https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pi+is+3.141&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-01T22:38:59Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_II_Problems/Problem_5&diff=123508 2001 AIME II Problems/Problem 5 2020-06-02T23:32:47Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> A [[set]] of positive numbers has the ''triangle property'' if it has three distinct elements that are the lengths of the sides of a [[triangle]] whose area is positive. Consider sets &lt;math&gt;\{4, 5, 6, \ldots, n\}&lt;/math&gt; of consecutive positive integers, all of whose ten-element subsets have the triangle property. What is the largest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> == Solution 1 ==<br /> Out of all ten-element subsets with distinct elements that do not possess the triangle property, we want to find the one with the smallest maximum element. Call this subset &lt;math&gt;\mathcal{S}&lt;/math&gt;. Without loss of generality, consider any &lt;math&gt;a, b, c \,\in \mathcal{S}&lt;/math&gt; with &lt;math&gt;a &lt; b &lt; c&lt;/math&gt;. &lt;math&gt;\,\mathcal{S}&lt;/math&gt; does not possess the [[triangle inequality|triangle property]], so &lt;math&gt;c \geq a + b&lt;/math&gt;. We use this property to build up &lt;math&gt;\mathcal{S}&lt;/math&gt; from the smallest possible &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\mathcal{S} = \{\, 4,\, 5,\, 4+5, \,5+(4+5),\, \ldots\,\} = \{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\mathcal{S}&lt;/math&gt; is the &quot;smallest&quot; ten-element subset without the triangle property, and since the set &lt;math&gt;\{4, 5, 6, \ldots, 253\}&lt;/math&gt; is the largest set of consecutive integers that does not contain this subset, it is also the largest set of consecutive integers in which all ten-element subsets possess the triangle property. Thus, our answer is &lt;math&gt;n = \fbox{253}&lt;/math&gt;.<br /> ==Solution 2==<br /> I claim that the answer is &lt;math&gt;253&lt;/math&gt;. We will show that no number greater than 253 works. Consider the subset&lt;math&gt; {4, 5, 9, 14, 23, 37, 60, 97, 157, 254}&lt;/math&gt;. This subset does not have the property. It is easy to see that any number less than or equal to &lt;math&gt;253&lt;/math&gt; works, and hence our answer is &lt;math&gt;253&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11&diff=123371 1983 AIME Problems/Problem 11 2020-05-31T23:24:25Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> The solid shown has a square base of side length &lt;math&gt;s&lt;/math&gt;. The upper edge is parallel to the base and has length &lt;math&gt;2s&lt;/math&gt;. All other edges have length &lt;math&gt;s&lt;/math&gt;. Given that &lt;math&gt;s=6\sqrt{2}&lt;/math&gt;, what is the volume of the solid?<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> label(&quot;A&quot;,A,W);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,SE);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> &lt;/asy&gt;&lt;/center&gt; &lt;!-- Asymptote replacement for Image:1983Number11.JPG by bpms --&gt;<br /> <br /> == Solutions ==<br /> <br /> === Solution 1 ===<br /> First, we find the height of the solid by dropping a perpendicular from the midpoint of &lt;math&gt;AD&lt;/math&gt; to &lt;math&gt;EF&lt;/math&gt;. The hypotenuse of the triangle formed is the [[median]] of equilateral triangle &lt;math&gt;ADE&lt;/math&gt;, and one of the legs is &lt;math&gt;3\sqrt{2}&lt;/math&gt;. We apply the Pythagorean Theorem to deduce that the height is &lt;math&gt;6&lt;/math&gt;.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5);<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);<br /> triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F); <br /> draw(B--Ba--Ca--C,dashed+d);<br /> draw(A--Aa--Da--D,dashed+d);<br /> draw(E--(E.x,E.y,0),dashed+l);<br /> draw(F--(F.x,F.y,0),dashed+l);<br /> draw(Aa--E--Da,dashed+d);<br /> draw(Ba--F--Ca,dashed+d);<br /> label(&quot;A&quot;,A,S);<br /> label(&quot;B&quot;,B,S);<br /> label(&quot;C&quot;,C,S);<br /> label(&quot;D&quot;,D,NE);<br /> label(&quot;E&quot;,E,N);<br /> label(&quot;F&quot;,F,N);<br /> label(&quot;$12\sqrt{2}$&quot;,(E+F)/2,N);<br /> label(&quot;$6\sqrt{2}$&quot;,(A+B)/2,S);<br /> label(&quot;6&quot;,(3*s/2,s/2,3),ENE);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Next, we complete the figure into a triangular prism, and find its volume, which is &lt;math&gt;\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432&lt;/math&gt;.<br /> <br /> Now, we subtract off the two extra [[pyramid]]s that we included, whose combined volume is &lt;math&gt;2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144&lt;/math&gt;.<br /> <br /> Thus, our answer is &lt;math&gt;432-144=\boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> &lt;center&gt;&lt;asy&gt;<br /> size(180);<br /> import three; pathpen = black+linewidth(0.65); pointpen = black;<br /> currentprojection = perspective(30,-20,10);<br /> real s = 6 * 2^.5;<br /> triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6);<br /> draw(A--B--C--D--A--E--D);<br /> draw(B--F--C);<br /> draw(E--F);<br /> draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed);<br /> label(&quot;A&quot;,A,(-1,-1,0));<br /> label(&quot;B&quot;,B,( 2,-1,0));<br /> label(&quot;C&quot;,C,( 1, 1,0));<br /> label(&quot;D&quot;,D,(-1, 1,0));<br /> label(&quot;E&quot;,E,(0,0,1));<br /> label(&quot;F&quot;,F,(0,0,1));<br /> label(&quot;G&quot;,G,(0,0,-1));<br /> label(&quot;H&quot;,H,(0,0,-1));<br /> &lt;/asy&gt;&lt;/center&gt;<br /> Extend &lt;math&gt;EA&lt;/math&gt; and &lt;math&gt;FB&lt;/math&gt; to meet at &lt;math&gt;G&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; and &lt;math&gt;FC&lt;/math&gt; to meet at &lt;math&gt;H&lt;/math&gt;. Now, we have a regular tetrahedron &lt;math&gt;EFGH&lt;/math&gt;, which by symmetry has twice the volume of our original solid. This tetrahedron has side length &lt;math&gt;2s = 12\sqrt{2}&lt;/math&gt;. Using the formula for the volume of a regular tetrahedron, which is &lt;math&gt;V = \frac{\sqrt{2}S^3}{12}&lt;/math&gt;, where S is the side length of the tetrahedron, the volume of our original solid is:<br /> <br /> &lt;math&gt;V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is &lt;math&gt;6&lt;/math&gt;; thus, we will integrate with respect to height from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;, noting that each cross section of height &lt;math&gt;dh&lt;/math&gt; is a rectangle. The volume is then &lt;math&gt;\int_0^h(wl) \ \text{d}h&lt;/math&gt;, where &lt;math&gt;w&lt;/math&gt; is the width of the rectangle and &lt;math&gt;l&lt;/math&gt; is the length. We can express &lt;math&gt;w&lt;/math&gt; in terms of &lt;math&gt;h&lt;/math&gt; as &lt;math&gt;w=6\sqrt{2}-\sqrt{2}h&lt;/math&gt; since it decreases linearly with respect to &lt;math&gt;h&lt;/math&gt;, and &lt;math&gt;l=6\sqrt{2}+\sqrt{2}h&lt;/math&gt; since it similarly increases linearly with respect to &lt;math&gt;h&lt;/math&gt;. Now we solve:&lt;cmath&gt;\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}&lt;/cmath&gt;.<br /> <br /> ==Solution 4==<br /> Draw an altitude from a vertex of the square base to the top edge. By using &lt;math&gt;30,60, 90&lt;/math&gt; triangle ratios, we obtain that the altitude has a lgenth of &lt;math&gt;3 \sqrt{6}&lt;/math&gt;, and that little portion that hangs out has a length of &lt;math&gt;\3sqrt2&lt;/math&gt;. This is a trianglular pyramid with a base of &lt;math&gt;3\sqrt6, 3\sqrt6, 3\sqrt2&lt;/math&gt;, and a height of &lt;math&gt;\3sqrt{2}&lt;/math&gt;. Since there are two of these, we can compute the sum of the volumes of these two to be &lt;math&gt;\72&lt;/math&gt;. Now we are left with a triangular prism with a base of dimensions &lt;math&gt;3\sqrt6, 3\sqrt6, 3\sqrt2&lt;/math&gt; and a height of &lt;math&gt;\6sqrt2&lt;/math&gt;. We can compute the volume of this to be 216, and thus our answer is &lt;math&gt;\boxed{288}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> == See Also ==<br /> {{AIME box|year=1983|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems&diff=122953 1997 AIME Problems 2020-05-26T01:53:02Z <p>Pi is 3.141: /* Problem 12 */</p> <hr /> <div>{{AIME Problems|year=1997}}<br /> <br /> == Problem 1 ==<br /> How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?<br /> <br /> [[1997 AIME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> The nine horizontal and nine vertical lines on an &lt;math&gt;8\times8&lt;/math&gt; checkerboard form &lt;math&gt;r&lt;/math&gt; rectangles, of which &lt;math&gt;s&lt;/math&gt; are squares. The number &lt;math&gt;s/r&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 /> [[1997 AIME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?<br /> <br /> [[1997 AIME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> Circles of radii 5, 5, 8, and &lt;math&gt;m/n&lt;/math&gt; are mutually externally tangent, 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 /> [[1997 AIME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> The number &lt;math&gt;r&lt;/math&gt; can be expressed as a four-place decimal &lt;math&gt;0.abcd,&lt;/math&gt; where &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; represent digits, any of which could be zero. It is desired to approximate &lt;math&gt;r&lt;/math&gt; by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\frac 27.&lt;/math&gt; What is the number of possible values for &lt;math&gt;r&lt;/math&gt;?<br /> <br /> [[1997 AIME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> Point &lt;math&gt;B&lt;/math&gt; is in the exterior of the regular &lt;math&gt;n&lt;/math&gt;-sided polygon &lt;math&gt;A_1A_2\cdots A_n&lt;/math&gt;, and &lt;math&gt;A_1A_2B&lt;/math&gt; is an equilateral triangle. What is the largest value of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;A_1&lt;/math&gt;, &lt;math&gt;A_n&lt;/math&gt;, and &lt;math&gt;B&lt;/math&gt; are consecutive vertices of a regular polygon?<br /> <br /> [[1997 AIME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> A car travels due east at &lt;math&gt;\frac 23&lt;/math&gt; miles per minute on a long, straight road. At the same time, a circular storm, whose radius is &lt;math&gt;51&lt;/math&gt; miles, moves southeast at &lt;math&gt;\frac 12\sqrt{2}&lt;/math&gt; mile per minute. At time &lt;math&gt;t=0&lt;/math&gt;, the center of the storm is &lt;math&gt;110&lt;/math&gt; miles due north of the car. At time &lt;math&gt;t=t_1&lt;/math&gt; minutes, the car enters the storm circle, and at time &lt;math&gt;t=t_2&lt;/math&gt; minutes, the car leaves the storm circle. Find &lt;math&gt;\frac 12(t_1+t_2)&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> How many different &lt;math&gt;4\times 4&lt;/math&gt; arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?<br /> <br /> [[1997 AIME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> Given a nonnegative real number &lt;math&gt;x&lt;/math&gt;, let &lt;math&gt;\langle x\rangle&lt;/math&gt; denote the fractional part of &lt;math&gt;x&lt;/math&gt;; that is, &lt;math&gt;\langle x\rangle=x-\lfloor x\rfloor&lt;/math&gt;, where &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;. Suppose that &lt;math&gt;a&lt;/math&gt; is positive, &lt;math&gt;\langle a^{-1}\rangle=\langle a^2\rangle&lt;/math&gt;, and &lt;math&gt;2&lt;a^2&lt;3&lt;/math&gt;. Find the value of &lt;math&gt;a^{12}-144a^{-1}&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:<br /> <br /> i. Either each of the three cards has a different shape or all three of the card have the same shape.<br /> <br /> ii. Either each of the three cards has a different color or all three of the cards have the same color.<br /> <br /> iii. Either each of the three cards has a different shade or all three of the cards have the same shade.<br /> <br /> How many different complementary three-card sets are there?<br /> <br /> [[1997 AIME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> Let &lt;math&gt;x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}&lt;/math&gt;. What is the greatest integer that does not exceed &lt;math&gt;100x&lt;/math&gt;?<br /> <br /> [[1997 AIME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> The function &lt;math&gt;f&lt;/math&gt; defined by &lt;math&gt;f(x)= \frac{ax+b}{cx+d}&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; and &lt;math&gt;d&lt;/math&gt; are nonzero real numbers, has the properties &lt;math&gt;f(19)=19&lt;/math&gt;, &lt;math&gt;f(97)=97&lt;/math&gt; and &lt;math&gt;f(f(x))=x&lt;/math&gt; for all values except &lt;math&gt;\frac{-d}{c}&lt;/math&gt;. Find the unique number that is not in the range of &lt;math&gt;f&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points 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 /> [[1997 AIME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> Let &lt;math&gt;v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; be distinct, randomly chosen roots of the equation &lt;math&gt;z^{1997}-1=0&lt;/math&gt;. Let &lt;math&gt;m/n&lt;/math&gt; be the probability that &lt;math&gt;\sqrt{2+\sqrt{3}}\le |v+w|&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 /> [[1997 AIME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> The sides of rectangle &lt;math&gt;ABCD&lt;/math&gt; have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt;. An equilateral triangle is drawn so that no point of the triangle lies outside &lt;math&gt;ABCD&lt;/math&gt;. The maximum possible area of such a triangle can be written in the form &lt;math&gt;p\sqrt{q}-r&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are positive integers, and &lt;math&gt;q&lt;/math&gt; is not divisible by the square of any prime number. Find &lt;math&gt;p+q+r&lt;/math&gt;.<br /> <br /> [[1997 AIME Problems/Problem 15|Solution]]<br /> <br /> == See also ==<br /> <br /> {{AIME box|year=1997|before=[[1996 AIME Problems]]|after=[[1998 AIME Problems]]}}<br /> <br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=122950 1990 AIME Problems/Problem 15 2020-05-25T22:02:41Z <p>Pi is 3.141: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt;.<br /> <br /> Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> <br /> <br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+\sqrt{87},-7-\sqrt{87})&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us &lt;math&gt;k= \boxed{020}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=122949 1990 AIME Problems/Problem 15 2020-05-25T22:02:23Z <p>Pi is 3.141: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt;.<br /> <br /> Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> <br /> <br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+\sqrt{87},-7-\sqrt{87})&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us &lt;math&gt;k= \boxed{020}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_14&diff=122853 1988 AIME Problems/Problem 14 2020-05-23T00:52:07Z <p>Pi is 3.141: /* Solution 4(IMO the BEST solution) */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;C&lt;/math&gt; be the [[graph]] of &lt;math&gt;xy = 1&lt;/math&gt;, and denote by &lt;math&gt;C^*&lt;/math&gt; the [[reflection]] of &lt;math&gt;C&lt;/math&gt; in the line &lt;math&gt;y = 2x&lt;/math&gt;. Let the [[equation]] of &lt;math&gt;C^*&lt;/math&gt; be written in the form<br /> <br /> &lt;cmath&gt;12x^2 + bxy + cy^2 + d = 0.&lt;/cmath&gt;<br /> <br /> Find the product &lt;math&gt;bc&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Given a point &lt;math&gt;P (x,y)&lt;/math&gt; on &lt;math&gt;C&lt;/math&gt;, we look to find a formula for &lt;math&gt;P' (x', y')&lt;/math&gt; on &lt;math&gt;C^*&lt;/math&gt;. Both points lie on a line that is [[perpendicular]] to &lt;math&gt;y=2x&lt;/math&gt;, so the slope of &lt;math&gt;\overline{PP'}&lt;/math&gt; is &lt;math&gt;\frac{-1}{2}&lt;/math&gt;. Thus &lt;math&gt;\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y&lt;/math&gt;. Also, the midpoint of &lt;math&gt;\overline{PP'}&lt;/math&gt;, &lt;math&gt;\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)&lt;/math&gt;, lies on the line &lt;math&gt;y = 2x&lt;/math&gt;. Therefore &lt;math&gt;\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x&lt;/math&gt;.<br /> <br /> Solving these two equations, we find &lt;math&gt;x = \frac{-3x' + 4y'}{5}&lt;/math&gt; and &lt;math&gt;y = \frac{4x' + 3y'}{5}&lt;/math&gt;. Substituting these points into the equation of &lt;math&gt;C&lt;/math&gt;, we get &lt;math&gt;\frac{(-3x'+4y')(4x'+3y')}{25}=1&lt;/math&gt;, which when expanded becomes &lt;math&gt;12x'^2-7x'y'-12y'^2+25=0&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;bc=(-7)(-12)=\boxed{084}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> The [[asymptote]]s of &lt;math&gt;C&lt;/math&gt; are given by &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt;. Now if we represent the line &lt;math&gt;y=2x&lt;/math&gt; by the complex number &lt;math&gt;1+2i&lt;/math&gt;, then we find the direction of the reflection of the asymptote &lt;math&gt;x=0&lt;/math&gt; by multiplying this by &lt;math&gt;2-i&lt;/math&gt;, getting &lt;math&gt;4+3i&lt;/math&gt;. Therefore, the asymptotes of &lt;math&gt;C^*&lt;/math&gt; are given by &lt;math&gt;4y-3x=0&lt;/math&gt; and &lt;math&gt;3y+4x=0&lt;/math&gt;.<br /> <br /> Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: &lt;math&gt;(3x-4y)(4x+3y)=12x^2-7xy-12y^2&lt;/math&gt;. At this point, the right hand side of the equation will be determined by plugging the point &lt;math&gt;(\frac{\sqrt{2}}{2},\sqrt{2})&lt;/math&gt;, which is unchanged by the reflection, into the expression. But this is not necessary. We see that &lt;math&gt;b=-7&lt;/math&gt;, &lt;math&gt;c=-12&lt;/math&gt;, so &lt;math&gt;bc=\boxed{084}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> The matrix for a reflection about the polar line &lt;math&gt;\theta = \alpha, \alpha+\pi&lt;/math&gt; is:<br /> &lt;cmath&gt;$\left[ \begin{array}{ccc}<br /> \cos(2\alpha) &amp; \sin(2\alpha) \\<br /> \sin(2\alpha) &amp; -\cos(2\alpha)<br /> \end{array} \right]$&lt;/cmath&gt;<br /> This is not hard to derive using a basic knowledge of linear transformations. You can refer here for more information: https://en.wikipedia.org/wiki/Orthogonal_matrix<br /> <br /> Let &lt;math&gt;\alpha = \arctan 2&lt;/math&gt;. Note that the line of reflection, &lt;math&gt;y = 2x&lt;/math&gt;, is the polar line &lt;math&gt;\theta = \alpha, \alpha+\pi&lt;/math&gt;. Then &lt;math&gt;2\alpha = \arctan\left(-\frac{4}{3}\right)&lt;/math&gt;, so &lt;math&gt;\cos(2\alpha) = -\frac{3}{5}&lt;/math&gt; and &lt;math&gt;\sin(2\alpha) = \frac{4}{5}&lt;/math&gt;.<br /> <br /> Therefore, if &lt;math&gt;(x', y')&lt;/math&gt; is mapped to &lt;math&gt;(x, y)&lt;/math&gt; under the reflection, then &lt;math&gt;x = -\frac{3}{5}x'+\frac{4}{5}y'&lt;/math&gt; and &lt;math&gt;y = \frac{4}{5}x'+\frac{3}{5}y'&lt;/math&gt;. Since the transformation matrix represents a reflection, it must be its own inverse; therefore, &lt;math&gt;x' = -\frac{3}{5}x+\frac{4}{5}y&lt;/math&gt; and &lt;math&gt;y' = \frac{4}{5}x+\frac{3}{5}y&lt;/math&gt;.<br /> <br /> The original coordinates &lt;math&gt;(x', y')&lt;/math&gt; must satisfy &lt;math&gt;x'y' = 1&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\left(-\frac{3}{5}x+\frac{4}{5}y\right)\left(\frac{4}{5}x+\frac{5}{5}y\right) = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;12x^2 - 7xy - 12y^2 + 25 = 0&lt;/cmath&gt;<br /> Thus, &lt;math&gt;b = -7&lt;/math&gt; and &lt;math&gt;c = -12&lt;/math&gt;, so &lt;math&gt;bc = 84&lt;/math&gt;. The answer is &lt;math&gt;\boxed{084}&lt;/math&gt;.<br /> <br /> ==Solution 4(the best solution)==<br /> Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that &lt;math&gt;b = -7&lt;/math&gt; and &lt;math&gt;c = -12&lt;/math&gt;, so &lt;math&gt;bc = 84&lt;/math&gt;. The answer is &lt;math&gt;\boxed{084}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_14&diff=122852 1988 AIME Problems/Problem 14 2020-05-23T00:51:44Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;C&lt;/math&gt; be the [[graph]] of &lt;math&gt;xy = 1&lt;/math&gt;, and denote by &lt;math&gt;C^*&lt;/math&gt; the [[reflection]] of &lt;math&gt;C&lt;/math&gt; in the line &lt;math&gt;y = 2x&lt;/math&gt;. Let the [[equation]] of &lt;math&gt;C^*&lt;/math&gt; be written in the form<br /> <br /> &lt;cmath&gt;12x^2 + bxy + cy^2 + d = 0.&lt;/cmath&gt;<br /> <br /> Find the product &lt;math&gt;bc&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Given a point &lt;math&gt;P (x,y)&lt;/math&gt; on &lt;math&gt;C&lt;/math&gt;, we look to find a formula for &lt;math&gt;P' (x', y')&lt;/math&gt; on &lt;math&gt;C^*&lt;/math&gt;. Both points lie on a line that is [[perpendicular]] to &lt;math&gt;y=2x&lt;/math&gt;, so the slope of &lt;math&gt;\overline{PP'}&lt;/math&gt; is &lt;math&gt;\frac{-1}{2}&lt;/math&gt;. Thus &lt;math&gt;\frac{y' - y}{x' - x} = \frac{-1}{2} \Longrightarrow x' + 2y' = x + 2y&lt;/math&gt;. Also, the midpoint of &lt;math&gt;\overline{PP'}&lt;/math&gt;, &lt;math&gt;\left(\frac{x + x'}{2}, \frac{y + y'}{2}\right)&lt;/math&gt;, lies on the line &lt;math&gt;y = 2x&lt;/math&gt;. Therefore &lt;math&gt;\frac{y + y'}{2} = x + x' \Longrightarrow 2x' - y' = y - 2x&lt;/math&gt;.<br /> <br /> Solving these two equations, we find &lt;math&gt;x = \frac{-3x' + 4y'}{5}&lt;/math&gt; and &lt;math&gt;y = \frac{4x' + 3y'}{5}&lt;/math&gt;. Substituting these points into the equation of &lt;math&gt;C&lt;/math&gt;, we get &lt;math&gt;\frac{(-3x'+4y')(4x'+3y')}{25}=1&lt;/math&gt;, which when expanded becomes &lt;math&gt;12x'^2-7x'y'-12y'^2+25=0&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;bc=(-7)(-12)=\boxed{084}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> The [[asymptote]]s of &lt;math&gt;C&lt;/math&gt; are given by &lt;math&gt;x=0&lt;/math&gt; and &lt;math&gt;y=0&lt;/math&gt;. Now if we represent the line &lt;math&gt;y=2x&lt;/math&gt; by the complex number &lt;math&gt;1+2i&lt;/math&gt;, then we find the direction of the reflection of the asymptote &lt;math&gt;x=0&lt;/math&gt; by multiplying this by &lt;math&gt;2-i&lt;/math&gt;, getting &lt;math&gt;4+3i&lt;/math&gt;. Therefore, the asymptotes of &lt;math&gt;C^*&lt;/math&gt; are given by &lt;math&gt;4y-3x=0&lt;/math&gt; and &lt;math&gt;3y+4x=0&lt;/math&gt;.<br /> <br /> Now to find the equation of the hyperbola, we multiply the two expressions together to get one side of the equation: &lt;math&gt;(3x-4y)(4x+3y)=12x^2-7xy-12y^2&lt;/math&gt;. At this point, the right hand side of the equation will be determined by plugging the point &lt;math&gt;(\frac{\sqrt{2}}{2},\sqrt{2})&lt;/math&gt;, which is unchanged by the reflection, into the expression. But this is not necessary. We see that &lt;math&gt;b=-7&lt;/math&gt;, &lt;math&gt;c=-12&lt;/math&gt;, so &lt;math&gt;bc=\boxed{084}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> The matrix for a reflection about the polar line &lt;math&gt;\theta = \alpha, \alpha+\pi&lt;/math&gt; is:<br /> &lt;cmath&gt;$\left[ \begin{array}{ccc}<br /> \cos(2\alpha) &amp; \sin(2\alpha) \\<br /> \sin(2\alpha) &amp; -\cos(2\alpha)<br /> \end{array} \right]$&lt;/cmath&gt;<br /> This is not hard to derive using a basic knowledge of linear transformations. You can refer here for more information: https://en.wikipedia.org/wiki/Orthogonal_matrix<br /> <br /> Let &lt;math&gt;\alpha = \arctan 2&lt;/math&gt;. Note that the line of reflection, &lt;math&gt;y = 2x&lt;/math&gt;, is the polar line &lt;math&gt;\theta = \alpha, \alpha+\pi&lt;/math&gt;. Then &lt;math&gt;2\alpha = \arctan\left(-\frac{4}{3}\right)&lt;/math&gt;, so &lt;math&gt;\cos(2\alpha) = -\frac{3}{5}&lt;/math&gt; and &lt;math&gt;\sin(2\alpha) = \frac{4}{5}&lt;/math&gt;.<br /> <br /> Therefore, if &lt;math&gt;(x', y')&lt;/math&gt; is mapped to &lt;math&gt;(x, y)&lt;/math&gt; under the reflection, then &lt;math&gt;x = -\frac{3}{5}x'+\frac{4}{5}y'&lt;/math&gt; and &lt;math&gt;y = \frac{4}{5}x'+\frac{3}{5}y'&lt;/math&gt;. Since the transformation matrix represents a reflection, it must be its own inverse; therefore, &lt;math&gt;x' = -\frac{3}{5}x+\frac{4}{5}y&lt;/math&gt; and &lt;math&gt;y' = \frac{4}{5}x+\frac{3}{5}y&lt;/math&gt;.<br /> <br /> The original coordinates &lt;math&gt;(x', y')&lt;/math&gt; must satisfy &lt;math&gt;x'y' = 1&lt;/math&gt;. Therefore,<br /> &lt;cmath&gt;\left(-\frac{3}{5}x+\frac{4}{5}y\right)\left(\frac{4}{5}x+\frac{5}{5}y\right) = 1&lt;/cmath&gt;<br /> &lt;cmath&gt;-\frac{12}{25}x^2 + \frac{7}{25}xy + \frac{12}{25}y^2 - 1 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;12x^2 - 7xy - 12y^2 + 25 = 0&lt;/cmath&gt;<br /> Thus, &lt;math&gt;b = -7&lt;/math&gt; and &lt;math&gt;c = -12&lt;/math&gt;, so &lt;math&gt;bc = 84&lt;/math&gt;. The answer is &lt;math&gt;\boxed{084}&lt;/math&gt;.<br /> <br /> ==Solution 4(IMO the BEST solution)==<br /> Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that &lt;math&gt;b = -7&lt;/math&gt; and &lt;math&gt;c = -12&lt;/math&gt;, so &lt;math&gt;bc = 84&lt;/math&gt;. The answer is &lt;math&gt;\boxed{084}&lt;/math&gt;.<br /> == See also ==<br /> {{AIME box|year=1988|num-b=13|num-a=15}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=120393 2020 AMC 12B Problems/Problem 22 2020-04-02T05:34:22Z <p>Pi is 3.141: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2}&lt;/math&gt; &lt;math&gt;\ge \sqrt((2^t-3t)(3t))&lt;/math&gt;. Thus, squaring gives us that &lt;math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;/math&gt;. Rembering what we want to find(divide by &lt;math&gt;4^t&lt;/math&gt;), we get the maximal values as &lt;math&gt;\frac{1}{12}&lt;/math&gt;, and we are done.<br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=120392 2020 AMC 12B Problems/Problem 22 2020-04-02T05:33:47Z <p>Pi is 3.141: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2}&lt;/math&gt; &lt;math&gt;\ge \sqrt((2^t-3t)(3t))&lt;/math&gt;. Thus, squaring gives us that &lt;math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;/math&gt;. Rembering that what we want to find(divide by &lt;math&gt;4^t&lt;/math&gt;), we get the maximal values as &lt;math&gt;\frac{1}{12}&lt;/math&gt;, and we are done.<br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=120391 2020 AMC 12B Problems/Problem 22 2020-04-02T05:33:33Z <p>Pi is 3.141: /* Solution 1 */</p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2}&lt;/math&gt; &lt;math&gt;\ge \sqrt((2^t-3t)(3t))&lt;/math&gt;. Thus, squaring gives us that &lt;math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;/math&gt;. Rembering that what we want to find(divide by &lt;math&gt;4^t), we get the maximal values as &lt;/math&gt;\frac{1}{12}, and we are done.<br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&diff=120390 2020 AMC 12B Problems/Problem 22 2020-04-02T05:33:04Z <p>Pi is 3.141: </p> <hr /> <div>==Problem 22==<br /> <br /> What is the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; for real values of &lt;math&gt;t?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed by using AM-GM. We get &lt;math&gt;\frac{(2^t-3t) + 3t}{2} &lt;/math&gt;\ge \sqrt((2^t-3t)(3t))&lt;math&gt;. Thus, squaring gives us that &lt;/math&gt;4^{t-1} \ge (2^t-3t)(3t)&lt;math&gt;. Rembering that what we want to find(divide by &lt;/math&gt;4^t), we get the maximal values as &lt;math&gt;\frac{1}{12}&lt;/math&gt;, and we are done. <br /> <br /> ==Solution 2==<br /> <br /> Set &lt;math&gt; u = t2^{-t}&lt;/math&gt;. Then the expression in the problem can be written as &lt;cmath&gt;R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .&lt;/cmath&gt; It is easy to see that &lt;math&gt;u =\frac{1}{6}&lt;/math&gt; is attained for some value of &lt;math&gt;t&lt;/math&gt; between &lt;math&gt;t = 0&lt;/math&gt; and &lt;math&gt;t = 1&lt;/math&gt;, thus the maximal value of &lt;math&gt;R&lt;/math&gt; is &lt;math&gt;\textbf{(C)}\ \frac{1}{12}&lt;/math&gt;.<br /> <br /> ==Solution 3 (Calculus Needed)==<br /> <br /> We want to maximize &lt;math&gt;f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}&lt;/math&gt;. We can use the first derivative test. Use quotient rule to get the following:<br /> &lt;cmath&gt;<br /> \frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2} <br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}<br /> &lt;/cmath&gt;<br /> &lt;cmath&gt;<br /> \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t<br /> &lt;/cmath&gt;Therefore, we plug this back into the original equation to get &lt;math&gt;\boxed{\textbf{(C)} \frac{1}{12}}&lt;/math&gt;<br /> <br /> ~awesome1st<br /> <br /> ==Solution 4==<br /> <br /> First, substitute &lt;math&gt;2^t = x (\log_2{x} = t)&lt;/math&gt; so that <br /> &lt;cmath&gt;<br /> \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}<br /> &lt;/cmath&gt;<br /> <br /> Notice that <br /> &lt;cmath&gt;<br /> \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.<br /> &lt;/cmath&gt;<br /> <br /> When seen as a function, &lt;math&gt;\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2&lt;/math&gt; is a synthesis function that has &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; as its inner function.<br /> <br /> If we substitute &lt;math&gt;\frac{\log_2{x}}{x} = p&lt;/math&gt;, the given function becomes a quadratic function that has a maximum value of &lt;math&gt;\frac{1}{12}&lt;/math&gt; when &lt;math&gt;p = \frac{1}{6}&lt;/math&gt;.<br /> <br /> <br /> Now we need to check if &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; can have the value of &lt;math&gt;\frac{1}{6}&lt;/math&gt; in the range of real numbers.<br /> <br /> In the range of (positive) real numbers, function &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; is a continuous function whose value gets infinitely smaller as &lt;math&gt;x&lt;/math&gt; gets closer to 0 (as &lt;math&gt;log_2{x}&lt;/math&gt; also diverges toward negative infinity in the same condition). When &lt;math&gt;x = 2&lt;/math&gt;, &lt;math&gt;\frac{\log_2{x}}{x} = \frac{1}{2}&lt;/math&gt;, which is larger than &lt;math&gt;\frac{1}{6}&lt;/math&gt;.<br /> <br /> Therefore, we can assume that &lt;math&gt;\frac{\log_2{x}}{x}&lt;/math&gt; equals to &lt;math&gt;\frac{1}{6}&lt;/math&gt; when &lt;math&gt;x&lt;/math&gt; is somewhere between 1 and 2 (at least), which means that the maximum value of &lt;math&gt;\frac{(2^t-3t)t}{4^t}&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ \frac{1}{12}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1982_USAMO_Problems&diff=119836 1982 USAMO Problems 2020-03-20T02:36:31Z <p>Pi is 3.141: /* Problem 1 */</p> <hr /> <div>Problems from the '''1982 [[United States of America Mathematical Olympiad | USAMO]]'''.<br /> <br /> ==Problem 1==<br /> In a party with &lt;math&gt;1982&lt;/math&gt; persons, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?<br /> <br /> [[1982 USAMO Problems/Problem 1 | Solution]]<br /> <br /> ==Problem 2==<br /> Let &lt;math&gt;S_r=x^r+y^r+z^r&lt;/math&gt; with &lt;math&gt;x,y,z&lt;/math&gt; real. It is known that if &lt;math&gt;S_1=0&lt;/math&gt;,<br /> <br /> &lt;math&gt;(*) &lt;/math&gt; &lt;math&gt;\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}&lt;/math&gt;<br /> <br /> for &lt;math&gt;(m,n)=(2,3),(3,2),(2,5)&lt;/math&gt;, or &lt;math&gt;(5,2)&lt;/math&gt;. Determine ''all'' other pairs of integers &lt;math&gt;(m,n)&lt;/math&gt; if any, so that &lt;math&gt;(*)&lt;/math&gt; holds for all real numbers &lt;math&gt;x,y,z&lt;/math&gt; such that &lt;math&gt;S_1=0&lt;/math&gt;.<br /> <br /> [[1982 USAMO Problems/Problem 2 | Solution]]<br /> <br /> ==Problem 3==<br /> If a point &lt;math&gt;A_1&lt;/math&gt; is in the interior of an equilateral triangle &lt;math&gt;ABC&lt;/math&gt; and point &lt;math&gt;A_2&lt;/math&gt; is in the interior of &lt;math&gt;\triangle{A_1BC}&lt;/math&gt;, prove that<br /> <br /> &lt;math&gt;I.Q. (A_1BC) &gt; I.Q.(A_2BC)&lt;/math&gt;,<br /> <br /> where the ''isoperimetric quotient'' of a figure &lt;math&gt;F&lt;/math&gt; is defined by<br /> <br /> &lt;math&gt;I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2}&lt;/math&gt;<br /> <br /> [[1982 USAMO Problems/Problem 3 | Solution]]<br /> <br /> ==Problem 4==<br /> Prove that there exists a positive integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;k\cdot2^n+1&lt;/math&gt; is composite for every positive integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[1982 USAMO Problems/Problem 4 | Solution]]<br /> <br /> ==Problem 5==<br /> &lt;math&gt;A,B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; are three interior points of a sphere &lt;math&gt;S&lt;/math&gt; such that &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt; are perpendicular to the diameter of &lt;math&gt;S&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt;, and so that two spheres can be constructed through &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; which are both tangent to &lt;math&gt;S&lt;/math&gt;. Prove that the sum of their radii is equal to the radius of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[1982 USAMO Problems/Problem 5 | Solution]]<br /> <br /> == See Also ==<br /> {{USAMO box|year=1982|before=[[1981 USAMO]]|after=[[1983 USAMO]]}}<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=118922 2015 AIME I Problems/Problem 3 2020-03-08T23:21:36Z <p>Pi is 3.141: /* Solution 5 */</p> <hr /> <div>==Problem==<br /> <br /> There is a prime number &lt;math&gt;p&lt;/math&gt; such that &lt;math&gt;16p+1&lt;/math&gt; is the cube of a positive integer. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let the positive integer mentioned be &lt;math&gt;a&lt;/math&gt;, so that &lt;math&gt;a^3 = 16p+1&lt;/math&gt;. Note that &lt;math&gt;a&lt;/math&gt; must be odd, because &lt;math&gt;16p+1&lt;/math&gt; is odd.<br /> <br /> Rearrange this expression and factor the left side (this factoring can be done using &lt;math&gt;(a^3-b^3) = (a-b)(a^2+a b+b^2)&lt;/math&gt; or synthetic divison once it is realized that &lt;math&gt;a = 1&lt;/math&gt; is a root):<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a^3-1 &amp;= 16p\\<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;a&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; is even and &lt;math&gt;a^2+a+1&lt;/math&gt; is odd. If &lt;math&gt;a^2+a+1&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; must be some multiple of &lt;math&gt;16&lt;/math&gt;. However, for &lt;math&gt;a-1&lt;/math&gt; to be any multiple of &lt;math&gt;16&lt;/math&gt; other than &lt;math&gt;16&lt;/math&gt; would mean &lt;math&gt;p&lt;/math&gt; is not a prime. Therefore, &lt;math&gt;a-1 = 16&lt;/math&gt; and &lt;math&gt;a = 17&lt;/math&gt;.<br /> <br /> Then our other factor, &lt;math&gt;a^2+a+1&lt;/math&gt;, is the prime &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> (17-1)(17^2+17+1) &amp;=16p\\<br /> p = 289+17+1 &amp;= \boxed{307}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since &lt;math&gt;16p+1&lt;/math&gt; is odd, let &lt;math&gt;16p+1 = (2a+1)^3&lt;/math&gt;. Therefore, &lt;math&gt;16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1&lt;/math&gt;. From this, we get &lt;math&gt;8p=a(4a^2+6a+3)&lt;/math&gt;. We know &lt;math&gt;p&lt;/math&gt; is a prime number and it is not an even number. Since &lt;math&gt;4a^2+6a+3&lt;/math&gt; is an odd number, we know that &lt;math&gt;a=8&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;16p+1=a^3&lt;/math&gt;. Realize that &lt;math&gt;a&lt;/math&gt; congruent to &lt;math&gt;1\mod 4&lt;/math&gt;, so let &lt;math&gt;a=4n+1&lt;/math&gt;. Expansion, then division by 4, gets &lt;math&gt;16n^3+12n^2+3n=4p&lt;/math&gt;. Clearly &lt;math&gt;n=4m&lt;/math&gt; for some &lt;math&gt;m&lt;/math&gt;. Substitution and another division by 4 gets &lt;math&gt;256m^3+48m^4+3m=p&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is prime and there is a factor of &lt;math&gt;m&lt;/math&gt; in the LHS, &lt;math&gt;m=1&lt;/math&gt;. Therefore, &lt;math&gt;p=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Notice that &lt;math&gt;16p+1&lt;/math&gt; must be in the form &lt;math&gt;(a+1)^3 = a^3 + 3a^2 + 3a + 1&lt;/math&gt;. Thus &lt;math&gt;16p = a^3 + 3a^2 + 3a&lt;/math&gt;, or &lt;math&gt;16p = a\cdot (a^2 + 3a + 3)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; must be prime, we either have &lt;math&gt;p = a&lt;/math&gt; or &lt;math&gt;a = 16&lt;/math&gt;. Upon further inspection and/or using the quadratic formula, we can deduce &lt;math&gt;p \neq a&lt;/math&gt;. Thus we have &lt;math&gt;a = 16&lt;/math&gt;, and &lt;math&gt;p = 16^2 + 3\cdot 16 + 3 = \boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br /> Case one: The cube is of the form 16k+1--&gt;Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br /> Case two: The cube is of the form 16k+15--&gt; Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br /> <br /> Hence, &lt;math&gt;p=\boxed{307}&lt;/math&gt; is our only answer<br /> <br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=118921 2015 AIME I Problems/Problem 3 2020-03-08T23:21:19Z <p>Pi is 3.141: </p> <hr /> <div>==Problem==<br /> <br /> There is a prime number &lt;math&gt;p&lt;/math&gt; such that &lt;math&gt;16p+1&lt;/math&gt; is the cube of a positive integer. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let the positive integer mentioned be &lt;math&gt;a&lt;/math&gt;, so that &lt;math&gt;a^3 = 16p+1&lt;/math&gt;. Note that &lt;math&gt;a&lt;/math&gt; must be odd, because &lt;math&gt;16p+1&lt;/math&gt; is odd.<br /> <br /> Rearrange this expression and factor the left side (this factoring can be done using &lt;math&gt;(a^3-b^3) = (a-b)(a^2+a b+b^2)&lt;/math&gt; or synthetic divison once it is realized that &lt;math&gt;a = 1&lt;/math&gt; is a root):<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a^3-1 &amp;= 16p\\<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;a&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; is even and &lt;math&gt;a^2+a+1&lt;/math&gt; is odd. If &lt;math&gt;a^2+a+1&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; must be some multiple of &lt;math&gt;16&lt;/math&gt;. However, for &lt;math&gt;a-1&lt;/math&gt; to be any multiple of &lt;math&gt;16&lt;/math&gt; other than &lt;math&gt;16&lt;/math&gt; would mean &lt;math&gt;p&lt;/math&gt; is not a prime. Therefore, &lt;math&gt;a-1 = 16&lt;/math&gt; and &lt;math&gt;a = 17&lt;/math&gt;.<br /> <br /> Then our other factor, &lt;math&gt;a^2+a+1&lt;/math&gt;, is the prime &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> (17-1)(17^2+17+1) &amp;=16p\\<br /> p = 289+17+1 &amp;= \boxed{307}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since &lt;math&gt;16p+1&lt;/math&gt; is odd, let &lt;math&gt;16p+1 = (2a+1)^3&lt;/math&gt;. Therefore, &lt;math&gt;16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1&lt;/math&gt;. From this, we get &lt;math&gt;8p=a(4a^2+6a+3)&lt;/math&gt;. We know &lt;math&gt;p&lt;/math&gt; is a prime number and it is not an even number. Since &lt;math&gt;4a^2+6a+3&lt;/math&gt; is an odd number, we know that &lt;math&gt;a=8&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;16p+1=a^3&lt;/math&gt;. Realize that &lt;math&gt;a&lt;/math&gt; congruent to &lt;math&gt;1\mod 4&lt;/math&gt;, so let &lt;math&gt;a=4n+1&lt;/math&gt;. Expansion, then division by 4, gets &lt;math&gt;16n^3+12n^2+3n=4p&lt;/math&gt;. Clearly &lt;math&gt;n=4m&lt;/math&gt; for some &lt;math&gt;m&lt;/math&gt;. Substitution and another division by 4 gets &lt;math&gt;256m^3+48m^4+3m=p&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is prime and there is a factor of &lt;math&gt;m&lt;/math&gt; in the LHS, &lt;math&gt;m=1&lt;/math&gt;. Therefore, &lt;math&gt;p=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Notice that &lt;math&gt;16p+1&lt;/math&gt; must be in the form &lt;math&gt;(a+1)^3 = a^3 + 3a^2 + 3a + 1&lt;/math&gt;. Thus &lt;math&gt;16p = a^3 + 3a^2 + 3a&lt;/math&gt;, or &lt;math&gt;16p = a\cdot (a^2 + 3a + 3)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; must be prime, we either have &lt;math&gt;p = a&lt;/math&gt; or &lt;math&gt;a = 16&lt;/math&gt;. Upon further inspection and/or using the quadratic formula, we can deduce &lt;math&gt;p \neq a&lt;/math&gt;. Thus we have &lt;math&gt;a = 16&lt;/math&gt;, and &lt;math&gt;p = 16^2 + 3\cdot 16 + 3 = \boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br /> Case one: The cube is of the form 16k+1--&gt;Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br /> Case two: The cube is of the form 16k+15--&gt; Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br /> <br /> Hence, &lt;math&gt;p=\boxed{307}&lt;/math&gt; is our only answer<br /> pi_is_3.141<br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_3&diff=118896 2015 AIME I Problems/Problem 3 2020-03-08T06:23:16Z <p>Pi is 3.141: </p> <hr /> <div>==Problem==<br /> <br /> There is a prime number &lt;math&gt;p&lt;/math&gt; such that &lt;math&gt;16p+1&lt;/math&gt; is the cube of a positive integer. Find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let the positive integer mentioned be &lt;math&gt;a&lt;/math&gt;, so that &lt;math&gt;a^3 = 16p+1&lt;/math&gt;. Note that &lt;math&gt;a&lt;/math&gt; must be odd, because &lt;math&gt;16p+1&lt;/math&gt; is odd.<br /> <br /> Rearrange this expression and factor the left side (this factoring can be done using &lt;math&gt;(a^3-b^3) = (a-b)(a^2+a b+b^2)&lt;/math&gt; or synthetic divison once it is realized that &lt;math&gt;a = 1&lt;/math&gt; is a root):<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a^3-1 &amp;= 16p\\<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;a&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; is even and &lt;math&gt;a^2+a+1&lt;/math&gt; is odd. If &lt;math&gt;a^2+a+1&lt;/math&gt; is odd, &lt;math&gt;a-1&lt;/math&gt; must be some multiple of &lt;math&gt;16&lt;/math&gt;. However, for &lt;math&gt;a-1&lt;/math&gt; to be any multiple of &lt;math&gt;16&lt;/math&gt; other than &lt;math&gt;16&lt;/math&gt; would mean &lt;math&gt;p&lt;/math&gt; is not a prime. Therefore, &lt;math&gt;a-1 = 16&lt;/math&gt; and &lt;math&gt;a = 17&lt;/math&gt;.<br /> <br /> Then our other factor, &lt;math&gt;a^2+a+1&lt;/math&gt;, is the prime &lt;math&gt;p&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> (a-1)(a^2+a+1) &amp;= 16p\\<br /> (17-1)(17^2+17+1) &amp;=16p\\<br /> p = 289+17+1 &amp;= \boxed{307}<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> <br /> Since &lt;math&gt;16p+1&lt;/math&gt; is odd, let &lt;math&gt;16p+1 = (2a+1)^3&lt;/math&gt;. Therefore, &lt;math&gt;16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1&lt;/math&gt;. From this, we get &lt;math&gt;8p=a(4a^2+6a+3)&lt;/math&gt;. We know &lt;math&gt;p&lt;/math&gt; is a prime number and it is not an even number. Since &lt;math&gt;4a^2+6a+3&lt;/math&gt; is an odd number, we know that &lt;math&gt;a=8&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;16p+1=a^3&lt;/math&gt;. Realize that &lt;math&gt;a&lt;/math&gt; congruent to &lt;math&gt;1\mod 4&lt;/math&gt;, so let &lt;math&gt;a=4n+1&lt;/math&gt;. Expansion, then division by 4, gets &lt;math&gt;16n^3+12n^2+3n=4p&lt;/math&gt;. Clearly &lt;math&gt;n=4m&lt;/math&gt; for some &lt;math&gt;m&lt;/math&gt;. Substitution and another division by 4 gets &lt;math&gt;256m^3+48m^4+3m=p&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is prime and there is a factor of &lt;math&gt;m&lt;/math&gt; in the LHS, &lt;math&gt;m=1&lt;/math&gt;. Therefore, &lt;math&gt;p=\boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Notice that &lt;math&gt;16p+1&lt;/math&gt; must be in the form &lt;math&gt;(a+1)^3 = a^3 + 3a^2 + 3a + 1&lt;/math&gt;. Thus &lt;math&gt;16p = a^3 + 3a^2 + 3a&lt;/math&gt;, or &lt;math&gt;16p = a\cdot (a^2 + 3a + 3)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; must be prime, we either have &lt;math&gt;p = a&lt;/math&gt; or &lt;math&gt;a = 16&lt;/math&gt;. Upon further inspection and/or using the quadratic formula, we can deduce &lt;math&gt;p \neq a&lt;/math&gt;. Thus we have &lt;math&gt;a = 16&lt;/math&gt;, and &lt;math&gt;p = 16^2 + 3\cdot 16 + 3 = \boxed{307}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. <br /> Case one: The cube is of the form 16k+1--&gt;Plugging this in, and taking note that p is prime and has only 1 factor gives p=307<br /> Case two: The cube is of the form 16k+15--&gt; Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here<br /> <br /> Hence, &lt;math&gt;p=\boxed{307}&lt;/math&gt; is our only answer<br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=2|num-a=4}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Number Theory Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&diff=118632 2011 AIME II Problems/Problem 13 2020-03-02T06:35:48Z <p>Pi is 3.141: /* Another Coordinate Solution!!! */</p> <hr /> <div>==Problem==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of [[square]] &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; be the [[circumcenter]]s of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_{1}PO_{2} = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{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. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Denote the [[midpoint]] of &lt;math&gt;\overline{DC}&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;F&lt;/math&gt;. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and these bisectors go through &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;.<br /> <br /> It is given that &lt;math&gt;\angle O_{1}PO_{2}=120^{\circ}&lt;/math&gt;. Because &lt;math&gt;O_{1}P&lt;/math&gt; and &lt;math&gt;O_{1}B&lt;/math&gt; are [[radius|radii]] of the same circle, the have the same length. This is also true of &lt;math&gt;O_{2}P&lt;/math&gt; and &lt;math&gt;O_{2}D&lt;/math&gt;. Because &lt;math&gt;m\angle CAB=m\angle ACD=45^{\circ}&lt;/math&gt;, &lt;math&gt;m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}&lt;/math&gt;. Thus, &lt;math&gt;O_{1}PB&lt;/math&gt; and &lt;math&gt;O_{2}PD&lt;/math&gt; are isosceles right triangles. Using the given information above and symmetry, &lt;math&gt;m\angle DPB = 120^{\circ}&lt;/math&gt;. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.<br /> <br /> Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; have measures of 30 degrees. Thus, both triangles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, &lt;math&gt;DO_{2}=BO_{1}=4\sqrt{3}&lt;/math&gt;. Because of 45-45-90 right triangles, &lt;math&gt;PB=PD=4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, letting &lt;math&gt;x = AP&lt;/math&gt; and using [[Law of Cosines]] on &lt;math&gt;\triangle ABP&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;96=144+x^{2}-24x\frac{\sqrt{2}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;0=x^{2}-12x\sqrt{2}+48&lt;/cmath&gt;<br /> <br /> Using the quadratic formula, we arrive at<br /> <br /> &lt;cmath&gt;x = \sqrt{72} \pm \sqrt{24}&lt;/cmath&gt;<br /> <br /> Taking the positive root, &lt;math&gt;AP=\sqrt{72}+ \sqrt{24}&lt;/math&gt; and the answer is thus &lt;math&gt;\framebox[1.5\width]{096.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This takes a slightly different route than Solution 1.<br /> <br /> Solution 1 proves that &lt;math&gt;\angle{DPB}=120^{\circ}&lt;/math&gt; and that &lt;math&gt;\overline{BP} = \overline{DP}&lt;/math&gt;.<br /> Construct diagonal &lt;math&gt;\overline{BD}&lt;/math&gt; and using the two statements above it quickly becomes clear that &lt;math&gt;\angle{BDP} = \angle{DBP} = 30^{\circ}&lt;/math&gt; by isosceles triangle base angles.<br /> Let the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;, and since the diagonals are perpendicular, both triangle &lt;math&gt;DMP&lt;/math&gt; and triangle &lt;math&gt;BMP&lt;/math&gt; are 30-60-90 right triangles.<br /> Since &lt;math&gt;\overline{AB} = 12&lt;/math&gt;, &lt;math&gt;\overline{AC} = \overline{BD} = 12\sqrt{2}&lt;/math&gt; and &lt;math&gt;\overline{BM} = \overline{DM} = 6\sqrt{2}&lt;/math&gt;.<br /> 30-60-90 triangles' sides are in the ratio &lt;math&gt;1 : \sqrt{3} : 2&lt;/math&gt;, so &lt;math&gt;\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}&lt;/math&gt;.<br /> &lt;math&gt;\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}&lt;/math&gt;.<br /> Hence, &lt;math&gt;72 + 24 = \framebox[1.5\width]{096}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Use vectors. In an &lt;math&gt;xy&lt;/math&gt; plane, let &lt;math&gt;(-s,0)&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;(0,s)&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;(s,0)&lt;/math&gt; be &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;(0,-s)&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;(p,0)&lt;/math&gt; be P, where &lt;math&gt;s=|AB|/\sqrt{2}=6\sqrt{2}&lt;/math&gt;. It remains to find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> The line &lt;math&gt;y=-x&lt;/math&gt; is the [[perpendicular bisector]] of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, so &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; lies on the line. Now compute the [[perpendicular bisector]] of &lt;math&gt;AP&lt;/math&gt;. The center has coordinate &lt;math&gt;(\frac{p-s}{2},0)&lt;/math&gt;, and the segment is part of the &lt;math&gt;x&lt;/math&gt;-axis, so the perpendicular bisector has equation &lt;math&gt;x=\frac{p-s}{2}&lt;/math&gt;. Since &lt;math&gt;O_1&lt;/math&gt; is the [[circumcenter]] of triangle &lt;math&gt;ABP&lt;/math&gt;, it lies on the perpendicular bisector of both &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> O_1=(\frac{p-s}{2},-\frac{p-s}{2})<br /> &lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;<br /> O_2=(\frac{p+s}{2},-\frac{p+s}{2})<br /> &lt;/cmath&gt;<br /> The relation &lt;math&gt;\angle O_1PO_2=120^\circ&lt;/math&gt; can now be written using [[Vectors|dot product]] as<br /> &lt;cmath&gt;<br /> \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|<br /> &lt;/cmath&gt;<br /> Computation of both sides yields<br /> &lt;cmath&gt;<br /> \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}<br /> &lt;/cmath&gt;<br /> Solve for &lt;math&gt;p&lt;/math&gt; gives &lt;math&gt;p=s/\sqrt{3}=2\sqrt{6}&lt;/math&gt;, so &lt;math&gt;AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}&lt;/math&gt;. The answer is 72+24&lt;math&gt;\Rightarrow\boxed{096}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Translate &lt;math&gt;\triangle{ABP}&lt;/math&gt; so that the image of &lt;math&gt;AB&lt;/math&gt; coincides &lt;math&gt;DC&lt;/math&gt;. Let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P’&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle{DPC}=\angle{CPB}&lt;/math&gt; by symmetry, and &lt;math&gt;\angle{APB}=\angle{DP’C}&lt;/math&gt; because translation preserves angles. Thus &lt;math&gt;\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ&lt;/math&gt;. Therefore, quadrilateral &lt;math&gt;CPDP’&lt;/math&gt; is cyclic. Thus the image of &lt;math&gt;O_1&lt;/math&gt; coincides with &lt;math&gt;O_2&lt;/math&gt;. <br /> <br /> &lt;math&gt;O_1P&lt;/math&gt; is parallel to &lt;math&gt;O_2P’&lt;/math&gt; so &lt;math&gt;\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ&lt;/math&gt;, so &lt;math&gt;\angle{PDP’}=60^\circ&lt;/math&gt; and &lt;math&gt;\angle{PDC}=15^\circ&lt;/math&gt;, thus &lt;math&gt;\angle{ADP}=75^{\circ}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;\triangle{AMD}&lt;/math&gt; is a 45-45-90 triangle and &lt;math&gt;\triangle{DMP}&lt;/math&gt; is a 30-60-90 triangle. Thus<br /> <br /> &lt;math&gt;AM=6\sqrt{2}&lt;/math&gt; and &lt;math&gt;MP=\frac{6\sqrt{2}}{\sqrt{3}}&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;AP=AM+MP=\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Reflect &lt;math&gt;O_1&lt;/math&gt; across &lt;math&gt;AP&lt;/math&gt; to &lt;math&gt;O_1'&lt;/math&gt;. By symmetry &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle{DO_1’P}&lt;/math&gt; = &lt;math&gt;2*\angle{DAP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_1’PD}=45^\circ&lt;/math&gt;<br /> <br /> similarly &lt;math&gt;\angle{DO_2P}&lt;/math&gt; = &lt;math&gt;2*\angle{DCP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_2PD}=45^\circ&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\angle{O_1’PO_2}=90^\circ&lt;/math&gt;, so that &lt;math&gt;\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ&lt;/math&gt;<br /> <br /> By symmetry, &lt;math&gt;\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ&lt;/math&gt;<br /> <br /> Therefore, since &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;\angle{ADP}&lt;/math&gt; = &lt;math&gt;0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ&lt;/math&gt;<br /> <br /> Using sine rule in &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> By Kris17<br /> <br /> ==The Simple Way of the Coordinate Plane Solution==<br /> Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set &lt;math&gt;A=(0, 12)&lt;/math&gt;, &lt;math&gt;B=(12,12)&lt;/math&gt;, &lt;math&gt;C=(12, 0)&lt;/math&gt;, &lt;math&gt;D=(0, 0)&lt;/math&gt;. Let this &lt;math&gt;P=(a, 12-a)&lt;/math&gt; for some &lt;math&gt;a&lt;/math&gt;. <br /> <br /> We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that &lt;math&gt;O_{1}&lt;/math&gt; is the intersection of &lt;math&gt;x=6&lt;/math&gt; and, knowing the midpoint of &lt;math&gt;AP&lt;/math&gt; to be &lt;math&gt;(a/2, 12-a/2)&lt;/math&gt; and thus the equation to be &lt;math&gt;y=x+(12-a)&lt;/math&gt;, we get &lt;math&gt;(6, 18-a)&lt;/math&gt;. Likewise for &lt;math&gt;O_{2}&lt;/math&gt; it's &lt;math&gt;(6, 6-a)&lt;/math&gt;. Now what do we see? &lt;math&gt;O_{1}P=O_{2}P&lt;/math&gt; (just look at the coordinates)! So both of those distances are &lt;math&gt;4\sqrt{3}&lt;/math&gt;. Solving for &lt;math&gt;a&lt;/math&gt; we get it to be &lt;math&gt;6+2\sqrt{3}&lt;/math&gt;, since &lt;math&gt;AP&gt;CP&lt;/math&gt;. Multiply by &lt;math&gt;\sqrt{2}&lt;/math&gt; because we are looking for &lt;math&gt;AP&lt;/math&gt; to get the answer of &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> ==Another Coordinate Solution==<br /> Okay set A(0,12), B(12, 12), and the the coordinates for C and D follow. P is thus some (x, -x+12) (remember it lies on the line y=-x+12). You can easily find the circumcenters of both desired triangles(remember perpendicular bisectors!!). Once you do this, you get Coordinates as O1(6, -x+18) and O2(6, -x+6). From here you can uses Law Of Cosines and get the answer as &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&diff=118631 2011 AIME II Problems/Problem 13 2020-03-02T06:35:25Z <p>Pi is 3.141: /* Another Coordinate Solution!!! */</p> <hr /> <div>==Problem==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of [[square]] &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; be the [[circumcenter]]s of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_{1}PO_{2} = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{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. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Denote the [[midpoint]] of &lt;math&gt;\overline{DC}&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;F&lt;/math&gt;. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and these bisectors go through &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;.<br /> <br /> It is given that &lt;math&gt;\angle O_{1}PO_{2}=120^{\circ}&lt;/math&gt;. Because &lt;math&gt;O_{1}P&lt;/math&gt; and &lt;math&gt;O_{1}B&lt;/math&gt; are [[radius|radii]] of the same circle, the have the same length. This is also true of &lt;math&gt;O_{2}P&lt;/math&gt; and &lt;math&gt;O_{2}D&lt;/math&gt;. Because &lt;math&gt;m\angle CAB=m\angle ACD=45^{\circ}&lt;/math&gt;, &lt;math&gt;m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}&lt;/math&gt;. Thus, &lt;math&gt;O_{1}PB&lt;/math&gt; and &lt;math&gt;O_{2}PD&lt;/math&gt; are isosceles right triangles. Using the given information above and symmetry, &lt;math&gt;m\angle DPB = 120^{\circ}&lt;/math&gt;. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.<br /> <br /> Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; have measures of 30 degrees. Thus, both triangles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, &lt;math&gt;DO_{2}=BO_{1}=4\sqrt{3}&lt;/math&gt;. Because of 45-45-90 right triangles, &lt;math&gt;PB=PD=4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, letting &lt;math&gt;x = AP&lt;/math&gt; and using [[Law of Cosines]] on &lt;math&gt;\triangle ABP&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;96=144+x^{2}-24x\frac{\sqrt{2}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;0=x^{2}-12x\sqrt{2}+48&lt;/cmath&gt;<br /> <br /> Using the quadratic formula, we arrive at<br /> <br /> &lt;cmath&gt;x = \sqrt{72} \pm \sqrt{24}&lt;/cmath&gt;<br /> <br /> Taking the positive root, &lt;math&gt;AP=\sqrt{72}+ \sqrt{24}&lt;/math&gt; and the answer is thus &lt;math&gt;\framebox[1.5\width]{096.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This takes a slightly different route than Solution 1.<br /> <br /> Solution 1 proves that &lt;math&gt;\angle{DPB}=120^{\circ}&lt;/math&gt; and that &lt;math&gt;\overline{BP} = \overline{DP}&lt;/math&gt;.<br /> Construct diagonal &lt;math&gt;\overline{BD}&lt;/math&gt; and using the two statements above it quickly becomes clear that &lt;math&gt;\angle{BDP} = \angle{DBP} = 30^{\circ}&lt;/math&gt; by isosceles triangle base angles.<br /> Let the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;, and since the diagonals are perpendicular, both triangle &lt;math&gt;DMP&lt;/math&gt; and triangle &lt;math&gt;BMP&lt;/math&gt; are 30-60-90 right triangles.<br /> Since &lt;math&gt;\overline{AB} = 12&lt;/math&gt;, &lt;math&gt;\overline{AC} = \overline{BD} = 12\sqrt{2}&lt;/math&gt; and &lt;math&gt;\overline{BM} = \overline{DM} = 6\sqrt{2}&lt;/math&gt;.<br /> 30-60-90 triangles' sides are in the ratio &lt;math&gt;1 : \sqrt{3} : 2&lt;/math&gt;, so &lt;math&gt;\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}&lt;/math&gt;.<br /> &lt;math&gt;\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}&lt;/math&gt;.<br /> Hence, &lt;math&gt;72 + 24 = \framebox[1.5\width]{096}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Use vectors. In an &lt;math&gt;xy&lt;/math&gt; plane, let &lt;math&gt;(-s,0)&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;(0,s)&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;(s,0)&lt;/math&gt; be &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;(0,-s)&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;(p,0)&lt;/math&gt; be P, where &lt;math&gt;s=|AB|/\sqrt{2}=6\sqrt{2}&lt;/math&gt;. It remains to find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> The line &lt;math&gt;y=-x&lt;/math&gt; is the [[perpendicular bisector]] of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, so &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; lies on the line. Now compute the [[perpendicular bisector]] of &lt;math&gt;AP&lt;/math&gt;. The center has coordinate &lt;math&gt;(\frac{p-s}{2},0)&lt;/math&gt;, and the segment is part of the &lt;math&gt;x&lt;/math&gt;-axis, so the perpendicular bisector has equation &lt;math&gt;x=\frac{p-s}{2}&lt;/math&gt;. Since &lt;math&gt;O_1&lt;/math&gt; is the [[circumcenter]] of triangle &lt;math&gt;ABP&lt;/math&gt;, it lies on the perpendicular bisector of both &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> O_1=(\frac{p-s}{2},-\frac{p-s}{2})<br /> &lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;<br /> O_2=(\frac{p+s}{2},-\frac{p+s}{2})<br /> &lt;/cmath&gt;<br /> The relation &lt;math&gt;\angle O_1PO_2=120^\circ&lt;/math&gt; can now be written using [[Vectors|dot product]] as<br /> &lt;cmath&gt;<br /> \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|<br /> &lt;/cmath&gt;<br /> Computation of both sides yields<br /> &lt;cmath&gt;<br /> \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}<br /> &lt;/cmath&gt;<br /> Solve for &lt;math&gt;p&lt;/math&gt; gives &lt;math&gt;p=s/\sqrt{3}=2\sqrt{6}&lt;/math&gt;, so &lt;math&gt;AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}&lt;/math&gt;. The answer is 72+24&lt;math&gt;\Rightarrow\boxed{096}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Translate &lt;math&gt;\triangle{ABP}&lt;/math&gt; so that the image of &lt;math&gt;AB&lt;/math&gt; coincides &lt;math&gt;DC&lt;/math&gt;. Let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P’&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle{DPC}=\angle{CPB}&lt;/math&gt; by symmetry, and &lt;math&gt;\angle{APB}=\angle{DP’C}&lt;/math&gt; because translation preserves angles. Thus &lt;math&gt;\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ&lt;/math&gt;. Therefore, quadrilateral &lt;math&gt;CPDP’&lt;/math&gt; is cyclic. Thus the image of &lt;math&gt;O_1&lt;/math&gt; coincides with &lt;math&gt;O_2&lt;/math&gt;. <br /> <br /> &lt;math&gt;O_1P&lt;/math&gt; is parallel to &lt;math&gt;O_2P’&lt;/math&gt; so &lt;math&gt;\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ&lt;/math&gt;, so &lt;math&gt;\angle{PDP’}=60^\circ&lt;/math&gt; and &lt;math&gt;\angle{PDC}=15^\circ&lt;/math&gt;, thus &lt;math&gt;\angle{ADP}=75^{\circ}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;\triangle{AMD}&lt;/math&gt; is a 45-45-90 triangle and &lt;math&gt;\triangle{DMP}&lt;/math&gt; is a 30-60-90 triangle. Thus<br /> <br /> &lt;math&gt;AM=6\sqrt{2}&lt;/math&gt; and &lt;math&gt;MP=\frac{6\sqrt{2}}{\sqrt{3}}&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;AP=AM+MP=\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Reflect &lt;math&gt;O_1&lt;/math&gt; across &lt;math&gt;AP&lt;/math&gt; to &lt;math&gt;O_1'&lt;/math&gt;. By symmetry &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle{DO_1’P}&lt;/math&gt; = &lt;math&gt;2*\angle{DAP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_1’PD}=45^\circ&lt;/math&gt;<br /> <br /> similarly &lt;math&gt;\angle{DO_2P}&lt;/math&gt; = &lt;math&gt;2*\angle{DCP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_2PD}=45^\circ&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\angle{O_1’PO_2}=90^\circ&lt;/math&gt;, so that &lt;math&gt;\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ&lt;/math&gt;<br /> <br /> By symmetry, &lt;math&gt;\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ&lt;/math&gt;<br /> <br /> Therefore, since &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;\angle{ADP}&lt;/math&gt; = &lt;math&gt;0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ&lt;/math&gt;<br /> <br /> Using sine rule in &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> By Kris17<br /> <br /> ==The Simple Way of the Coordinate Plane Solution==<br /> Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set &lt;math&gt;A=(0, 12)&lt;/math&gt;, &lt;math&gt;B=(12,12)&lt;/math&gt;, &lt;math&gt;C=(12, 0)&lt;/math&gt;, &lt;math&gt;D=(0, 0)&lt;/math&gt;. Let this &lt;math&gt;P=(a, 12-a)&lt;/math&gt; for some &lt;math&gt;a&lt;/math&gt;. <br /> <br /> We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that &lt;math&gt;O_{1}&lt;/math&gt; is the intersection of &lt;math&gt;x=6&lt;/math&gt; and, knowing the midpoint of &lt;math&gt;AP&lt;/math&gt; to be &lt;math&gt;(a/2, 12-a/2)&lt;/math&gt; and thus the equation to be &lt;math&gt;y=x+(12-a)&lt;/math&gt;, we get &lt;math&gt;(6, 18-a)&lt;/math&gt;. Likewise for &lt;math&gt;O_{2}&lt;/math&gt; it's &lt;math&gt;(6, 6-a)&lt;/math&gt;. Now what do we see? &lt;math&gt;O_{1}P=O_{2}P&lt;/math&gt; (just look at the coordinates)! So both of those distances are &lt;math&gt;4\sqrt{3}&lt;/math&gt;. Solving for &lt;math&gt;a&lt;/math&gt; we get it to be &lt;math&gt;6+2\sqrt{3}&lt;/math&gt;, since &lt;math&gt;AP&gt;CP&lt;/math&gt;. Multiply by &lt;math&gt;\sqrt{2}&lt;/math&gt; because we are looking for &lt;math&gt;AP&lt;/math&gt; to get the answer of &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> ==Another Coordinate Solution!!!==<br /> Okay set A(0,12), B(12, 12), and the the coordinates for C and D follow. P is thus some (x, -x+12) (remember it lies on the line y=-x+12). You can easily find the circumcenters of both desired triangles(remember perpendicular bisectors!!). Once you do this, you get Coordinates as O1(6, -x+18) and O2(6, -x+6). From here you can uses Law Of Cosines and get the answer as &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_13&diff=118629 2011 AIME II Problems/Problem 13 2020-03-02T06:34:54Z <p>Pi is 3.141: </p> <hr /> <div>==Problem==<br /> Point &lt;math&gt;P&lt;/math&gt; lies on the diagonal &lt;math&gt;AC&lt;/math&gt; of [[square]] &lt;math&gt;ABCD&lt;/math&gt; with &lt;math&gt;AP &gt; CP&lt;/math&gt;. Let &lt;math&gt;O_{1}&lt;/math&gt; and &lt;math&gt;O_{2}&lt;/math&gt; be the [[circumcenter]]s of triangles &lt;math&gt;ABP&lt;/math&gt; and &lt;math&gt;CDP&lt;/math&gt; respectively. Given that &lt;math&gt;AB = 12&lt;/math&gt; and &lt;math&gt;\angle O_{1}PO_{2} = 120^{\circ}&lt;/math&gt;, then &lt;math&gt;AP = \sqrt{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. Find &lt;math&gt;a + b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Denote the [[midpoint]] of &lt;math&gt;\overline{DC}&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt; be &lt;math&gt;F&lt;/math&gt;. Because they are the circumcenters, both Os lie on the [[perpendicular bisector]]s of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; and these bisectors go through &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;.<br /> <br /> It is given that &lt;math&gt;\angle O_{1}PO_{2}=120^{\circ}&lt;/math&gt;. Because &lt;math&gt;O_{1}P&lt;/math&gt; and &lt;math&gt;O_{1}B&lt;/math&gt; are [[radius|radii]] of the same circle, the have the same length. This is also true of &lt;math&gt;O_{2}P&lt;/math&gt; and &lt;math&gt;O_{2}D&lt;/math&gt;. Because &lt;math&gt;m\angle CAB=m\angle ACD=45^{\circ}&lt;/math&gt;, &lt;math&gt;m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}&lt;/math&gt;. Thus, &lt;math&gt;O_{1}PB&lt;/math&gt; and &lt;math&gt;O_{2}PD&lt;/math&gt; are isosceles right triangles. Using the given information above and symmetry, &lt;math&gt;m\angle DPB = 120^{\circ}&lt;/math&gt;. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.<br /> <br /> Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; have measures of 30 degrees. Thus, both triangles &lt;math&gt;O_{1}BF&lt;/math&gt; and &lt;math&gt;O_{2}DE&lt;/math&gt; are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, &lt;math&gt;DO_{2}=BO_{1}=4\sqrt{3}&lt;/math&gt;. Because of 45-45-90 right triangles, &lt;math&gt;PB=PD=4\sqrt{6}&lt;/math&gt;.<br /> <br /> Now, letting &lt;math&gt;x = AP&lt;/math&gt; and using [[Law of Cosines]] on &lt;math&gt;\triangle ABP&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;96=144+x^{2}-24x\frac{\sqrt{2}}{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;0=x^{2}-12x\sqrt{2}+48&lt;/cmath&gt;<br /> <br /> Using the quadratic formula, we arrive at<br /> <br /> &lt;cmath&gt;x = \sqrt{72} \pm \sqrt{24}&lt;/cmath&gt;<br /> <br /> Taking the positive root, &lt;math&gt;AP=\sqrt{72}+ \sqrt{24}&lt;/math&gt; and the answer is thus &lt;math&gt;\framebox[1.5\width]{096.}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> This takes a slightly different route than Solution 1.<br /> <br /> Solution 1 proves that &lt;math&gt;\angle{DPB}=120^{\circ}&lt;/math&gt; and that &lt;math&gt;\overline{BP} = \overline{DP}&lt;/math&gt;.<br /> Construct diagonal &lt;math&gt;\overline{BD}&lt;/math&gt; and using the two statements above it quickly becomes clear that &lt;math&gt;\angle{BDP} = \angle{DBP} = 30^{\circ}&lt;/math&gt; by isosceles triangle base angles.<br /> Let the midpoint of diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;, and since the diagonals are perpendicular, both triangle &lt;math&gt;DMP&lt;/math&gt; and triangle &lt;math&gt;BMP&lt;/math&gt; are 30-60-90 right triangles.<br /> Since &lt;math&gt;\overline{AB} = 12&lt;/math&gt;, &lt;math&gt;\overline{AC} = \overline{BD} = 12\sqrt{2}&lt;/math&gt; and &lt;math&gt;\overline{BM} = \overline{DM} = 6\sqrt{2}&lt;/math&gt;.<br /> 30-60-90 triangles' sides are in the ratio &lt;math&gt;1 : \sqrt{3} : 2&lt;/math&gt;, so &lt;math&gt;\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}&lt;/math&gt;.<br /> &lt;math&gt;\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}&lt;/math&gt;.<br /> Hence, &lt;math&gt;72 + 24 = \framebox[1.5\width]{096}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Use vectors. In an &lt;math&gt;xy&lt;/math&gt; plane, let &lt;math&gt;(-s,0)&lt;/math&gt; be &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;(0,s)&lt;/math&gt; be &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;(s,0)&lt;/math&gt; be &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;(0,-s)&lt;/math&gt; be &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;(p,0)&lt;/math&gt; be P, where &lt;math&gt;s=|AB|/\sqrt{2}=6\sqrt{2}&lt;/math&gt;. It remains to find &lt;math&gt;p&lt;/math&gt;.<br /> <br /> The line &lt;math&gt;y=-x&lt;/math&gt; is the [[perpendicular bisector]] of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt;, so &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; lies on the line. Now compute the [[perpendicular bisector]] of &lt;math&gt;AP&lt;/math&gt;. The center has coordinate &lt;math&gt;(\frac{p-s}{2},0)&lt;/math&gt;, and the segment is part of the &lt;math&gt;x&lt;/math&gt;-axis, so the perpendicular bisector has equation &lt;math&gt;x=\frac{p-s}{2}&lt;/math&gt;. Since &lt;math&gt;O_1&lt;/math&gt; is the [[circumcenter]] of triangle &lt;math&gt;ABP&lt;/math&gt;, it lies on the perpendicular bisector of both &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;, so<br /> &lt;cmath&gt;<br /> O_1=(\frac{p-s}{2},-\frac{p-s}{2})<br /> &lt;/cmath&gt;<br /> Similarly, <br /> &lt;cmath&gt;<br /> O_2=(\frac{p+s}{2},-\frac{p+s}{2})<br /> &lt;/cmath&gt;<br /> The relation &lt;math&gt;\angle O_1PO_2=120^\circ&lt;/math&gt; can now be written using [[Vectors|dot product]] as<br /> &lt;cmath&gt;<br /> \vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|<br /> &lt;/cmath&gt;<br /> Computation of both sides yields<br /> &lt;cmath&gt;<br /> \frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}<br /> &lt;/cmath&gt;<br /> Solve for &lt;math&gt;p&lt;/math&gt; gives &lt;math&gt;p=s/\sqrt{3}=2\sqrt{6}&lt;/math&gt;, so &lt;math&gt;AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}&lt;/math&gt;. The answer is 72+24&lt;math&gt;\Rightarrow\boxed{096}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Translate &lt;math&gt;\triangle{ABP}&lt;/math&gt; so that the image of &lt;math&gt;AB&lt;/math&gt; coincides &lt;math&gt;DC&lt;/math&gt;. Let the image of &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;P’&lt;/math&gt;. <br /> <br /> &lt;math&gt;\angle{DPC}=\angle{CPB}&lt;/math&gt; by symmetry, and &lt;math&gt;\angle{APB}=\angle{DP’C}&lt;/math&gt; because translation preserves angles. Thus &lt;math&gt;\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ&lt;/math&gt;. Therefore, quadrilateral &lt;math&gt;CPDP’&lt;/math&gt; is cyclic. Thus the image of &lt;math&gt;O_1&lt;/math&gt; coincides with &lt;math&gt;O_2&lt;/math&gt;. <br /> <br /> &lt;math&gt;O_1P&lt;/math&gt; is parallel to &lt;math&gt;O_2P’&lt;/math&gt; so &lt;math&gt;\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ&lt;/math&gt;, so &lt;math&gt;\angle{PDP’}=60^\circ&lt;/math&gt; and &lt;math&gt;\angle{PDC}=15^\circ&lt;/math&gt;, thus &lt;math&gt;\angle{ADP}=75^{\circ}&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Then &lt;math&gt;\triangle{AMD}&lt;/math&gt; is a 45-45-90 triangle and &lt;math&gt;\triangle{DMP}&lt;/math&gt; is a 30-60-90 triangle. Thus<br /> <br /> &lt;math&gt;AM=6\sqrt{2}&lt;/math&gt; and &lt;math&gt;MP=\frac{6\sqrt{2}}{\sqrt{3}}&lt;/math&gt;.<br /> <br /> This gives us &lt;math&gt;AP=AM+MP=\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> <br /> Reflect &lt;math&gt;O_1&lt;/math&gt; across &lt;math&gt;AP&lt;/math&gt; to &lt;math&gt;O_1'&lt;/math&gt;. By symmetry &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;<br /> <br /> &lt;math&gt;\angle{DO_1’P}&lt;/math&gt; = &lt;math&gt;2*\angle{DAP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_1’PD}=45^\circ&lt;/math&gt;<br /> <br /> similarly &lt;math&gt;\angle{DO_2P}&lt;/math&gt; = &lt;math&gt;2*\angle{DCP} = 90^\circ&lt;/math&gt;, so &lt;math&gt;\angle{O_2PD}=45^\circ&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;\angle{O_1’PO_2}=90^\circ&lt;/math&gt;, so that &lt;math&gt;\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ&lt;/math&gt;<br /> <br /> By symmetry, &lt;math&gt;\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ&lt;/math&gt;<br /> <br /> Therefore, since &lt;math&gt;O_1’&lt;/math&gt; is the circumcenter of &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;\angle{ADP}&lt;/math&gt; = &lt;math&gt;0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ&lt;/math&gt;<br /> <br /> Therefore &lt;math&gt;\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ&lt;/math&gt;<br /> <br /> Using sine rule in &lt;math&gt;\triangle{ADP}&lt;/math&gt;, &lt;math&gt;AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}&lt;/math&gt;, and the answer is &lt;math&gt;72+24=\boxed{096}.&lt;/math&gt;<br /> <br /> By Kris17<br /> <br /> ==The Simple Way of the Coordinate Plane Solution==<br /> Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set &lt;math&gt;A=(0, 12)&lt;/math&gt;, &lt;math&gt;B=(12,12)&lt;/math&gt;, &lt;math&gt;C=(12, 0)&lt;/math&gt;, &lt;math&gt;D=(0, 0)&lt;/math&gt;. Let this &lt;math&gt;P=(a, 12-a)&lt;/math&gt; for some &lt;math&gt;a&lt;/math&gt;. <br /> <br /> We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that &lt;math&gt;O_{1}&lt;/math&gt; is the intersection of &lt;math&gt;x=6&lt;/math&gt; and, knowing the midpoint of &lt;math&gt;AP&lt;/math&gt; to be &lt;math&gt;(a/2, 12-a/2)&lt;/math&gt; and thus the equation to be &lt;math&gt;y=x+(12-a)&lt;/math&gt;, we get &lt;math&gt;(6, 18-a)&lt;/math&gt;. Likewise for &lt;math&gt;O_{2}&lt;/math&gt; it's &lt;math&gt;(6, 6-a)&lt;/math&gt;. Now what do we see? &lt;math&gt;O_{1}P=O_{2}P&lt;/math&gt; (just look at the coordinates)! So both of those distances are &lt;math&gt;4\sqrt{3}&lt;/math&gt;. Solving for &lt;math&gt;a&lt;/math&gt; we get it to be &lt;math&gt;6+2\sqrt{3}&lt;/math&gt;, since &lt;math&gt;AP&gt;CP&lt;/math&gt;. Multiply by &lt;math&gt;\sqrt{2}&lt;/math&gt; because we are looking for &lt;math&gt;AP&lt;/math&gt; to get the answer of &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> ==Another Coordinate Solution!!!==<br /> Okay set A(0,12), B(12, 12), and the rest follows. P is thus some (x, -x+12) (remember it lies on the line y=-x+12). You can easily find the circumcenters of both desired triangles(remember perpendicular bisectors!!). Once you do this, you get Coordinates as O1(6, -x+18) and O2(6, -x+6). From here you can uses Law Of Cosines and get the answer as &lt;math&gt;\boxed{096}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AIME box|year=2011|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_12&diff=117842 1985 AIME Problems/Problem 12 2020-02-17T03:19:59Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; be the [[vertex | vertices]] of a regular [[tetrahedron]] each of whose [[edge]]s measures 1 meter. A bug, starting from vertex &lt;math&gt;A&lt;/math&gt;, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let &lt;math&gt;p = \frac n{729}&lt;/math&gt; be the [[probability]] that the bug is at vertex &lt;math&gt;A&lt;/math&gt; when it has crawled exactly 7 meters. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;P(n)&lt;/math&gt; denote the probability that the bug is at &lt;math&gt;A&lt;/math&gt; after it has crawled &lt;math&gt;n&lt;/math&gt; meters. Since the bug can only be at vertex &lt;math&gt;A&lt;/math&gt; if it just left a vertex which is not &lt;math&gt;A&lt;/math&gt;, we have &lt;math&gt;P(n + 1) = \frac13 (1 - P(n))&lt;/math&gt;. We also know &lt;math&gt;P(0) = 1&lt;/math&gt;, so we can quickly compute &lt;math&gt;P(1)=0&lt;/math&gt;, &lt;math&gt;P(2) = \frac 13&lt;/math&gt;, &lt;math&gt;P(3) = \frac29&lt;/math&gt;, &lt;math&gt;P(4) = \frac7{27}&lt;/math&gt;, &lt;math&gt;P(5) = \frac{20}{81}&lt;/math&gt;, &lt;math&gt;P(6) = \frac{61}{243}&lt;/math&gt; and &lt;math&gt;P(7) = \frac{182}{729}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{182}&lt;/math&gt;. One can solve this [[recursion]] fairly easily to determine a closed-form expression for &lt;math&gt;P(n)&lt;/math&gt;. <br /> <br /> === Solution 2 ===<br /> We can find the number of different times the bug reaches vertex &lt;math&gt;A&lt;/math&gt; before the 7th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(x)&lt;/math&gt; to be the number of paths of length &lt;math&gt;x&lt;/math&gt; which start and end at &lt;math&gt;A&lt;/math&gt; but do not pass through &lt;math&gt;A&lt;/math&gt; otherwise. Obviously &lt;math&gt;f(1) = 0&lt;/math&gt;. In general for &lt;math&gt;f(x)&lt;/math&gt;, the bug has three initial edges to pick from. From there, since the bug cannot return to &lt;math&gt;A&lt;/math&gt; by definition, the bug has exactly two choices. This continues from the 2nd move up to the &lt;math&gt;(x-1)&lt;/math&gt;th move. The last move must be a return to &lt;math&gt;A&lt;/math&gt;, so this move is determined. So &lt;math&gt;f(x) = 2^{x-2}3&lt;/math&gt;.<br /> <br /> Now we need to find the number of cycles by which the bug can reach &lt;math&gt;A&lt;/math&gt; at the end. Since &lt;math&gt;f(1) = 0&lt;/math&gt;, &lt;math&gt;f(6)&lt;/math&gt; cannot be used since on the 7th move the bug cannot move from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;. So we need to find the number of [[partition]]s of 7 using only 2,3,4,5, and 7. These are &lt;math&gt;f(2)f(2)f(3)&lt;/math&gt;, &lt;math&gt;f(2)f(5)&lt;/math&gt;, &lt;math&gt;f(3)f(4)&lt;/math&gt;, and &lt;math&gt;f(7)&lt;/math&gt;. We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7)&lt;/math&gt;&lt;br /&gt;&lt;math&gt;= 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53)&lt;/math&gt;&lt;br /&gt;&lt;math&gt;= 546&lt;/math&gt;&lt;/div&gt;<br /> <br /> Finally, this is a [[probability]] question, so we divide by &lt;math&gt;3^7&lt;/math&gt;: &lt;math&gt;\frac{546}{3^7} = \frac{182}{3^6}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> There exists a simple heuristic method to arrive at the answer to this question, due to [[User: ComplexZeta | Simon Rubinstein-Salzedo]], as follows: after a couple of moves, the randomness of movement of the bug and smallness of the system ensures that we should expect its [[probability distribution]] to be very close to [[uniform distribution | uniform]]. In particular, we would expect &lt;math&gt;P(n)&lt;/math&gt; to be very close to &lt;math&gt;\frac 14&lt;/math&gt; for decently-sized &lt;math&gt;n&lt;/math&gt;, for example &lt;math&gt;n = 7&lt;/math&gt;. (In fact, from looking at the previous solution we can see that it is already close when &lt;math&gt;n = 3&lt;/math&gt;, and in fact, the earlier values are also the best possible approximations given the restraints on where the bug can be.) Since we know the answer is of the form &lt;math&gt;\frac n{729}&lt;/math&gt;, we realize that &lt;math&gt;n&lt;/math&gt; must be very close to &lt;math&gt;\frac{729}{4} = 182.25&lt;/math&gt;, as indeed it is.<br /> <br /> === Solution 4 (Cheap Solution) ===<br /> Only do this if you don't know how to solve the problem and want to make a good guess. Since there are 4 vertices of a tetrahedron, there is approximately a 1/4 probability of coming back to A after 7 moves. Dividing 729 by 4 gives a number between 182 and 183. If the ant continuously alternates his location from A to some other vertice, in the end, it will not be at A. Therefore we chose the smaller number, 182.<br /> - Rocket123<br /> <br /> <br /> === Solution 5 ===<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; denotes the number of ways that the bug arrives at &lt;math&gt;A&lt;/math&gt; after crawling &lt;math&gt;n&lt;/math&gt; meters, then we have &lt;math&gt;a_1=0&lt;/math&gt;.<br /> <br /> Notice that there is respectively &lt;math&gt;1&lt;/math&gt; way to arrive at &lt;math&gt;A&lt;/math&gt; for each of the different routes after the previous &lt;math&gt;n-1&lt;/math&gt; crawls, excluding the possibility that the bug ends up at &lt;math&gt;A&lt;/math&gt; after the &lt;math&gt;(n-1)&lt;/math&gt;th crawl (as it will be forced to move somewhere else.). Thus we get the recurrence relation<br /> &lt;cmath&gt;a_n=3^{n-1}-a_{n-1}&lt;/cmath&gt;<br /> Quick calculations yield<br /> &lt;cmath&gt;\begin{align*}<br /> a_7&amp;=3^6-a_6\\<br /> &amp;=3^6-3^5+3^4-...+a_1\\<br /> &amp;=546<br /> \end{align*}&lt;/cmath&gt;<br /> Thus &lt;math&gt;p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> === Solution 6 ===<br /> Draw an organized table and keep track of how many ways are there to get to each vertex after 1, 2, 3, ..., 7 steps. Thus you get the answer of &lt;math&gt;p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}&lt;/math&gt;.<br /> -pi_is_3.141<br /> == See also ==<br /> *[[2003 AIME II Problems/Problem 13]] - very similar problem with an equilateral triangle<br /> <br /> {{AIME box|year=1985|num-b=11|num-a=13}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_12&diff=117841 1985 AIME Problems/Problem 12 2020-02-17T03:19:37Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; be the [[vertex | vertices]] of a regular [[tetrahedron]] each of whose [[edge]]s measures 1 meter. A bug, starting from vertex &lt;math&gt;A&lt;/math&gt;, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let &lt;math&gt;p = \frac n{729}&lt;/math&gt; be the [[probability]] that the bug is at vertex &lt;math&gt;A&lt;/math&gt; when it has crawled exactly 7 meters. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;P(n)&lt;/math&gt; denote the probability that the bug is at &lt;math&gt;A&lt;/math&gt; after it has crawled &lt;math&gt;n&lt;/math&gt; meters. Since the bug can only be at vertex &lt;math&gt;A&lt;/math&gt; if it just left a vertex which is not &lt;math&gt;A&lt;/math&gt;, we have &lt;math&gt;P(n + 1) = \frac13 (1 - P(n))&lt;/math&gt;. We also know &lt;math&gt;P(0) = 1&lt;/math&gt;, so we can quickly compute &lt;math&gt;P(1)=0&lt;/math&gt;, &lt;math&gt;P(2) = \frac 13&lt;/math&gt;, &lt;math&gt;P(3) = \frac29&lt;/math&gt;, &lt;math&gt;P(4) = \frac7{27}&lt;/math&gt;, &lt;math&gt;P(5) = \frac{20}{81}&lt;/math&gt;, &lt;math&gt;P(6) = \frac{61}{243}&lt;/math&gt; and &lt;math&gt;P(7) = \frac{182}{729}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{182}&lt;/math&gt;. One can solve this [[recursion]] fairly easily to determine a closed-form expression for &lt;math&gt;P(n)&lt;/math&gt;. <br /> <br /> === Solution 2 ===<br /> We can find the number of different times the bug reaches vertex &lt;math&gt;A&lt;/math&gt; before the 7th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at &lt;math&gt;A&lt;/math&gt;.<br /> <br /> Define &lt;math&gt;f(x)&lt;/math&gt; to be the number of paths of length &lt;math&gt;x&lt;/math&gt; which start and end at &lt;math&gt;A&lt;/math&gt; but do not pass through &lt;math&gt;A&lt;/math&gt; otherwise. Obviously &lt;math&gt;f(1) = 0&lt;/math&gt;. In general for &lt;math&gt;f(x)&lt;/math&gt;, the bug has three initial edges to pick from. From there, since the bug cannot return to &lt;math&gt;A&lt;/math&gt; by definition, the bug has exactly two choices. This continues from the 2nd move up to the &lt;math&gt;(x-1)&lt;/math&gt;th move. The last move must be a return to &lt;math&gt;A&lt;/math&gt;, so this move is determined. So &lt;math&gt;f(x) = 2^{x-2}3&lt;/math&gt;.<br /> <br /> Now we need to find the number of cycles by which the bug can reach &lt;math&gt;A&lt;/math&gt; at the end. Since &lt;math&gt;f(1) = 0&lt;/math&gt;, &lt;math&gt;f(6)&lt;/math&gt; cannot be used since on the 7th move the bug cannot move from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt;. So we need to find the number of [[partition]]s of 7 using only 2,3,4,5, and 7. These are &lt;math&gt;f(2)f(2)f(3)&lt;/math&gt;, &lt;math&gt;f(2)f(5)&lt;/math&gt;, &lt;math&gt;f(3)f(4)&lt;/math&gt;, and &lt;math&gt;f(7)&lt;/math&gt;. We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7)&lt;/math&gt;&lt;br /&gt;&lt;math&gt;= 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53)&lt;/math&gt;&lt;br /&gt;&lt;math&gt;= 546&lt;/math&gt;&lt;/div&gt;<br /> <br /> Finally, this is a [[probability]] question, so we divide by &lt;math&gt;3^7&lt;/math&gt;: &lt;math&gt;\frac{546}{3^7} = \frac{182}{3^6}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> There exists a simple heuristic method to arrive at the answer to this question, due to [[User: ComplexZeta | Simon Rubinstein-Salzedo]], as follows: after a couple of moves, the randomness of movement of the bug and smallness of the system ensures that we should expect its [[probability distribution]] to be very close to [[uniform distribution | uniform]]. In particular, we would expect &lt;math&gt;P(n)&lt;/math&gt; to be very close to &lt;math&gt;\frac 14&lt;/math&gt; for decently-sized &lt;math&gt;n&lt;/math&gt;, for example &lt;math&gt;n = 7&lt;/math&gt;. (In fact, from looking at the previous solution we can see that it is already close when &lt;math&gt;n = 3&lt;/math&gt;, and in fact, the earlier values are also the best possible approximations given the restraints on where the bug can be.) Since we know the answer is of the form &lt;math&gt;\frac n{729}&lt;/math&gt;, we realize that &lt;math&gt;n&lt;/math&gt; must be very close to &lt;math&gt;\frac{729}{4} = 182.25&lt;/math&gt;, as indeed it is.<br /> <br /> === Solution 4 (Cheap Solution) ===<br /> Only do this if you don't know how to solve the problem and want to make a good guess. Since there are 4 vertices of a tetrahedron, there is approximately a 1/4 probability of coming back to A after 7 moves. Dividing 729 by 4 gives a number between 182 and 183. If the ant continuously alternates his location from A to some other vertice, in the end, it will not be at A. Therefore we chose the smaller number, 182.<br /> - Rocket123<br /> <br /> <br /> === Solution 5 ===<br /> <br /> Let &lt;math&gt;a_n&lt;/math&gt; denotes the number of ways that the bug arrives at &lt;math&gt;A&lt;/math&gt; after crawling &lt;math&gt;n&lt;/math&gt; meters, then we have &lt;math&gt;a_1=0&lt;/math&gt;.<br /> <br /> Notice that there is respectively &lt;math&gt;1&lt;/math&gt; way to arrive at &lt;math&gt;A&lt;/math&gt; for each of the different routes after the previous &lt;math&gt;n-1&lt;/math&gt; crawls, excluding the possibility that the bug ends up at &lt;math&gt;A&lt;/math&gt; after the &lt;math&gt;(n-1)&lt;/math&gt;th crawl (as it will be forced to move somewhere else.). Thus we get the recurrence relation<br /> &lt;cmath&gt;a_n=3^{n-1}-a_{n-1}&lt;/cmath&gt;<br /> Quick calculations yield<br /> &lt;cmath&gt;\begin{align*}<br /> a_7&amp;=3^6-a_6\\<br /> &amp;=3^6-3^5+3^4-...+a_1\\<br /> &amp;=546<br /> \end{align*}&lt;/cmath&gt;<br /> Thus &lt;math&gt;p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}&lt;/math&gt;.<br /> <br /> ~ Nafer<br /> === Solution 6 ===<br /> Draw an organized table and keep track of how many ways are there to get to each vertex after 1, 2, 3, ..., 7 steps. Thus you get the answer of &lt;math&gt;p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}&lt;/math&gt;.<br /> == See also ==<br /> *[[2003 AIME II Problems/Problem 13]] - very similar problem with an equilateral triangle<br /> <br /> {{AIME box|year=1985|num-b=11|num-a=13}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&diff=114008 1997 AIME Problems/Problem 8 2020-01-02T06:55:30Z <p>Pi is 3.141: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> How many different &lt;math&gt;4\times 4&lt;/math&gt; arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> :''For more detailed explanations, see related [[2007 AIME I Problems/Problem 10|problem (AIME I 2007, 10)]].''<br /> The problem is asking us for all configurations of &lt;math&gt;4\times 4&lt;/math&gt; grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:<br /> <br /> *The first two columns share no two numbers in the same row. There are &lt;math&gt;{4\choose2} = 6&lt;/math&gt; ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are &lt;math&gt;{4\choose 2}&lt;/math&gt; ways. This gives &lt;math&gt;6^2 = 36&lt;/math&gt;.<br /> *The first two columns share one number in the same row. There are &lt;math&gt;{4\choose 1} = 4&lt;/math&gt; ways to pick the position of the shared 1, then &lt;math&gt;{3\choose 2} = 3&lt;/math&gt; ways to pick the locations for the next two 1s, and then &lt;math&gt;2&lt;/math&gt; ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in &lt;math&gt;2&lt;/math&gt; ways. This gives &lt;math&gt;4 \cdot 3 \cdot 2 \cdot 2 = 48&lt;/math&gt;.<br /> *The first two columns share two numbers in the same row. There are &lt;math&gt;{4\choose 2} = 6&lt;/math&gt; ways to pick the position of the shared 1s. Everything is then fixed.<br /> <br /> Adding these cases up, we get &lt;math&gt;36 + 48 + 6 = \boxed{090}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of &lt;math&gt;6&lt;/math&gt; ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is &lt;math&gt;6*3*(2+1+2)&lt;/math&gt; = &lt;math&gt; \boxed{090}&lt;/math&gt;.<br /> <br /> -pi_is_3.141<br /> <br /> == See also ==<br /> *[[2007 AIME I Problems/Problem 10]] - the same problem, but with a &lt;math&gt;4\times 6&lt;/math&gt;<br /> <br /> {{AIME box|year=1997|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&diff=114007 1997 AIME Problems/Problem 8 2020-01-02T06:54:52Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> How many different &lt;math&gt;4\times 4&lt;/math&gt; arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> :''For more detailed explanations, see related [[2007 AIME I Problems/Problem 10|problem (AIME I 2007, 10)]].''<br /> The problem is asking us for all configurations of &lt;math&gt;4\times 4&lt;/math&gt; grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:<br /> <br /> *The first two columns share no two numbers in the same row. There are &lt;math&gt;{4\choose2} = 6&lt;/math&gt; ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are &lt;math&gt;{4\choose 2}&lt;/math&gt; ways. This gives &lt;math&gt;6^2 = 36&lt;/math&gt;.<br /> *The first two columns share one number in the same row. There are &lt;math&gt;{4\choose 1} = 4&lt;/math&gt; ways to pick the position of the shared 1, then &lt;math&gt;{3\choose 2} = 3&lt;/math&gt; ways to pick the locations for the next two 1s, and then &lt;math&gt;2&lt;/math&gt; ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in &lt;math&gt;2&lt;/math&gt; ways. This gives &lt;math&gt;4 \cdot 3 \cdot 2 \cdot 2 = 48&lt;/math&gt;.<br /> *The first two columns share two numbers in the same row. There are &lt;math&gt;{4\choose 2} = 6&lt;/math&gt; ways to pick the position of the shared 1s. Everything is then fixed.<br /> <br /> Adding these cases up, we get &lt;math&gt;36 + 48 + 6 = \boxed{090}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of &lt;math&gt;6&lt;/math&gt; ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange one 1 and 2 -1's. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is &lt;math&gt;6*3*(2+1+2)&lt;/math&gt; = &lt;math&gt; \boxed{090}&lt;/math&gt;.<br /> <br /> -pi_is_3.141<br /> <br /> == See also ==<br /> *[[2007 AIME I Problems/Problem 10]] - the same problem, but with a &lt;math&gt;4\times 6&lt;/math&gt;<br /> <br /> {{AIME box|year=1997|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&diff=114006 1997 AIME Problems/Problem 8 2020-01-02T06:54:29Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> How many different &lt;math&gt;4\times 4&lt;/math&gt; arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> :''For more detailed explanations, see related [[2007 AIME I Problems/Problem 10|problem (AIME I 2007, 10)]].''<br /> The problem is asking us for all configurations of &lt;math&gt;4\times 4&lt;/math&gt; grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:<br /> <br /> *The first two columns share no two numbers in the same row. There are &lt;math&gt;{4\choose2} = 6&lt;/math&gt; ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are &lt;math&gt;{4\choose 2}&lt;/math&gt; ways. This gives &lt;math&gt;6^2 = 36&lt;/math&gt;.<br /> *The first two columns share one number in the same row. There are &lt;math&gt;{4\choose 1} = 4&lt;/math&gt; ways to pick the position of the shared 1, then &lt;math&gt;{3\choose 2} = 3&lt;/math&gt; ways to pick the locations for the next two 1s, and then &lt;math&gt;2&lt;/math&gt; ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in &lt;math&gt;2&lt;/math&gt; ways. This gives &lt;math&gt;4 \cdot 3 \cdot 2 \cdot 2 = 48&lt;/math&gt;.<br /> *The first two columns share two numbers in the same row. There are &lt;math&gt;{4\choose 2} = 6&lt;/math&gt; ways to pick the position of the shared 1s. Everything is then fixed.<br /> <br /> Adding these cases up, we get &lt;math&gt;36 + 48 + 6 = \boxed{090}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of &lt;math&gt;6&lt;/math&gt; ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange one 1 and 2 -1's. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is &lt;math&gt;6*3*(2+1+2)&lt;/math&gt; = &lt;math&gt; \boxed{090}&lt;/math&gt;.<br /> <br /> == See also ==<br /> *[[2007 AIME I Problems/Problem 10]] - the same problem, but with a &lt;math&gt;4\times 6&lt;/math&gt;<br /> <br /> {{AIME box|year=1997|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_46&diff=113102 1954 AHSME Problems/Problem 46 2019-12-20T22:51:22Z <p>Pi is 3.141: Created page with &quot;E&quot;</p> <hr /> <div>E</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=113065 1990 AIME Problems/Problem 15 2019-12-19T02:57:37Z <p>Pi is 3.141: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt; Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+sqrt(87),-7-sqrt(87))&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us k= &lt;math&gt;\boxed{020}&lt;/math&gt;.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=113064 1990 AIME Problems/Problem 15 2019-12-19T02:57:20Z <p>Pi is 3.141: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt; Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+sqrt(87),-7-sqrt(87))&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us k= boxed{020}.<br /> <br /> pi_is_3.141<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_15&diff=113063 1990 AIME Problems/Problem 15 2019-12-19T02:56:58Z <p>Pi is 3.141: </p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a_{}^{}x^5 + b_{}y^5&lt;/math&gt; if the [[real number]]s &lt;math&gt;a_{}^{}&lt;/math&gt;, &lt;math&gt;b_{}^{}&lt;/math&gt;, &lt;math&gt;x_{}^{}&lt;/math&gt;, and &lt;math&gt;y_{}^{}&lt;/math&gt; satisfy the [[equation]]s<br /> &lt;cmath&gt;ax + by = 3^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^2 + by^2 = 7^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^3 + by^3 = 16^{}_{},&lt;/cmath&gt;<br /> &lt;cmath&gt;ax^4 + by^4 = 42^{}_{}.&lt;/cmath&gt;<br /> <br /> == Solution 1 ==<br /> Set &lt;math&gt;S = (x + y)&lt;/math&gt; and &lt;math&gt;P = xy&lt;/math&gt;. Then the relationship<br /> <br /> &lt;cmath&gt;(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})&lt;/cmath&gt;<br /> <br /> can be exploited:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^2 + by^2)(x + y) &amp; = &amp; (ax^3 + by^3) + (xy)(ax + by) \\<br /> (ax^3 + by^3)(x + y) &amp; = &amp; (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Therefore:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}7S &amp; = &amp; 16 + 3P \\<br /> 16S &amp; = &amp; 42 + 7P\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> Consequently, &lt;math&gt;S = - 14&lt;/math&gt; and &lt;math&gt;P = - 38&lt;/math&gt;. Finally:<br /> <br /> &lt;cmath&gt;\begin{eqnarray*}(ax^4 + by^4)(x + y) &amp; = &amp; (ax^5 + by^5) + (xy)(ax^3 + by^3) \\<br /> (42)(S) &amp; = &amp; (ax^5 + by^5) + (P)(16) \\<br /> (42)( - 14) &amp; = &amp; (ax^5 + by^5) + ( - 38)(16) \\<br /> ax^5 + by^5 &amp; = &amp; \boxed{020}\end{eqnarray*}&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> <br /> A [[linear recurrence | recurrence]] of the form &lt;math&gt;T_n=AT_{n-1}+BT_{n-2}&lt;/math&gt; will have the closed form &lt;math&gt;T_n=ax^n+by^n&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are the values of the starting term that make the sequence geometric, and &lt;math&gt;a,b&lt;/math&gt; are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.<br /> <br /> Suppose we have such a recurrence with &lt;math&gt;T_1=3&lt;/math&gt; and &lt;math&gt;T_2=7&lt;/math&gt;. Then &lt;math&gt;T_3=ax^3+by^3=16=7A+3B&lt;/math&gt;, and &lt;math&gt;T_4=ax^4+by^4=42=16A+7B&lt;/math&gt;. <br /> <br /> Solving these simultaneous equations for &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, we see that &lt;math&gt;A=-14&lt;/math&gt; and &lt;math&gt;B=38&lt;/math&gt;. So, &lt;math&gt;ax^5+by^5=T_5=-14(42)+38(16)= \boxed{020}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:<br /> <br /> &lt;math&gt;a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16&lt;/math&gt;.<br /> <br /> Similarly take the first two terms, yielding:<br /> <br /> &lt;math&gt;ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10&lt;/math&gt;.<br /> <br /> Lastly take an alternating three-term sum,<br /> <br /> &lt;math&gt;a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12&lt;/math&gt;<br /> &lt;math&gt;ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12&lt;/math&gt;.<br /> <br /> Now to get the solution, let the answer be &lt;math&gt;k&lt;/math&gt;, so <br /> <br /> &lt;math&gt;ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68&lt;/math&gt;.<br /> <br /> Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated<br /> as done in the first solution.<br /> <br /> == Solution 4 ==<br /> We first let the answer to this problem be k. Multiplying the first equation by &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt;ax^2 + bxy=3x&lt;/math&gt; Subtracting this equation from the second equation gives &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;. Similarily, doing the same for the other equations, we obtain:<br /> &lt;math&gt;by^2-bxy=7-3x&lt;/math&gt;, &lt;math&gt;by^3-bxy^2=16-7x&lt;/math&gt;, &lt;math&gt;by^4-bxy^3=42-16x&lt;/math&gt;, and &lt;math&gt;by^5-bxy^4=k-42x&lt;/math&gt;<br /> Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us &lt;math&gt;by^3-bxy^2=(7-3x)y&lt;/math&gt;. We can also obtain &lt;math&gt;by^4-bxy^3=(16-7x)y&lt;/math&gt;. <br /> Now we can solve for x and y! &lt;math&gt;(7-3x)y = 16-7x&lt;/math&gt; and &lt;math&gt;(16-7x)y=42-16x&lt;/math&gt;. Solving for x and y gives us &lt;math&gt;(-7+sqrt(87),-7-sqrt(87))&lt;/math&gt;.(It can be switched, but since the given equations are symmetric, it doesn't matter). &lt;math&gt;k-42x= (42-16x)y&lt;/math&gt;, and solving for k gives us k=\boxed{020}\$.<br /> <br /> pi_is_3.141<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pi is 3.141