3=20&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=2&lt;/math&gt; and &lt;math&gt;c=2&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{2}\cdot 1^2\cdot 2^2=120&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=3&lt;/math&gt; and &lt;math&gt;c=1&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{3}\cdot 1^3\cdot 3^1=60&lt;/math&gt;<br /> <br /> For &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=4&lt;/math&gt; and &lt;math&gt;c=0&lt;/math&gt;, the number of &lt;math&gt;f&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\cdot \binom{4}{4}\cdot 1^4\cdot 4^0=5&lt;/math&gt;<br /> <br /> Finally, we get the total number of function &lt;math&gt;f&lt;/math&gt;, the number is &lt;math&gt;1+20+60+90+60+240+80+20+120+60+5=\boxed{756}&lt;/math&gt;<br /> <br /> ~Solution by &lt;math&gt;BladeRunnerAUG&lt;/math&gt; (Frank FYC)<br /> <br /> ==Solution 3==<br /> <br /> Note that there are &lt;math&gt;5^5&lt;/math&gt; possible functions &lt;math&gt;f(x)&lt;/math&gt;. <br /> <br /> Then:<br /> <br /> &lt;math&gt;P(x = f(x) = f(f(x)) = f(f(f(x)))) = \frac{1}{5^5}&lt;/math&gt;<br /> ==Note (fun fact)==<br /> This exact problem showed up earlier on the 2011 Stanford Math Tournament, Advanced Topics Test.<br /> This problem also showed up on the 2010 Mock AIME 2 here: https://artofproblemsolving.com/wiki/index.php/Mock_AIME_2_2010_Problems<br /> {{AIME box|year=2018|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=121421 2011 AIME II Problems/Problem 15 2020-04-21T19:15:20Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;P(x) = x^2 - 3x - 9&lt;/math&gt;. A real number &lt;math&gt;x&lt;/math&gt; is chosen at random from the interval &lt;math&gt;5 \le x \le 15&lt;/math&gt;. The probability that &lt;math&gt;\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; is equal to &lt;math&gt;\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}&lt;/math&gt; , where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b + c + d + e&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Table of values of &lt;math&gt;P(x)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> P(5) &amp;= 1 \\<br /> P(6) &amp;= 9 \\<br /> P(7) &amp;= 19 \\<br /> P(8) &amp;= 31 \\<br /> P(9) &amp;= 45 \\<br /> P(10) &amp;= 61 \\<br /> P(11) &amp;= 79 \\<br /> P(12) &amp;= 99 \\<br /> P(13) &amp;= 121 \\<br /> P(14) &amp;= 145 \\<br /> P(15) &amp;= 171 \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> In order for &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; to hold, &lt;math&gt;\sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; must be an integer and hence &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; must be a perfect square. This limits &lt;math&gt;x&lt;/math&gt; to &lt;math&gt;5 \le x &lt; 6&lt;/math&gt; or &lt;math&gt;6 \le x &lt; 7&lt;/math&gt; or &lt;math&gt;13 \le x &lt; 14&lt;/math&gt; since, from the table above, those are the only values of &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; is an perfect square. However, in order for &lt;math&gt;\sqrt{P(x)}&lt;/math&gt; to be rounded down to &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt;, &lt;math&gt;P(x)&lt;/math&gt; must be less than the next perfect square after &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; (for the said intervals). Now, we consider the three cases:<br /> <br /> <br /> Case &lt;math&gt;5 \le x &lt; 6&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;1&lt;/math&gt;, which is &lt;math&gt;4&lt;/math&gt;, ''i.e.'':<br /> <br /> &lt;math&gt;1 \le P(x) &lt; 4&lt;/math&gt; (because &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = 1&lt;/math&gt; implies &lt;math&gt;1 \le \sqrt{P(x)} &lt; 2&lt;/math&gt;)<br /> <br /> Since &lt;math&gt;P(x)&lt;/math&gt; is increasing for &lt;math&gt;x \ge 5&lt;/math&gt;, we just need to find the value &lt;math&gt;v \ge 5&lt;/math&gt; where &lt;math&gt;P(v) = 4&lt;/math&gt;, which will give us the working range &lt;math&gt;5 \le x &lt; v&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 4 \\<br /> v &amp;= \frac{3 + \sqrt{61}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;5 \le x &lt; \frac{3 + \sqrt{61}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;6 \le x &lt; 7&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;9&lt;/math&gt;, which is &lt;math&gt;16&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 16 \\<br /> v &amp;= \frac{3 + \sqrt{109}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;6 \le x &lt; \frac{3 + \sqrt{109}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;13 \le x &lt; 14&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;121&lt;/math&gt;, which is &lt;math&gt;144&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 144 \\<br /> v &amp;= \frac{3 + \sqrt{621}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;13 \le x &lt; \frac{3 + \sqrt{621}}{2}&lt;/math&gt;.<br /> <br /> Now, we find the length of the working intervals and divide it by the length of the total interval, &lt;math&gt;15 - 5 = 10&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\<br /> &amp;= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus, the answer is &lt;math&gt;61 + 109 + 621 + 39 + 20 = \fbox{850}&lt;/math&gt;.<br /> <br /> P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.<br /> <br /> == Solution 2 == <br /> <br /> Make the substitution &lt;math&gt;y=2x-3&lt;/math&gt;, so &lt;math&gt;P(x)=\frac{y^2-45}{4}.&lt;/math&gt; We're looking for solutions to &lt;cmath&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}&lt;/cmath&gt;with the new bounds &lt;math&gt;y\in{[7,27]}&lt;/math&gt;. Since the right side is an integer, it must be that &lt;math&gt;\frac{\lfloor{y\rfloor}^2-45}{4}&lt;/math&gt; is a perfect square. For simplicity, write &lt;math&gt;\lfloor{y\rfloor}=a&lt;/math&gt; and &lt;cmath&gt;a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.&lt;/cmath&gt;Since &lt;math&gt;a-2b&lt;a+2b&lt;/math&gt;, it must be that &lt;math&gt;(a-2b,a+2b)=(1,45),(3,15),(5,9)&lt;/math&gt;, which gives solutions &lt;math&gt;(23,11),(9,3),(7,1)&lt;/math&gt;, respectively. But this gives us three cases to check:<br /> <br /> Case 1: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;11\leq{\sqrt{\frac{y^2-45}{4}}}&lt;12\implies{y\in{[23,\sqrt{621})}}.&lt;/cmath&gt;<br /> Case 2: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;3\leq{\sqrt{\frac{y^2-45}{4}}}&lt;4\implies{y\in{[9,\sqrt{109})}}.&lt;/cmath&gt;<br /> Case 3: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1&lt;/math&gt;<br /> <br /> In this case, we have &lt;cmath&gt;1\leq{\sqrt{\frac{y^2-45}{4}}}&lt;2\implies{y\in{[7,\sqrt{61})}}.&lt;/cmath&gt;<br /> To finish, the total length of the interval from which we choose &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;27-7=20&lt;/math&gt;. The total length of the success intervals is &lt;cmath&gt;(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,&lt;/cmath&gt;which means the probability is &lt;cmath&gt;\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.&lt;/cmath&gt;AIMEifying the answer gives &lt;math&gt;\boxed{850}&lt;/math&gt;.<br /> <br /> == Solution 3 (Graphing) == <br /> <br /> It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let &lt;math&gt;A = \lfloor\sqrt{P(x)}\rfloor&lt;/math&gt; and &lt;math&gt;B = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt;. The graph of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will look like this, with &lt;math&gt;A&lt;/math&gt; having only integral y-values and &lt;math&gt;B&lt;/math&gt; having only integral x-values:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 1.png|400px]]<br /> <br /> As both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; consist of a bunch of line segments, the probability that &lt;math&gt;A = B&lt;/math&gt; is the &quot;length&quot; of the overlap between the segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; divided by the total length of the segments of &lt;math&gt;B&lt;/math&gt;. <br /> <br /> Looking at the graph, we see that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will overlap only when &lt;math&gt;B&lt;/math&gt; is an integer. Specifically, each region of overlap will begin when &lt;math&gt;\sqrt{P(x)}\ = k, 5 \le x \le 15&lt;/math&gt; has solutions for integral &lt;math&gt;k&lt;/math&gt; in the range of &lt;math&gt;A&lt;/math&gt;, which consists of the integers &lt;math&gt;1-13&lt;/math&gt;, and end when &lt;math&gt;A&lt;/math&gt; jumps up to its next y-value.<br /> <br /> Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of &lt;math&gt;\frac{3 + \sqrt{45 + 4k^2}}{2}&lt;/math&gt; for &lt;math&gt;k&lt;/math&gt; in the specified range, meaning &lt;math&gt;45 + 4k^2&lt;/math&gt; must be a perfect square. Plugging in all the possible values of &lt;math&gt;k&lt;/math&gt;, we get &lt;math&gt;k = 1, 3, 11&lt;/math&gt;, corresponding to start points of &lt;math&gt;x = 5, 6, 13&lt;/math&gt;. As already stated, the endpoints will occur when &lt;math&gt;A&lt;/math&gt; jumps up to the next integer &lt;math&gt;k&lt;/math&gt; at each of these segments, at which point the x-value will be &lt;math&gt;\frac{3 + \sqrt{45 + 4k^2}}{2}&lt;/math&gt;. On the graph, the overlapping segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; would be represented by the highlighted green segments below:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 2.png|400px]]<br /> <br /> <br /> Taking the difference between this second x-value and the start point for each of our start points &lt;math&gt;x = 5, 6, 13&lt;/math&gt; and summing them will give us the total length of these green segments. We can then divide this value by ten (the total length of the segments of &lt;math&gt;B&lt;/math&gt;) to give us the probability of overlap between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> Doing so gives us:<br /> <br /> &lt;math&gt;\frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies{61 + 109 + 621 + 39 + 20 = \fbox{850}}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=14 | after=Last Problem}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_15&diff=121420 2011 AIME II Problems/Problem 15 2020-04-21T19:13:56Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;P(x) = x^2 - 3x - 9&lt;/math&gt;. A real number &lt;math&gt;x&lt;/math&gt; is chosen at random from the interval &lt;math&gt;5 \le x \le 15&lt;/math&gt;. The probability that &lt;math&gt;\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; is equal to &lt;math&gt;\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}&lt;/math&gt; , where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; are positive integers. Find &lt;math&gt;a + b + c + d + e&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Table of values of &lt;math&gt;P(x)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> P(5) &amp;= 1 \\<br /> P(6) &amp;= 9 \\<br /> P(7) &amp;= 19 \\<br /> P(8) &amp;= 31 \\<br /> P(9) &amp;= 45 \\<br /> P(10) &amp;= 61 \\<br /> P(11) &amp;= 79 \\<br /> P(12) &amp;= 99 \\<br /> P(13) &amp;= 121 \\<br /> P(14) &amp;= 145 \\<br /> P(15) &amp;= 171 \\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> In order for &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; to hold, &lt;math&gt;\sqrt{P(\lfloor x \rfloor)}&lt;/math&gt; must be an integer and hence &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; must be a perfect square. This limits &lt;math&gt;x&lt;/math&gt; to &lt;math&gt;5 \le x &lt; 6&lt;/math&gt; or &lt;math&gt;6 \le x &lt; 7&lt;/math&gt; or &lt;math&gt;13 \le x &lt; 14&lt;/math&gt; since, from the table above, those are the only values of &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; is an perfect square. However, in order for &lt;math&gt;\sqrt{P(x)}&lt;/math&gt; to be rounded down to &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt;, &lt;math&gt;P(x)&lt;/math&gt; must be less than the next perfect square after &lt;math&gt;P(\lfloor x \rfloor)&lt;/math&gt; (for the said intervals). Now, we consider the three cases:<br /> <br /> <br /> Case &lt;math&gt;5 \le x &lt; 6&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;1&lt;/math&gt;, which is &lt;math&gt;4&lt;/math&gt;, ''i.e.'':<br /> <br /> &lt;math&gt;1 \le P(x) &lt; 4&lt;/math&gt; (because &lt;math&gt;\lfloor \sqrt{P(x)} \rfloor = 1&lt;/math&gt; implies &lt;math&gt;1 \le \sqrt{P(x)} &lt; 2&lt;/math&gt;)<br /> <br /> Since &lt;math&gt;P(x)&lt;/math&gt; is increasing for &lt;math&gt;x \ge 5&lt;/math&gt;, we just need to find the value &lt;math&gt;v \ge 5&lt;/math&gt; where &lt;math&gt;P(v) = 4&lt;/math&gt;, which will give us the working range &lt;math&gt;5 \le x &lt; v&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 4 \\<br /> v &amp;= \frac{3 + \sqrt{61}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;5 \le x &lt; \frac{3 + \sqrt{61}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;6 \le x &lt; 7&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;9&lt;/math&gt;, which is &lt;math&gt;16&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 16 \\<br /> v &amp;= \frac{3 + \sqrt{109}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;6 \le x &lt; \frac{3 + \sqrt{109}}{2}&lt;/math&gt;.<br /> <br /> Case &lt;math&gt;13 \le x &lt; 14&lt;/math&gt;:<br /> <br /> &lt;math&gt;P(x)&lt;/math&gt; must be less than the first perfect square after &lt;math&gt;121&lt;/math&gt;, which is &lt;math&gt;144&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> v^2 - 3v - 9 &amp;= 144 \\<br /> v &amp;= \frac{3 + \sqrt{621}}{2}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> So in this case, the only values that will work are &lt;math&gt;13 \le x &lt; \frac{3 + \sqrt{621}}{2}&lt;/math&gt;.<br /> <br /> Now, we find the length of the working intervals and divide it by the length of the total interval, &lt;math&gt;15 - 5 = 10&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> \frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} \\<br /> &amp;= \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Thus, the answer is &lt;math&gt;61 + 109 + 621 + 39 + 20 = \fbox{850}&lt;/math&gt;.<br /> <br /> P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.<br /> <br /> == Solution 2 == <br /> <br /> Make the substitution &lt;math&gt;y=2x-3&lt;/math&gt;, so &lt;math&gt;P(x)=\frac{y^2-45}{4}.&lt;/math&gt; We're looking for solutions to &lt;cmath&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}&lt;/cmath&gt;with the new bounds &lt;math&gt;y\in{[7,27]}&lt;/math&gt;. Since the right side is an integer, it must be that &lt;math&gt;\frac{\lfloor{y\rfloor}^2-45}{4}&lt;/math&gt; is a perfect square. For simplicity, write &lt;math&gt;\lfloor{y\rfloor}=a&lt;/math&gt; and &lt;cmath&gt;a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.&lt;/cmath&gt;Since &lt;math&gt;a-2b&lt;a+2b&lt;/math&gt;, it must be that &lt;math&gt;(a-2b,a+2b)=(1,45),(3,15),(5,9)&lt;/math&gt;, which gives solutions &lt;math&gt;(23,11),(9,3),(7,1)&lt;/math&gt;, respectively. But this gives us three cases to check:<br /> <br /> Case 1: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;11\leq{\sqrt{\frac{y^2-45}{4}}}&lt;12\implies{y\in{[23,\sqrt{621})}}.&lt;/cmath&gt;<br /> Case 2: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3&lt;/math&gt;.<br /> <br /> In this case, we have &lt;cmath&gt;3\leq{\sqrt{\frac{y^2-45}{4}}}&lt;4\implies{y\in{[9,\sqrt{109})}}.&lt;/cmath&gt;<br /> Case 3: &lt;math&gt;\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1&lt;/math&gt;<br /> <br /> In this case, we have &lt;cmath&gt;1\leq{\sqrt{\frac{y^2-45}{4}}}&lt;2\implies{y\in{[7,\sqrt{61})}}.&lt;/cmath&gt;<br /> To finish, the total length of the interval from which we choose &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;27-7=20&lt;/math&gt;. The total length of the success intervals is &lt;cmath&gt;(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,&lt;/cmath&gt;which means the probability is &lt;cmath&gt;\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.&lt;/cmath&gt;AIMEifying the answer gives &lt;math&gt;\boxed{850}&lt;/math&gt;.<br /> <br /> == Solution 3 (Graphing) == <br /> <br /> It's definitely possible to conceptualize this problem visually, if that's your thing, since it is a geometric probability problem. Let &lt;math&gt;A = \lfloor\sqrt{P(x)}\rfloor&lt;/math&gt; and &lt;math&gt;B = \sqrt{P(\lfloor x \rfloor)}&lt;/math&gt;. The graph of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will look like this, with &lt;math&gt;A&lt;/math&gt; having only integral y-values and &lt;math&gt;B&lt;/math&gt; having only integral x-values:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 1.png|400px]]<br /> <br /> As both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; consist of a bunch of line segments, the probability that &lt;math&gt;A = B&lt;/math&gt; is the &quot;length&quot; of the overlap between the segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; divided by the total length of the segments of &lt;math&gt;B&lt;/math&gt;. <br /> <br /> Looking at the graph, we see that &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; will overlap only when &lt;math&gt;B&lt;/math&gt; is an integer. Specifically, each region of overlap will begin when &lt;math&gt;\sqrt{P(x)}\ = k, 5 \le x \le 15&lt;/math&gt; has solutions for integral &lt;math&gt;k&lt;/math&gt; in the range of &lt;math&gt;A&lt;/math&gt;, which consists of the integers &lt;math&gt;1-13&lt;/math&gt;, and end when &lt;math&gt;A&lt;/math&gt; jumps up to its next y-value.<br /> <br /> Using the quadratic formula, we see that the start point of each of these overlapping segments will be the integral values of &lt;math&gt;\frac{3 + \sqrt{45 + 4k^2}}{2}&lt;/math&gt; for &lt;math&gt;k&lt;/math&gt; in the specified range, meaning &lt;math&gt;45 + 4k^2&lt;/math&gt; must be a perfect square. Plugging in all the possible values of &lt;math&gt;k&lt;/math&gt;, we get &lt;math&gt;k = 1, 3, 11&lt;/math&gt;, corresponding to start points of &lt;math&gt;x = 5, 6, 13&lt;/math&gt;. As already stated, the endpoints will occur when &lt;math&gt;A&lt;/math&gt; jumps up to the next integer &lt;math&gt;k&lt;/math&gt; at each of these segments, at which point the x-value will be &lt;math&gt;\frac{3 + \sqrt{45 + 4k^2}}{2}&lt;/math&gt;. On the graph, the overlapping segments of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; would be represented by the highlighted green segments below:<br /> <br /> [[File:2011 AIME II Problem 15 Graph 2.png|400px]]<br /> <br /> <br /> Taking the difference between this second x-value and the start point for each of our start points &lt;math&gt;x = 5, 6, 13&lt;/math&gt; and summing them will give us the total length of these overlapping segments. We can then divide this value by ten (the total length of the segments of &lt;math&gt;B&lt;/math&gt;) to give us the probability of overlap between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> Doing so gives us:<br /> <br /> &lt;math&gt;\frac{\left( \frac{3 + \sqrt{61}}{2} - 5 \right) + \left( \frac{3 + \sqrt{109}}{2} - 6 \right) + \left( \frac{3 + \sqrt{621}}{2} - 13 \right)}{10} = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20}&lt;/math&gt;<br /> <br /> &lt;math&gt;\implies{61 + 109 + 621 + 39 + 20 = \fbox{850}}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> ==See also==<br /> {{AIME box | year = 2011 | n = II | num-b=14 | after=Last Problem}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=File:2011_AIME_II_Problem_15_Graph_2.png&diff=121417 File:2011 AIME II Problem 15 Graph 2.png 2020-04-21T18:57:21Z <p>Anellipticcurveoverq: </p> <hr /> <div></div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=File:2011_AIME_II_Problem_15_Graph_1.png&diff=121405 File:2011 AIME II Problem 15 Graph 1.png 2020-04-21T18:23:07Z <p>Anellipticcurveoverq: Graph of A and B from the problem (see Solution 3).</p> <hr /> <div>== Summary ==<br /> Graph of A and B from the problem (see Solution 3).</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121051 2007 AIME I Problems/Problem 14 2020-04-16T16:02:10Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the numerator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the second-largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, whose &lt;math&gt;\lim_{k\to\infty}=&lt;/math&gt; &lt;math&gt;\boxed{224}&lt;/math&gt;.<br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121050 2007 AIME I Problems/Problem 14 2020-04-16T16:00:52Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the numerator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the second-largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, whose &lt;math&gt;\lim_{k\to\infty}&lt;/math&gt; is &lt;math&gt;\boxed{224}&lt;/math&gt;.<br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121049 2007 AIME I Problems/Problem 14 2020-04-16T15:57:09Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the numerator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the second-largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, an expression whose limit is &lt;math&gt;\boxed{224}&lt;/math&gt;.<br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121048 2007 AIME I Problems/Problem 14 2020-04-16T15:55:47Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the numerator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the second-largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, an expression whose limit is &lt;math&gt;\boxed{224}&lt;/math&gt;, which will certainly be its floor when &lt;math&gt;n = 2006&lt;/math&gt;. <br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121047 2007 AIME I Problems/Problem 14 2020-04-16T15:54:45Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the numerator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the second-largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, an expression whose eventual limit is &lt;math&gt;\boxed{224}&lt;/math&gt;, which will be its floor when &lt;math&gt;n = 2006&lt;/math&gt;. <br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121046 2007 AIME I Problems/Problem 14 2020-04-16T15:53:04Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the denominator, and that &lt;math&gt;224^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, an expression whose eventual limit is &lt;math&gt;\boxed{224}&lt;/math&gt;, which will be its floor when &lt;math&gt;n = 2006&lt;/math&gt;. <br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_14&diff=121045 2007 AIME I Problems/Problem 14 2020-04-16T15:52:03Z <p>Anellipticcurveoverq: /* Solution 5 (using limits) */</p> <hr /> <div>== Problem ==<br /> A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: &lt;math&gt;a_{0}=a_{1}=3&lt;/math&gt;, &lt;math&gt;a_{n+1}a_{n-1}=a_{n}^{2}+2007&lt;/math&gt;.<br /> <br /> Find the [[floor function|greatest integer]] that does not exceed &lt;math&gt;\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.&lt;/math&gt;<br /> <br /> === Solution 1 ===<br /> We are given that<br /> <br /> &lt;math&gt;a_{n+1}a_{n-1}= a_{n}^{2}+2007&lt;/math&gt;,<br /> &lt;math&gt;a_{n-1}^{2}+2007 = a_{n}a_{n-2}&lt;/math&gt;.<br /> <br /> Add these two equations to get<br /> <br /> :&lt;math&gt;a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})&lt;/math&gt;<br /> <br /> :&lt;math&gt;\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}&lt;/math&gt;.<br /> <br /> This is an [[invariant]]. Defining &lt;math&gt;b_{i}= \frac{a_{i}}{a_{i-1}}&lt;/math&gt; for each &lt;math&gt;i \ge 2&lt;/math&gt;, the above equation means<br /> <br /> &lt;math&gt;b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}&lt;/math&gt;.<br /> <br /> We can thus calculate that &lt;math&gt;b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225&lt;/math&gt;. Using the equation &lt;math&gt;a_{2007}a_{2005}=a_{2006}^{2}+2007&lt;/math&gt; and dividing both sides by &lt;math&gt;a_{2006}a_{2005}&lt;/math&gt;, notice that &lt;math&gt;b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}&gt; \frac{a_{2006}}{a_{2005}}= b_{2006}&lt;/math&gt;. This means that<br /> <br /> &lt;math&gt;b_{2007}+\frac{1}{b_{2007}}&lt; b_{2007}+\frac{1}{b_{2006}}= 225&lt;/math&gt;. It is only a tiny bit less because all the &lt;math&gt;b_i&lt;/math&gt; are greater than &lt;math&gt;1&lt;/math&gt;, so we conclude that the floor of &lt;math&gt;\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}&lt;/math&gt; is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> The equation &lt;math&gt;a_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt; looks like the determinant &lt;cmath&gt;\left|\begin{array}{cc}a_{n+1}&amp;a_n\\a_n&amp;a_{n-1}\end{array}\right|=2007.&lt;/cmath&gt;<br /> Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence &lt;math&gt;b_n&lt;/math&gt; defined by &lt;math&gt;b_1=b_2=3&lt;/math&gt; and &lt;math&gt;b_{n+1}=\alpha b_n+\beta b_{n-1}&lt;/math&gt; for &lt;math&gt;n\ge 2&lt;/math&gt;. We wish to find &lt;math&gt;\alpha&lt;/math&gt; and &lt;math&gt;\beta&lt;/math&gt; such that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. To do this, we use the following matrix form of a linear recurrence relation<br /> <br /> &lt;cmath&gt;\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt; <br /> <br /> When we take determinants, this equation becomes<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We want &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=2007&lt;/cmath&gt; for all &lt;math&gt;n&lt;/math&gt;. Therefore, we replace the two matrices by &lt;math&gt;2007&lt;/math&gt; to find that <br /> <br /> &lt;cmath&gt;2007=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\cdot 2007&lt;/cmath&gt;<br /> &lt;cmath&gt;1=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)=-\beta.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\beta=-1&lt;/math&gt;. Computing that &lt;math&gt;a_3=672&lt;/math&gt;, and using the fact that &lt;math&gt;b_3=\alpha b_2-b_1&lt;/math&gt;, we conclude that &lt;math&gt;\alpha=225&lt;/math&gt;. Clearly, &lt;math&gt;a_1=b_1&lt;/math&gt;, &lt;math&gt;a_2=b_2&lt;/math&gt;, and &lt;math&gt;a_3=b_3&lt;/math&gt;. We claim that &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We proceed by [[induction]]. If &lt;math&gt;a_k=b_k&lt;/math&gt; for all &lt;math&gt;k\le n&lt;/math&gt;, then clearly, &lt;cmath&gt;b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.&lt;/cmath&gt; We also know by the definition of &lt;math&gt;b_{n+1}&lt;/math&gt; that<br /> <br /> &lt;cmath&gt;\text{det}\left(\begin{array}{cc}b_{n+1}&amp;b_n\\b_n&amp;b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&amp;\beta\\1&amp;0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&amp;b_{n-1}\\b_{n-1}&amp;b_{n-2}\end{array}\right).&lt;/cmath&gt;<br /> <br /> We know that the RHS is &lt;math&gt;2007&lt;/math&gt; by previous work. Therefore, &lt;math&gt;b_{n+1}b_{n-1}-b_n^2=2007&lt;/math&gt;. After substuting in the values we know, this becomes &lt;math&gt;b_{n+1}a_{n-1}-a_n^2=2007&lt;/math&gt;. Thinking of this as a linear equation in the variable &lt;math&gt;b_{n+1}&lt;/math&gt;, we already know that this has the solution &lt;math&gt;b_{n+1}=a_{n+1}&lt;/math&gt;. Therefore, by induction, &lt;math&gt;a_n=b_n&lt;/math&gt; for all &lt;math&gt;n\ge 1&lt;/math&gt;. We conclude that &lt;math&gt;a_n&lt;/math&gt; satisfies the linear recurrence &lt;math&gt;a_{n+1}=225a_n-a_{n-1}&lt;/math&gt;.<br /> <br /> It's easy to prove that &lt;math&gt;a_n&lt;/math&gt; is a strictly increasing sequence of integers for &lt;math&gt;n\ge 3&lt;/math&gt;. Now <br /> <br /> &lt;cmath&gt;\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;=225-\frac{2007}{a_{2005}a_{2006}}.&lt;/cmath&gt;<br /> <br /> The sequence certainly grows fast enough such that &lt;math&gt;\frac{2007}{a_{2005}a_{2006}}&lt;1&lt;/math&gt;. Therefore, the largest integer less than or equal to this value is &lt;math&gt;224&lt;/math&gt;.<br /> <br /> ===Solution 3 ( generalized )===<br /> This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to <br /> <br /> &lt;cmath&gt;<br /> a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)<br /> &lt;/cmath&gt;<br /> where &lt;math&gt; k &lt;/math&gt; is a positive integer and &lt;math&gt; a_0 = a_1 = 3. &lt;/math&gt;<br /> <br /> Lemma 1 : For &lt;math&gt;n \geq 1&lt;/math&gt;,<br /> &lt;cmath&gt;<br /> a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)<br /> &lt;/cmath&gt;<br /> We shall prove by induction. From (1), &lt;math&gt; a_2 = 3k + 3 &lt;/math&gt;. From the lemma, &lt;math&gt;a_2 = (k + 2) 3 - 3 = 3k + 3.&lt;/math&gt; Base case proven. Assume that the lemma is true for some &lt;math&gt; t \geq 1 &lt;/math&gt;. Then, eliminating the &lt;math&gt;a_{t-1}&lt;/math&gt; using (1) and (2) gives<br /> <br /> &lt;cmath&gt;<br /> (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)<br /> &lt;/cmath&gt;<br /> <br /> It follows from (2) that <br /> <br /> &lt;cmath&gt;<br /> (k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},<br /> &lt;/cmath&gt;<br /> <br /> where the last line followed from (1) for case &lt;math&gt; n = t+1 &lt;/math&gt;.<br /> <br /> Lemma 2 : For &lt;math&gt;n \geq 0,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_{n+1} \geq a_{n}.<br /> &lt;/cmath&gt;<br /> Base case is obvious. Assume that &lt;math&gt;a_{t+1} \geq a_{t}&lt;/math&gt; for some &lt;math&gt;t \geq 0&lt;/math&gt;. Then it follows that <br /> &lt;cmath&gt;<br /> a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k &gt; a_{t+1}.<br /> &lt;/cmath&gt;<br /> <br /> This completes the induction.<br /> <br /> Lemma 3 : For &lt;math&gt;n \geq 1,&lt;/math&gt;<br /> &lt;cmath&gt;<br /> a_n a_{n+1} &gt; 9k<br /> &lt;/cmath&gt;<br /> <br /> Using (1) and Lemma 2, for &lt;math&gt;n \geq 1,&lt;/math&gt; <br /> &lt;cmath&gt;<br /> a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k &gt; 9k <br /> &lt;/cmath&gt;<br /> <br /> Finally, using (3), for &lt;math&gt;n \geq 1, &lt;/math&gt;<br /> &lt;cmath&gt;<br /> \frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.<br /> &lt;/cmath&gt;<br /> Using lemma 3, the largest integer less than or equal to this value would be &lt;math&gt;k + 1&lt;/math&gt;.<br /> <br /> === Solution 4 (pure algebra)===<br /> We will try to manipulate &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1}&lt;/math&gt; to get &lt;math&gt;\frac{a_1^2+a_2^2}{a_1a_2}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}&lt;/math&gt;<br /> Using the recurrence relation, <br /> &lt;math&gt;= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}&lt;/math&gt;<br /> Applying the relation to &lt;math&gt;a_0a_2&lt;/math&gt;, <br /> &lt;math&gt;= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}&lt;/math&gt;<br /> <br /> We can keep on using this method to get that <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> This telescopes to <br /> &lt;math&gt;\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}&lt;/math&gt;<br /> <br /> or <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;<br /> <br /> Finding the first few values, we notice that they increase rapidly, so &lt;math&gt;\frac{2007}{a_{2006}a_{2007}} &lt; 1&lt;/math&gt;. Calculating the other values, <br /> &lt;math&gt;\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}&lt;/math&gt;.<br /> <br /> The greatest number that does not exceed this is &lt;math&gt;224&lt;/math&gt;<br /> <br /> === Solution 5 (using limits) ===<br /> <br /> Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that &lt;math&gt;a_0 = a_1 = 0&lt;/math&gt;, and solving for &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_3&lt;/math&gt; using the given relation we get &lt;math&gt;a_2 = 672 = 3(224)&lt;/math&gt; and &lt;math&gt;a_3 = 3(224^{2} + 223)&lt;/math&gt;, respectively. It will be clear why I decided to factor these expressions as I did momentarily. <br /> <br /> Next, let's see what the expression &lt;math&gt;\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}&lt;/math&gt; looks like for small values of &lt;math&gt;n&lt;/math&gt;. For &lt;math&gt;n = 1&lt;/math&gt;, we get &lt;math&gt;\frac{1 + 224^2}{224}&lt;/math&gt;, the floor of which is clearly &lt;math&gt;224&lt;/math&gt; because the &lt;math&gt;1&lt;/math&gt; in the numerator is insignificant. Repeating the procedure for &lt;math&gt;n + 1&lt;/math&gt; is somewhat messier, but we end up getting &lt;math&gt;\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}&lt;/math&gt;. It's not too hard to see that &lt;math&gt;224^4&lt;/math&gt; is much larger than the sum of the remaining terms in the denominator, and that &lt;math&gt;223^3&lt;/math&gt; is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than &lt;math&gt;224^3&lt;/math&gt;, while the largest term in the denominator is smaller than &lt;math&gt;224^2&lt;/math&gt;. Thus, the floor of this expression will come out to be &lt;math&gt;224&lt;/math&gt; as well.<br /> <br /> Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time &lt;math&gt;n&lt;/math&gt; increases by &lt;math&gt;1&lt;/math&gt;, the degrees of both the numerator and denominator increase by &lt;math&gt;2&lt;/math&gt;, because we are squaring the &lt;math&gt;n+1th&lt;/math&gt; term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation &lt;math&gt;a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2&lt;/math&gt;). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the &lt;math&gt;\approx224:1&lt;/math&gt; ratio between the two. <br /> <br /> For the non-greatest terms in the expression to offset this ratio for values of &lt;math&gt;n&lt;/math&gt; in the ballpark of &lt;math&gt;2006&lt;/math&gt;, they would have to have massive coefficients, because or else they are dwarfed by the additional &lt;math&gt;224&lt;/math&gt; attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to &lt;math&gt;\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }&lt;/math&gt; for all &lt;math&gt;k\geq2&lt;/math&gt;, an expression whose eventual limit is &lt;math&gt;\boxed{224}&lt;/math&gt;, which will be its floor when &lt;math&gt;n = 2006&lt;/math&gt;. <br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121044 1997 AIME Problems/Problem 13 2020-04-16T15:18:34Z <p>Anellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy &lt;center&gt;&lt;math&gt;\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> :''This solution is non-rigorous.''<br /> Let &lt;math&gt;f(x) = \Big|\big||x|-2\big|-1\Big|&lt;/math&gt;, &lt;math&gt;f(x) \ge 0&lt;/math&gt;. Then &lt;math&gt;f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4&lt;/math&gt;. We only have a &lt;math&gt;4\times 4&lt;/math&gt; area, so guessing points and graphing won't be too bad of an idea. Since &lt;math&gt;f(x) = f(-x)&lt;/math&gt;, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;background:none;&quot;<br /> |-<br /> | &lt;math&gt;f(1) = 0&lt;/math&gt; || &lt;math&gt;f(0.1) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(2) = 1&lt;/math&gt; || &lt;math&gt;f(0.9) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(3) = 0&lt;/math&gt; || &lt;math&gt;f(1.1) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(4) = 1&lt;/math&gt; || &lt;math&gt;f(1.9) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(0.5) = 0.5&lt;/math&gt; || &lt;math&gt;f(2.1) = 0.9&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(1.5) = 1.5&lt;/math&gt; || &lt;math&gt;f(2.9) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(2.5) = 2.5&lt;/math&gt; || &lt;math&gt;f(3.1) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(3.5) = 3.5&lt;/math&gt; || &lt;math&gt;f(3.9) = 0.9&lt;/math&gt;<br /> |}<br /> <br /> We can now graph the pairs of coordinates which add up to &lt;math&gt;1&lt;/math&gt;. Just using the first column of information gives us an interesting [[lattice]] pattern:<br /> <br /> [[Image:1997_AIME-13a.png]]<br /> <br /> Plotting the remaining points and connecting lines, the graph looks like:<br /> <br /> [[Image:1997_AIME-13b.png]]<br /> <br /> Calculating the lengths is now easy; each rectangle has sides of &lt;math&gt;\sqrt{2}, 3\sqrt{2}&lt;/math&gt;, so the answer is &lt;math&gt;4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}&lt;/math&gt;. For all four quadrants, this is &lt;math&gt;64\sqrt{2}&lt;/math&gt;, and &lt;math&gt;a+b=\boxed{066}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Since &lt;math&gt;0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1&lt;/math&gt; and &lt;math&gt;0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 1 \le \big||x| - 2\big| - 1 \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;0 \le \big||x| - 2\big| \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 2 \le |x| - 2 \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 4 \le x \le 4&lt;/math&gt;&lt;br /&gt;<br /> Also &lt;math&gt;- 4 \le y \le 4&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big|&lt;/math&gt;.<br /> *If &lt;math&gt;0 \le a \le 1&lt;/math&gt;:<br /> :&lt;math&gt;f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 + a) = f(3 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 2&lt;/math&gt;:<br /> :&lt;math&gt;f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 + a) = f(2 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 4&lt;/math&gt;:<br /> :&lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f(a) = f( - a)&lt;/math&gt;<br /> <br /> *So the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;3 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 3&lt;/math&gt; (reflected over the line x=3)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 2&lt;/math&gt; (reflected over the line x=2)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;- 4 \le x \le 0&lt;/math&gt; (reflected over the line x=0)<br /> [this is also true for horizontal reflection, with &lt;math&gt;3 \le y \le 4&lt;/math&gt;, etc]<br /> <br /> So it is only necessary to find the length of the function at &lt;math&gt;3 \le x \le 4&lt;/math&gt; and &lt;math&gt;3 \le y \le 4&lt;/math&gt;:<br /> &lt;math&gt;\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;x - 3 + y - 3 = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;y = - x + 7&lt;/math&gt;<br /> (Length = &lt;math&gt;\sqrt {2}&lt;/math&gt;)<br /> <br /> This graph is reflected over the line y=3, the quantity of which is reflected over y=2,<br /> :the quantity of which is reflected over y=0,<br /> :the quantity of which is reflected over x=3,<br /> :the quantity of which is reflected over x=2,<br /> :the quantity of which is reflected over x=0..<br /> <br /> So a total of &lt;math&gt;6&lt;/math&gt; doublings = &lt;math&gt;2^6&lt;/math&gt; = &lt;math&gt;64&lt;/math&gt;, the total length = &lt;math&gt;64 \cdot \sqrt {2} = a\sqrt {b}&lt;/math&gt;, and &lt;math&gt;a + b = 64 + 2 = \boxed{066}&lt;/math&gt;.<br /> ===Solution 3 (FASTEST)===<br /> We make use of several consecutive substitutions. <br /> Let &lt;math&gt;||x| - 2|= x_1&lt;/math&gt; and similarly with &lt;math&gt;y&lt;/math&gt;.<br /> Therefore, our graph is &lt;math&gt;|x_1 - 1| + |y_1 - 1| = 1&lt;/math&gt;. This is a diamond with perimeter &lt;math&gt;4\sqrt{2}&lt;/math&gt;. Now, we make use of the following fact for a function of two variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;: Suppose we have &lt;math&gt;f(x, y) = c&lt;/math&gt;. Then &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is equal to the graph of &lt;math&gt;f(x, y)&lt;/math&gt; reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is 4 times the perimeter of &lt;math&gt;f(x, y)&lt;/math&gt;. Now, we continue making substitutions at each absolute value sign (&lt;math&gt;|x| - 2 = x_2&lt;/math&gt; and finally &lt;math&gt;x = x_3&lt;/math&gt;, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is &lt;math&gt;4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{066}&lt;/math&gt;.<br /> <br /> - whatRthose<br /> <br /> ===Solution 4 (Exploiting intuitions about absolute value with linear functions)===<br /> In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting &lt;math&gt;a = \big||x|-2\big|-1&lt;/math&gt; and &lt;math&gt;b = \big||y|-2\big|-1&lt;/math&gt;. We then get that &lt;math&gt;b = 1 - a&lt;/math&gt; (obviously the other way would work too). One specific &quot;case&quot; here is the one where all the numbers in the absolute value bars are positive, and thus where &lt;math&gt;b = 1 - x - 2 = -x - 1&lt;/math&gt;. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in &lt;math&gt;b&lt;/math&gt;. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. <br /> Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function's graph at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are &lt;math&gt;2^6&lt;/math&gt; of these pieces of length &lt;math&gt;\sqrt{2}&lt;/math&gt;, making our answer &lt;math&gt;2 + 64 = \boxed{066}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=12|num-a=14}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121043 1997 AIME Problems/Problem 13 2020-04-16T15:17:53Z <p>Anellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy &lt;center&gt;&lt;math&gt;\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> :''This solution is non-rigorous.''<br /> Let &lt;math&gt;f(x) = \Big|\big||x|-2\big|-1\Big|&lt;/math&gt;, &lt;math&gt;f(x) \ge 0&lt;/math&gt;. Then &lt;math&gt;f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4&lt;/math&gt;. We only have a &lt;math&gt;4\times 4&lt;/math&gt; area, so guessing points and graphing won't be too bad of an idea. Since &lt;math&gt;f(x) = f(-x)&lt;/math&gt;, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;background:none;&quot;<br /> |-<br /> | &lt;math&gt;f(1) = 0&lt;/math&gt; || &lt;math&gt;f(0.1) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(2) = 1&lt;/math&gt; || &lt;math&gt;f(0.9) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(3) = 0&lt;/math&gt; || &lt;math&gt;f(1.1) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(4) = 1&lt;/math&gt; || &lt;math&gt;f(1.9) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(0.5) = 0.5&lt;/math&gt; || &lt;math&gt;f(2.1) = 0.9&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(1.5) = 1.5&lt;/math&gt; || &lt;math&gt;f(2.9) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(2.5) = 2.5&lt;/math&gt; || &lt;math&gt;f(3.1) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(3.5) = 3.5&lt;/math&gt; || &lt;math&gt;f(3.9) = 0.9&lt;/math&gt;<br /> |}<br /> <br /> We can now graph the pairs of coordinates which add up to &lt;math&gt;1&lt;/math&gt;. Just using the first column of information gives us an interesting [[lattice]] pattern:<br /> <br /> [[Image:1997_AIME-13a.png]]<br /> <br /> Plotting the remaining points and connecting lines, the graph looks like:<br /> <br /> [[Image:1997_AIME-13b.png]]<br /> <br /> Calculating the lengths is now easy; each rectangle has sides of &lt;math&gt;\sqrt{2}, 3\sqrt{2}&lt;/math&gt;, so the answer is &lt;math&gt;4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}&lt;/math&gt;. For all four quadrants, this is &lt;math&gt;64\sqrt{2}&lt;/math&gt;, and &lt;math&gt;a+b=\boxed{066}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Since &lt;math&gt;0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1&lt;/math&gt; and &lt;math&gt;0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 1 \le \big||x| - 2\big| - 1 \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;0 \le \big||x| - 2\big| \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 2 \le |x| - 2 \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 4 \le x \le 4&lt;/math&gt;&lt;br /&gt;<br /> Also &lt;math&gt;- 4 \le y \le 4&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big|&lt;/math&gt;.<br /> *If &lt;math&gt;0 \le a \le 1&lt;/math&gt;:<br /> :&lt;math&gt;f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 + a) = f(3 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 2&lt;/math&gt;:<br /> :&lt;math&gt;f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 + a) = f(2 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 4&lt;/math&gt;:<br /> :&lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f(a) = f( - a)&lt;/math&gt;<br /> <br /> *So the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;3 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 3&lt;/math&gt; (reflected over the line x=3)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 2&lt;/math&gt; (reflected over the line x=2)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;- 4 \le x \le 0&lt;/math&gt; (reflected over the line x=0)<br /> [this is also true for horizontal reflection, with &lt;math&gt;3 \le y \le 4&lt;/math&gt;, etc]<br /> <br /> So it is only necessary to find the length of the function at &lt;math&gt;3 \le x \le 4&lt;/math&gt; and &lt;math&gt;3 \le y \le 4&lt;/math&gt;:<br /> &lt;math&gt;\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;x - 3 + y - 3 = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;y = - x + 7&lt;/math&gt;<br /> (Length = &lt;math&gt;\sqrt {2}&lt;/math&gt;)<br /> <br /> This graph is reflected over the line y=3, the quantity of which is reflected over y=2,<br /> :the quantity of which is reflected over y=0,<br /> :the quantity of which is reflected over x=3,<br /> :the quantity of which is reflected over x=2,<br /> :the quantity of which is reflected over x=0..<br /> <br /> So a total of &lt;math&gt;6&lt;/math&gt; doublings = &lt;math&gt;2^6&lt;/math&gt; = &lt;math&gt;64&lt;/math&gt;, the total length = &lt;math&gt;64 \cdot \sqrt {2} = a\sqrt {b}&lt;/math&gt;, and &lt;math&gt;a + b = 64 + 2 = \boxed{066}&lt;/math&gt;.<br /> ===Solution 3 (FASTEST)===<br /> We make use of several consecutive substitutions. <br /> Let &lt;math&gt;||x| - 2|= x_1&lt;/math&gt; and similarly with &lt;math&gt;y&lt;/math&gt;.<br /> Therefore, our graph is &lt;math&gt;|x_1 - 1| + |y_1 - 1| = 1&lt;/math&gt;. This is a diamond with perimeter &lt;math&gt;4\sqrt{2}&lt;/math&gt;. Now, we make use of the following fact for a function of two variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;: Suppose we have &lt;math&gt;f(x, y) = c&lt;/math&gt;. Then &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is equal to the graph of &lt;math&gt;f(x, y)&lt;/math&gt; reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is 4 times the perimeter of &lt;math&gt;f(x, y)&lt;/math&gt;. Now, we continue making substitutions at each absolute value sign (&lt;math&gt;|x| - 2 = x_2&lt;/math&gt; and finally &lt;math&gt;x = x_3&lt;/math&gt;, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is &lt;math&gt;4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{066}&lt;/math&gt;.<br /> <br /> - whatRthose<br /> <br /> ===Solution 4 (Exploiting intuitions about absolute value with linear functions)===<br /> In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting &lt;math&gt;a = \big||x|-2\big|-1&lt;/math&gt; and &lt;math&gt;b = \big||y|-2\big|-1&lt;/math&gt;. We then get that &lt;math&gt;b = 1 - a&lt;/math&gt; (obviously the other way would work too). One specific &quot;case&quot; here is the one where all the numbers in the absolute value bars are positive, and thus where &lt;math&gt;b = 1 - x - 2 = -x - 1&lt;/math&gt;. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in &lt;math&gt;b&lt;/math&gt;. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. <br /> Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are &lt;math&gt;2^6&lt;/math&gt; of these pieces of length &lt;math&gt;\sqrt{2}&lt;/math&gt;, making our answer &lt;math&gt;2 + 64 = \boxed{066}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=12|num-a=14}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121042 1997 AIME Problems/Problem 13 2020-04-16T15:14:56Z <p>Anellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy &lt;center&gt;&lt;math&gt;\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> :''This solution is non-rigorous.''<br /> Let &lt;math&gt;f(x) = \Big|\big||x|-2\big|-1\Big|&lt;/math&gt;, &lt;math&gt;f(x) \ge 0&lt;/math&gt;. Then &lt;math&gt;f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4&lt;/math&gt;. We only have a &lt;math&gt;4\times 4&lt;/math&gt; area, so guessing points and graphing won't be too bad of an idea. Since &lt;math&gt;f(x) = f(-x)&lt;/math&gt;, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;background:none;&quot;<br /> |-<br /> | &lt;math&gt;f(1) = 0&lt;/math&gt; || &lt;math&gt;f(0.1) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(2) = 1&lt;/math&gt; || &lt;math&gt;f(0.9) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(3) = 0&lt;/math&gt; || &lt;math&gt;f(1.1) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(4) = 1&lt;/math&gt; || &lt;math&gt;f(1.9) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(0.5) = 0.5&lt;/math&gt; || &lt;math&gt;f(2.1) = 0.9&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(1.5) = 1.5&lt;/math&gt; || &lt;math&gt;f(2.9) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(2.5) = 2.5&lt;/math&gt; || &lt;math&gt;f(3.1) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(3.5) = 3.5&lt;/math&gt; || &lt;math&gt;f(3.9) = 0.9&lt;/math&gt;<br /> |}<br /> <br /> We can now graph the pairs of coordinates which add up to &lt;math&gt;1&lt;/math&gt;. Just using the first column of information gives us an interesting [[lattice]] pattern:<br /> <br /> [[Image:1997_AIME-13a.png]]<br /> <br /> Plotting the remaining points and connecting lines, the graph looks like:<br /> <br /> [[Image:1997_AIME-13b.png]]<br /> <br /> Calculating the lengths is now easy; each rectangle has sides of &lt;math&gt;\sqrt{2}, 3\sqrt{2}&lt;/math&gt;, so the answer is &lt;math&gt;4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}&lt;/math&gt;. For all four quadrants, this is &lt;math&gt;64\sqrt{2}&lt;/math&gt;, and &lt;math&gt;a+b=\boxed{066}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Since &lt;math&gt;0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1&lt;/math&gt; and &lt;math&gt;0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 1 \le \big||x| - 2\big| - 1 \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;0 \le \big||x| - 2\big| \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 2 \le |x| - 2 \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 4 \le x \le 4&lt;/math&gt;&lt;br /&gt;<br /> Also &lt;math&gt;- 4 \le y \le 4&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big|&lt;/math&gt;.<br /> *If &lt;math&gt;0 \le a \le 1&lt;/math&gt;:<br /> :&lt;math&gt;f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 + a) = f(3 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 2&lt;/math&gt;:<br /> :&lt;math&gt;f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 + a) = f(2 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 4&lt;/math&gt;:<br /> :&lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f(a) = f( - a)&lt;/math&gt;<br /> <br /> *So the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;3 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 3&lt;/math&gt; (reflected over the line x=3)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 2&lt;/math&gt; (reflected over the line x=2)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;- 4 \le x \le 0&lt;/math&gt; (reflected over the line x=0)<br /> [this is also true for horizontal reflection, with &lt;math&gt;3 \le y \le 4&lt;/math&gt;, etc]<br /> <br /> So it is only necessary to find the length of the function at &lt;math&gt;3 \le x \le 4&lt;/math&gt; and &lt;math&gt;3 \le y \le 4&lt;/math&gt;:<br /> &lt;math&gt;\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;x - 3 + y - 3 = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;y = - x + 7&lt;/math&gt;<br /> (Length = &lt;math&gt;\sqrt {2}&lt;/math&gt;)<br /> <br /> This graph is reflected over the line y=3, the quantity of which is reflected over y=2,<br /> :the quantity of which is reflected over y=0,<br /> :the quantity of which is reflected over x=3,<br /> :the quantity of which is reflected over x=2,<br /> :the quantity of which is reflected over x=0..<br /> <br /> So a total of &lt;math&gt;6&lt;/math&gt; doublings = &lt;math&gt;2^6&lt;/math&gt; = &lt;math&gt;64&lt;/math&gt;, the total length = &lt;math&gt;64 \cdot \sqrt {2} = a\sqrt {b}&lt;/math&gt;, and &lt;math&gt;a + b = 64 + 2 = \boxed{066}&lt;/math&gt;.<br /> ===Solution 3 (FASTEST)===<br /> We make use of several consecutive substitutions. <br /> Let &lt;math&gt;||x| - 2|= x_1&lt;/math&gt; and similarly with &lt;math&gt;y&lt;/math&gt;.<br /> Therefore, our graph is &lt;math&gt;|x_1 - 1| + |y_1 - 1| = 1&lt;/math&gt;. This is a diamond with perimeter &lt;math&gt;4\sqrt{2}&lt;/math&gt;. Now, we make use of the following fact for a function of two variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;: Suppose we have &lt;math&gt;f(x, y) = c&lt;/math&gt;. Then &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is equal to the graph of &lt;math&gt;f(x, y)&lt;/math&gt; reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is 4 times the perimeter of &lt;math&gt;f(x, y)&lt;/math&gt;. Now, we continue making substitutions at each absolute value sign (&lt;math&gt;|x| - 2 = x_2&lt;/math&gt; and finally &lt;math&gt;x = x_3&lt;/math&gt;, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is &lt;math&gt;4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{066}&lt;/math&gt;.<br /> <br /> - whatRthose<br /> <br /> ===Solution 4 (Exploiting intuitions about absolute value with linear functions)===<br /> In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting &lt;math&gt;a = \big||x|-2\big|-1&lt;/math&gt; and &lt;math&gt;b = \big||y|-2\big|-1&lt;/math&gt;. We then get that &lt;math&gt;b = 1 - a&lt;/math&gt; (obviously the other way would work too). One specific &quot;case&quot; here is the one where all the numbers in the absolute value bars are positive, and thus where &lt;math&gt;b = 1 - x - 2 = -x - 1&lt;/math&gt;. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. <br /> Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are &lt;math&gt;2^6&lt;/math&gt; of these pieces of length &lt;math&gt;\sqrt{2}&lt;/math&gt;, making our answer &lt;math&gt;2 + 64 = \boxed{066}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=12|num-a=14}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=121041 1997 AIME Problems/Problem 13 2020-04-16T15:14:29Z <p>Anellipticcurveoverq: /* Solution 4 (Exploiting intuitions about absolute value with linear functions) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy &lt;center&gt;&lt;math&gt;\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> :''This solution is non-rigorous.''<br /> Let &lt;math&gt;f(x) = \Big|\big||x|-2\big|-1\Big|&lt;/math&gt;, &lt;math&gt;f(x) \ge 0&lt;/math&gt;. Then &lt;math&gt;f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4&lt;/math&gt;. We only have a &lt;math&gt;4\times 4&lt;/math&gt; area, so guessing points and graphing won't be too bad of an idea. Since &lt;math&gt;f(x) = f(-x)&lt;/math&gt;, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;background:none;&quot;<br /> |-<br /> | &lt;math&gt;f(1) = 0&lt;/math&gt; || &lt;math&gt;f(0.1) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(2) = 1&lt;/math&gt; || &lt;math&gt;f(0.9) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(3) = 0&lt;/math&gt; || &lt;math&gt;f(1.1) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(4) = 1&lt;/math&gt; || &lt;math&gt;f(1.9) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(0.5) = 0.5&lt;/math&gt; || &lt;math&gt;f(2.1) = 0.9&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(1.5) = 1.5&lt;/math&gt; || &lt;math&gt;f(2.9) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(2.5) = 2.5&lt;/math&gt; || &lt;math&gt;f(3.1) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(3.5) = 3.5&lt;/math&gt; || &lt;math&gt;f(3.9) = 0.9&lt;/math&gt;<br /> |}<br /> <br /> We can now graph the pairs of coordinates which add up to &lt;math&gt;1&lt;/math&gt;. Just using the first column of information gives us an interesting [[lattice]] pattern:<br /> <br /> [[Image:1997_AIME-13a.png]]<br /> <br /> Plotting the remaining points and connecting lines, the graph looks like:<br /> <br /> [[Image:1997_AIME-13b.png]]<br /> <br /> Calculating the lengths is now easy; each rectangle has sides of &lt;math&gt;\sqrt{2}, 3\sqrt{2}&lt;/math&gt;, so the answer is &lt;math&gt;4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}&lt;/math&gt;. For all four quadrants, this is &lt;math&gt;64\sqrt{2}&lt;/math&gt;, and &lt;math&gt;a+b=\boxed{066}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Since &lt;math&gt;0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1&lt;/math&gt; and &lt;math&gt;0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 1 \le \big||x| - 2\big| - 1 \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;0 \le \big||x| - 2\big| \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 2 \le |x| - 2 \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 4 \le x \le 4&lt;/math&gt;&lt;br /&gt;<br /> Also &lt;math&gt;- 4 \le y \le 4&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big|&lt;/math&gt;.<br /> *If &lt;math&gt;0 \le a \le 1&lt;/math&gt;:<br /> :&lt;math&gt;f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 + a) = f(3 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 2&lt;/math&gt;:<br /> :&lt;math&gt;f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 + a) = f(2 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 4&lt;/math&gt;:<br /> :&lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f(a) = f( - a)&lt;/math&gt;<br /> <br /> *So the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;3 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 3&lt;/math&gt; (reflected over the line x=3)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 2&lt;/math&gt; (reflected over the line x=2)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;- 4 \le x \le 0&lt;/math&gt; (reflected over the line x=0)<br /> [this is also true for horizontal reflection, with &lt;math&gt;3 \le y \le 4&lt;/math&gt;, etc]<br /> <br /> So it is only necessary to find the length of the function at &lt;math&gt;3 \le x \le 4&lt;/math&gt; and &lt;math&gt;3 \le y \le 4&lt;/math&gt;:<br /> &lt;math&gt;\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;x - 3 + y - 3 = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;y = - x + 7&lt;/math&gt;<br /> (Length = &lt;math&gt;\sqrt {2}&lt;/math&gt;)<br /> <br /> This graph is reflected over the line y=3, the quantity of which is reflected over y=2,<br /> :the quantity of which is reflected over y=0,<br /> :the quantity of which is reflected over x=3,<br /> :the quantity of which is reflected over x=2,<br /> :the quantity of which is reflected over x=0..<br /> <br /> So a total of &lt;math&gt;6&lt;/math&gt; doublings = &lt;math&gt;2^6&lt;/math&gt; = &lt;math&gt;64&lt;/math&gt;, the total length = &lt;math&gt;64 \cdot \sqrt {2} = a\sqrt {b}&lt;/math&gt;, and &lt;math&gt;a + b = 64 + 2 = \boxed{066}&lt;/math&gt;.<br /> ===Solution 3 (FASTEST)===<br /> We make use of several consecutive substitutions. <br /> Let &lt;math&gt;||x| - 2|= x_1&lt;/math&gt; and similarly with &lt;math&gt;y&lt;/math&gt;.<br /> Therefore, our graph is &lt;math&gt;|x_1 - 1| + |y_1 - 1| = 1&lt;/math&gt;. This is a diamond with perimeter &lt;math&gt;4\sqrt{2}&lt;/math&gt;. Now, we make use of the following fact for a function of two variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;: Suppose we have &lt;math&gt;f(x, y) = c&lt;/math&gt;. Then &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is equal to the graph of &lt;math&gt;f(x, y)&lt;/math&gt; reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is 4 times the perimeter of &lt;math&gt;f(x, y)&lt;/math&gt;. Now, we continue making substitutions at each absolute value sign (&lt;math&gt;|x| - 2 = x_2&lt;/math&gt; and finally &lt;math&gt;x = x_3&lt;/math&gt;, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is &lt;math&gt;4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{066}&lt;/math&gt;.<br /> <br /> - whatRthose<br /> <br /> ===Solution 4 (Exploiting intuitions about absolute value with linear functions)===<br /> In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It is possible to go right for the smallest possible piece that'll help us by letting &lt;math&gt;a = \big||x|-2\big|-1&lt;/math&gt; and &lt;math&gt;b = \big||y|-2\big|-1&lt;/math&gt;. We then get that &lt;math&gt;b = 1 - a&lt;/math&gt; (obviously the other way would work too). One specific &quot;case&quot; here is the one where all the numbers in the absolute value bars are positive, and thus where &lt;math&gt;b = 1 - x - 2 = -x - 1&lt;/math&gt;. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. <br /> Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are &lt;math&gt;2^6&lt;/math&gt; of these pieces of length &lt;math&gt;\sqrt{2}&lt;/math&gt;, making our answer &lt;math&gt;2 + 64 = \boxed{066}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=12|num-a=14}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_13&diff=120871 1997 AIME Problems/Problem 13 2020-04-12T21:16:22Z <p>Anellipticcurveoverq: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of [[point]]s in the [[Cartesian plane]] that satisfy &lt;center&gt;&lt;math&gt;\Big|\big||x|-2\big|-1\Big|+\Big|\big||y|-2\big|-1\Big|=1.&lt;/math&gt;&lt;/center&gt; If a model of &lt;math&gt;S&lt;/math&gt; were built from wire of negligible thickness, then the total length of wire required would be &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> :''This solution is non-rigorous.''<br /> Let &lt;math&gt;f(x) = \Big|\big||x|-2\big|-1\Big|&lt;/math&gt;, &lt;math&gt;f(x) \ge 0&lt;/math&gt;. Then &lt;math&gt;f(x) + f(y) = 1 \Longrightarrow f(x), f(y) \le 1 \Longrightarrow x, y \le 4&lt;/math&gt;. We only have a &lt;math&gt;4\times 4&lt;/math&gt; area, so guessing points and graphing won't be too bad of an idea. Since &lt;math&gt;f(x) = f(-x)&lt;/math&gt;, there's a symmetry about all four [[quadrant]]s, so just consider the first quadrant. We now gather some points:<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;background:none;&quot;<br /> |-<br /> | &lt;math&gt;f(1) = 0&lt;/math&gt; || &lt;math&gt;f(0.1) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(2) = 1&lt;/math&gt; || &lt;math&gt;f(0.9) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(3) = 0&lt;/math&gt; || &lt;math&gt;f(1.1) = 0.1&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(4) = 1&lt;/math&gt; || &lt;math&gt;f(1.9) = 0.9&lt;/math&gt;<br /> |-<br /> | &lt;math&gt;f(0.5) = 0.5&lt;/math&gt; || &lt;math&gt;f(2.1) = 0.9&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(1.5) = 1.5&lt;/math&gt; || &lt;math&gt;f(2.9) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(2.5) = 2.5&lt;/math&gt; || &lt;math&gt;f(3.1) = 0.1&lt;/math&gt;<br /> |- <br /> | &lt;math&gt;f(3.5) = 3.5&lt;/math&gt; || &lt;math&gt;f(3.9) = 0.9&lt;/math&gt;<br /> |}<br /> <br /> We can now graph the pairs of coordinates which add up to &lt;math&gt;1&lt;/math&gt;. Just using the first column of information gives us an interesting [[lattice]] pattern:<br /> <br /> [[Image:1997_AIME-13a.png]]<br /> <br /> Plotting the remaining points and connecting lines, the graph looks like:<br /> <br /> [[Image:1997_AIME-13b.png]]<br /> <br /> Calculating the lengths is now easy; each rectangle has sides of &lt;math&gt;\sqrt{2}, 3\sqrt{2}&lt;/math&gt;, so the answer is &lt;math&gt;4(\sqrt{2} + 3\sqrt{2}) = 16\sqrt{2}&lt;/math&gt;. For all four quadrants, this is &lt;math&gt;64\sqrt{2}&lt;/math&gt;, and &lt;math&gt;a+b=\boxed{066}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Since &lt;math&gt;0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1&lt;/math&gt; and &lt;math&gt;0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 1 \le \big||x| - 2\big| - 1 \le 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;0 \le \big||x| - 2\big| \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 2 \le |x| - 2 \le 2&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;- 4 \le x \le 4&lt;/math&gt;&lt;br /&gt;<br /> Also &lt;math&gt;- 4 \le y \le 4&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big|&lt;/math&gt;.<br /> *If &lt;math&gt;0 \le a \le 1&lt;/math&gt;:<br /> :&lt;math&gt;f(3 + a) = \Big|\big||3 + a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 - a) = \Big|\big||3 - a| - 2\big| - 1\Big| = a&lt;/math&gt;<br /> :&lt;math&gt;f(3 + a) = f(3 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 2&lt;/math&gt;:<br /> :&lt;math&gt;f(2 + a) = \Big|\big||2 + a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 - a) = \Big|\big||2 - a| - 2\big| - 1\Big| = a - 1&lt;/math&gt;<br /> :&lt;math&gt;f(2 + a) = f(2 - a)&lt;/math&gt;<br /> <br /> *If &lt;math&gt;0 \le a \le 4&lt;/math&gt;:<br /> :&lt;math&gt;f(a) = \Big|\big||a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f( - a) = \Big|\big|| - a| - 2\big| - 1\Big| = a - 3&lt;/math&gt;<br /> :&lt;math&gt;f(a) = f( - a)&lt;/math&gt;<br /> <br /> *So the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;3 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 3&lt;/math&gt; (reflected over the line x=3)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;2 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 2&lt;/math&gt; (reflected over the line x=2)<br /> *And the graph of &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;0 \le x \le 4&lt;/math&gt; is symmetric to &lt;math&gt;y(x)&lt;/math&gt; at &lt;math&gt;- 4 \le x \le 0&lt;/math&gt; (reflected over the line x=0)<br /> [this is also true for horizontal reflection, with &lt;math&gt;3 \le y \le 4&lt;/math&gt;, etc]<br /> <br /> So it is only necessary to find the length of the function at &lt;math&gt;3 \le x \le 4&lt;/math&gt; and &lt;math&gt;3 \le y \le 4&lt;/math&gt;:<br /> &lt;math&gt;\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;x - 3 + y - 3 = 1&lt;/math&gt;&lt;br /&gt;<br /> &lt;math&gt;y = - x + 7&lt;/math&gt;<br /> (Length = &lt;math&gt;\sqrt {2}&lt;/math&gt;)<br /> <br /> This graph is reflected over the line y=3, the quantity of which is reflected over y=2,<br /> :the quantity of which is reflected over y=0,<br /> :the quantity of which is reflected over x=3,<br /> :the quantity of which is reflected over x=2,<br /> :the quantity of which is reflected over x=0..<br /> <br /> So a total of &lt;math&gt;6&lt;/math&gt; doublings = &lt;math&gt;2^6&lt;/math&gt; = &lt;math&gt;64&lt;/math&gt;, the total length = &lt;math&gt;64 \cdot \sqrt {2} = a\sqrt {b}&lt;/math&gt;, and &lt;math&gt;a + b = 64 + 2 = \boxed{066}&lt;/math&gt;.<br /> ===Solution 3 (FASTEST)===<br /> We make use of several consecutive substitutions. <br /> Let &lt;math&gt;||x| - 2|= x_1&lt;/math&gt; and similarly with &lt;math&gt;y&lt;/math&gt;.<br /> Therefore, our graph is &lt;math&gt;|x_1 - 1| + |y_1 - 1| = 1&lt;/math&gt;. This is a diamond with perimeter &lt;math&gt;4\sqrt{2}&lt;/math&gt;. Now, we make use of the following fact for a function of two variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;: Suppose we have &lt;math&gt;f(x, y) = c&lt;/math&gt;. Then &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is equal to the graph of &lt;math&gt;f(x, y)&lt;/math&gt; reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of of &lt;math&gt;f(|x|, |y|)&lt;/math&gt; is 4 times the perimeter of &lt;math&gt;f(x, y)&lt;/math&gt;. Now, we continue making substitutions at each absolute value sign (&lt;math&gt;|x| - 2 = x_2&lt;/math&gt; and finally &lt;math&gt;x = x_3&lt;/math&gt;, similarly for y as well. ), noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is &lt;math&gt;4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{066}&lt;/math&gt;.<br /> <br /> - whatRthose<br /> <br /> ===Solution 4 (Exploiting intuitions about absolute value with linear functions)===<br /> In a similar manner to the other posted solutions, we want to just find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It is possible to go right for the smallest possible piece that'll help us by letting &lt;math&gt;a = \big||x|-2\big|-1&lt;/math&gt; and &lt;math&gt;b = \big||y|-2\big|-1&lt;/math&gt;. We then get that &lt;math&gt;b = 1 - a&lt;/math&gt; (obviously the other way would work too). One specific &quot;case&quot; here is the one where all the numbers in the absolute value bars are positive, and thus where &lt;math&gt;b = 1 - x - 2 = -x - 1&lt;/math&gt;. Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes. Using the Pythagorean Theorem / distance formula, that tiny length comes out to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. <br /> Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are &lt;math&gt;2^6&lt;/math&gt; of these pieces of length &lt;math&gt;\sqrt{2}&lt;/math&gt;, making our answer &lt;math&gt;2 + 64 = \boxed{066}&lt;/math&gt;.<br /> <br /> ~ anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1997|num-b=12|num-a=14}}<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=120454 1984 AIME Problems/Problem 15 2020-04-03T02:02:33Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. We can move the expression &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt; to the left hand side to obtain the difference of the polynomials: &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt; and &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt;<br /> <br /> Since the polynomials are equal at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we can express the difference of the two polynomials with a quartic polynomial that has roots at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, so<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Note the leading coefficient of the RHS is &lt;math&gt;-1&lt;/math&gt; because it must match the leading coefficient of the LHS, which is &lt;math&gt;-1&lt;/math&gt;. <br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic. Because &lt;math&gt;A(t) - B(t) = 0, A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; are equivalent and must be equal for all &lt;math&gt;t&lt;/math&gt;.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=120429 1984 AIME Problems/Problem 15 2020-04-02T19:38:27Z <p>Anellipticcurveoverq: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. We can move the expression &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt; to the left hand side to obtain the difference of the polynomials: &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt; and &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt;<br /> <br /> Since the polynomials are equal at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we can express the difference of the two polynomials with a quartic polynomial that has roots at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, so<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Note the leading coefficient of the RHS is &lt;math&gt;-1&lt;/math&gt; because it must match the leading coefficient of the LHS, which is &lt;math&gt;-1&lt;/math&gt;. <br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic. Because &lt;math&gt;A(t) - B(t) = 0, A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; are equivalent and must be equal for all &lt;math&gt;t&lt;/math&gt;.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == Solution 3 == <br /> <br /> Let &lt;math&gt;a + b + c + d = \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2} = 1&lt;/math&gt;, where each of &lt;math&gt;a, b, c, d&lt;/math&gt; corresponds to the term in the same position in this original first equation. Dividing the &lt;math&gt;n&lt;/math&gt;th term in each equation by the &lt;math&gt;n&lt;/math&gt;th term in the equation before it, we find that the ratio between the two is dependent only on the first number in the denominator of each term (namely &lt;math&gt;2, 4, 6,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;), and can be written as a product of linear factors. Computing these ratios, we can rewrite our system linearly as the following: <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;a + b + c + d = 1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{5}a-\frac{5}{7}b+\frac{7}{3}c+\frac{15}{11}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{3}{35}a-\frac{5}{27}b-\frac{21}{11}c+\frac{45}{13}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{21}a-\frac{1}{11}b-\frac{7}{13}c-3d=1&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Solving this painstaking system, you'll find each of &lt;math&gt;a, b, c, d&lt;/math&gt; can be written in reduced form with a denominator of &lt;math&gt;1024&lt;/math&gt;, as shown in Solution 1. Substituting these values back in and solving for &lt;math&gt;x^2, y^2, z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; and summing them, we get a final value of &lt;math&gt;\boxed{036}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=120412 1984 AIME Problems/Problem 15 2020-04-02T17:41:20Z <p>Anellipticcurveoverq: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. We can move the expression &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt; to the left hand side to obtain the difference of the polynomials: &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt; and &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt;<br /> <br /> Since the polynomials are equal at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we can express the difference of the two polynomials with a quartic polynomial that has roots at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, so<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Note the leading coefficient of the RHS is &lt;math&gt;-1&lt;/math&gt; because it must match the leading coefficient of the LHS, which is &lt;math&gt;-1&lt;/math&gt;. <br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic. Because &lt;math&gt;A(t) - B(t) = 0, A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; are equivalent and must be equal for all &lt;math&gt;t&lt;/math&gt;.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == Solution 3 == <br /> <br /> Let &lt;math&gt;a + b + c + d = \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2} = 1&lt;/math&gt;, where each of &lt;math&gt;a, b, c, d&lt;/math&gt; corresponds to the term in the same position in this original first equation. Dividing the &lt;math&gt;n&lt;/math&gt;th term in each equation by the &lt;math&gt;n&lt;/math&gt;th term in the equation before it, we find that the ratio between the two is dependent only on the first number in the denominator of each term (namely &lt;math&gt;2, 4, 6,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;), and can be written as a product of linear factors. Computing these ratios, we can rewrite our system linearly as the following: <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;a + b + c + d = 1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{5}a-\frac{5}{7}b+\frac{7}{3}c+\frac{15}{11}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{3}{35}a-\frac{5}{27}b-\frac{21}{11}c+\frac{45}{13}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{21}a-\frac{1}{11}b-\frac{7}{13}c-3d=1&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Solving this painstaking system, you'll find each of &lt;math&gt;a, b, c, d&lt;/math&gt; can be written in reduced form with a denominator of &lt;math&gt;1024&lt;/math&gt;, as shown in Solution 1. Substituting these values back in and solving for &lt;math&gt;x^2, y^2, z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; and summing them, we get a final value of &lt;math&gt;\boxed{036}&lt;/math&gt;.<br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Anellipticcurveoverq https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&diff=120411 1984 AIME Problems/Problem 15 2020-04-02T17:22:16Z <p>Anellipticcurveoverq: </p> <hr /> <div>== Problem ==<br /> Determine &lt;math&gt;w^2+x^2+y^2+z^2&lt;/math&gt; if<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt; \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 &lt;/math&gt;&lt;/div&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the system of equations as &lt;math&gt; \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. &lt;/math&gt; This equation is satisfied when &lt;math&gt;t = 4,16,36,64&lt;/math&gt;, as then the equation is equivalent to the given equations.<br /> After clearing fractions, for each of the values &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we have the [[equation]] &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)&lt;/math&gt;. We can move the expression &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt; to the left hand side to obtain the difference of the polynomials: &lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)&lt;/math&gt; and &lt;math&gt;(t-1)(t-9)(t-25)(t-49)&lt;/math&gt;<br /> <br /> Since the polynomials are equal at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, we can express the difference of the two polynomials with a quartic polynomial that has roots at &lt;math&gt;t=4,16,36,64&lt;/math&gt;, so<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) &lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Note the leading coefficient of the RHS is &lt;math&gt;-1&lt;/math&gt; because it must match the leading coefficient of the LHS, which is &lt;math&gt;-1&lt;/math&gt;. <br /> <br /> Now we can plug in &lt;math&gt;t=1&lt;/math&gt; into the polynomial equation. Most terms drop, and we end up with<br /> <br /> &lt;cmath&gt;x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)&lt;/cmath&gt;<br /> <br /> so that<br /> <br /> &lt;cmath&gt;x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}&lt;/cmath&gt;<br /> <br /> Similarly, we can plug in &lt;math&gt;t=9,25,49&lt;/math&gt; and get<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> y^2&amp;=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\<br /> z^2&amp;=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\<br /> w^2&amp;=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}&lt;/cmath&gt;<br /> <br /> Now adding them up,<br /> <br /> &lt;cmath&gt;\begin{align*}z^2+w^2&amp;=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\<br /> x^2+y^2&amp;=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}&lt;/cmath&gt;<br /> <br /> with a sum of<br /> <br /> &lt;cmath&gt;\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.&lt;/cmath&gt;<br /> <br /> /*Lengthy proof that any two cubic polynomials in &lt;math&gt;t&lt;/math&gt; which are equal at 4 values of &lt;math&gt;t&lt;/math&gt; are themselves equivalent:<br /> Let the two polynomials be &lt;math&gt;A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; and let them be equal at &lt;math&gt;t=a,b,c,d&lt;/math&gt;. Thus we have &lt;math&gt;A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0&lt;/math&gt;. Also the polynomial &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic, but it equals 0 at 4 values of &lt;math&gt;t&lt;/math&gt;. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into &lt;math&gt;(t-a)(t-b)(t-c)(t-d)(&lt;/math&gt;some nonzero polynomial&lt;math&gt;)&lt;/math&gt; which would have a degree greater than or equal to 4, contradicting the statement that &lt;math&gt;A(t) - B(t)&lt;/math&gt; is cubic. Because &lt;math&gt;A(t) - B(t) = 0, A(t)&lt;/math&gt; and &lt;math&gt;B(t)&lt;/math&gt; are equivalent and must be equal for all &lt;math&gt;t&lt;/math&gt;.<br /> <br /> '''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; separately before adding them to obtain the final answer is appealing because it gives the individual values of &lt;math&gt;x^2,y^2,z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; which can be plugged into the given equations to check.<br /> <br /> == Solution 2 ==<br /> As in Solution 1, we have <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)&lt;/math&gt; &lt;math&gt;-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)&lt;/math&gt;<br /> &lt;math&gt;=(t-4)(t-16)(t-36)(t-64)&lt;/math&gt;<br /> &lt;/div&gt;<br /> Now the coefficient of &lt;math&gt;t^3&lt;/math&gt; on both sides must be equal. Therefore we have &lt;math&gt;1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}&lt;/math&gt;.<br /> <br /> == Solution 3 == <br /> <br /> Let &lt;math&gt;a + b + c + d = \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2} = 1&lt;/math&gt;, where each of &lt;math&gt;a, b, c, d&lt;/math&gt; corresponds to the term in the same position in this original first equation. Dividing the &lt;math&gt;n&lt;/math&gt;th term in each equation by the &lt;math&gt;n&lt;/math&gt;th term in the equation before it, we find that the ratio between the two is dependent only on the first number in the denominator of each term (namely &lt;math&gt;2, 4, 6,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;), and can be written as a product of linear factors. Computing these ratios, we can rewrite our system linearly as the following: <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;a + b + c + d = 1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{5}a-\frac{5}{7}b+\frac{7}{3}c+\frac{15}{11}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{3}{35}a-\frac{5}{27}b-\frac{21}{11}c+\frac{45}{13}d=1 &lt;/math&gt;&lt;br /&gt;&lt;math&gt; \frac{1}{21}a-\frac{1}{11}b-\frac{7}{13}c-3d=1&lt;/math&gt;<br /> &lt;/div&gt;<br /> <br /> Solving this painstaking system, you'll find each of &lt;math&gt;a, b, c, d&lt;/math&gt; can be written in reduced form with a denominator of &lt;math&gt;1024&lt;/math&gt;, as shown above. Substituting these values back in and solving for &lt;math&gt;x^2, y^2, z^2,&lt;/math&gt; and &lt;math&gt;w^2&lt;/math&gt; and summing them, we get a final value of &lt;math&gt;\boxed{036}&lt;/math&gt;.<br /> <br /> ~anellipticcurveoverq<br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Anellipticcurveoverq