https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hi13&feedformat=atom AoPS Wiki - User contributions [en] 2020-10-25T00:13:31Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_12&diff=130218 2003 AIME II Problems/Problem 12 2020-08-02T03:52:03Z <p>Hi13: /* Solution */</p> <hr /> <div>== Problem ==<br /> The members of a distinguished committee were choosing a president, and each member gave one vote to one of the 27 candidates. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What was the smallest possible number of members of the committee?<br /> <br /> == Solution ==<br /> Let &lt;math&gt;v_i&lt;/math&gt; be the number of votes candidate &lt;math&gt;i&lt;/math&gt; received, and let &lt;math&gt;s=v_1+\cdots+v_{27}&lt;/math&gt; be the total number of votes cast. Our goal is to determine the smallest possible &lt;math&gt;s&lt;/math&gt;.<br /> <br /> Candidate &lt;math&gt;i&lt;/math&gt; got &lt;math&gt;\frac{v_i}s&lt;/math&gt; of the votes, hence the percentage of votes they received is &lt;math&gt;\frac{100v_i}s&lt;/math&gt;. The condition in the problem statement says that &lt;math&gt;\forall i: \frac{100v_i}s + 1 \leq v_i&lt;/math&gt;. <br /> <br /> Obviously, if some &lt;math&gt;v_i&lt;/math&gt; would be &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;1&lt;/math&gt;, the condition would be false. Thus &lt;math&gt;\forall i: v_i\geq 2&lt;/math&gt;. We can then rewrite the above inequality as &lt;math&gt;\forall i: s\geq\frac{100v_i}{v_i-1}&lt;/math&gt;. <br /> <br /> If for some &lt;math&gt;i&lt;/math&gt; we have &lt;math&gt;v_i=2&lt;/math&gt;, then from the inequality we just derived we would have &lt;math&gt;s\geq 200&lt;/math&gt;. If for some &lt;math&gt;i&lt;/math&gt; we have &lt;math&gt;v_i=3&lt;/math&gt;, then &lt;math&gt;s\geq 150&lt;/math&gt;. And if for some &lt;math&gt;i&lt;/math&gt; we have &lt;math&gt;v_i=4&lt;/math&gt;, then &lt;math&gt;s\geq \frac{400}3 = 133\frac13&lt;/math&gt;, and hence &lt;math&gt;s\geq 134&lt;/math&gt;.<br /> <br /> Is it possible to have &lt;math&gt;s&lt;134&lt;/math&gt;? We just proved that to have such &lt;math&gt;s&lt;/math&gt;, all &lt;math&gt;v_i&lt;/math&gt; have to be at least &lt;math&gt;5&lt;/math&gt;. But then &lt;math&gt;s=v_1+\cdots+v_{27}\geq 27\cdot 5 = 135&lt;/math&gt;, which is a contradiction. Hence the smallest possible &lt;math&gt;s&lt;/math&gt; is at least &lt;math&gt;134&lt;/math&gt;.<br /> <br /> Now consider a situation where &lt;math&gt;26&lt;/math&gt; candidates got &lt;math&gt;5&lt;/math&gt; votes each, and one candidate got &lt;math&gt;4&lt;/math&gt; votes. In this situation, the total number of votes is exactly &lt;math&gt;134&lt;/math&gt;, and for each candidate the above inequality is satisfied. Hence the minimum number of committee members is &lt;math&gt;s=\boxed{134}&lt;/math&gt;.<br /> <br /> Note: Each of the &lt;math&gt;26&lt;/math&gt; candidates received &lt;math&gt;\simeq 3.63\%&lt;/math&gt; votes, and the last candidate received &lt;math&gt;\simeq 2.985\%&lt;/math&gt; votes.<br /> <br /> == Solution 2 ==<br /> Let there be &lt;math&gt;N&lt;/math&gt; members of the committee.<br /> Suppose candidate &lt;math&gt;n&lt;/math&gt; gets &lt;math&gt;a_n&lt;/math&gt; votes.<br /> Then &lt;math&gt;a_n&lt;/math&gt; as a percentage out of &lt;math&gt;N&lt;/math&gt; is <br /> &lt;math&gt;100\frac{a_n}{N}&lt;/math&gt;. Setting up the inequality <br /> &lt;math&gt;a_n \geq 1 + 100\frac{a_n}{N}&lt;/math&gt; and simplifying,<br /> &lt;math&gt;a_n \geq \lceil(\frac{N}{N - 100})\rceil&lt;/math&gt; (the ceiling function is there because &lt;math&gt;a_n&lt;/math&gt; is an integer.<br /> Note that if we set all &lt;math&gt;a_i&lt;/math&gt; equal to &lt;math&gt;\lceil(\frac{N}{100 - N})\rceil&lt;/math&gt; we have &lt;math&gt;N \geq 27\lceil(\frac{N}{100 - N})\rceil&lt;/math&gt;. Clearly &lt;math&gt;N = 134&lt;/math&gt; is the least such number that satisfies this inequality. Now we must show that we can find suitable &lt;math&gt;a_i&lt;/math&gt;. We can let 26 of them equal to &lt;math&gt;5&lt;/math&gt; and one of them equal to &lt;math&gt;4&lt;/math&gt;. Therefore, &lt;math&gt;N = \boxed{134}&lt;/math&gt; is the answer.<br /> - whatRthose<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_13&diff=129322 2008 AIME II Problems/Problem 13 2020-07-26T04:59:19Z <p>Hi13: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[regular polygon|regular]] [[hexagon]] with center at the [[origin]] in the [[complex plane]] has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let &lt;math&gt;R&lt;/math&gt; be the region outside the hexagon, and let &lt;math&gt;S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace&lt;/math&gt;. Then the area of &lt;math&gt;S&lt;/math&gt; has the form &lt;math&gt;a\pi + \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 /> If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.<br /> <br /> == Solution 2==<br /> If a point &lt;math&gt;z = r\text{cis}\,\theta&lt;/math&gt; is in &lt;math&gt;R&lt;/math&gt;, then the point &lt;math&gt;\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; (where [[cis]] denotes &lt;math&gt;\text{cis}\, \theta = \cos \theta + i \sin \theta&lt;/math&gt;). Since &lt;math&gt;R&lt;/math&gt; is symmetric every &lt;math&gt;60^{\circ}&lt;/math&gt; about the origin, it suffices to consider the area of the result of the transformation when &lt;math&gt;-30 \le \theta \le 30&lt;/math&gt;, and then to multiply by &lt;math&gt;6&lt;/math&gt; to account for the entire area.<br /> <br /> We note that if the region &lt;math&gt;S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace&lt;/math&gt;, where &lt;math&gt;R_2&lt;/math&gt; is the region (in green below) outside the circle of radius &lt;math&gt;1/\sqrt{3}&lt;/math&gt; centered at the origin, then &lt;math&gt;S_2&lt;/math&gt; is simply the region inside a circle of radius &lt;math&gt;\sqrt{3}&lt;/math&gt; centered at the origin. It now suffices to find what happens to the mapping of the region &lt;math&gt;R-R_2&lt;/math&gt; (in blue below). <br /> <br /> The equation of the hexagon side in that region is &lt;math&gt;x = r \cos \theta = \frac{1}{2}&lt;/math&gt;, which is transformed to &lt;math&gt;\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta = &lt;/math&gt;2 . Let &lt;math&gt;r\text{cis}\,\theta = a+bi&lt;/math&gt; where &lt;math&gt;a,b \in \mathbb{R}&lt;/math&gt;; then &lt;math&gt;r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}&lt;/math&gt;, so the equation becomes &lt;math&gt;a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1&lt;/math&gt;. Hence the side is sent to an arc of the unit circle centered at &lt;math&gt;(1,0)&lt;/math&gt;, after considering the restriction that the side of the hexagon is a segment of length &lt;math&gt;1/\sqrt{3}&lt;/math&gt;. <br /> <br /> Including &lt;math&gt;S_2&lt;/math&gt;, we find that &lt;math&gt;S&lt;/math&gt; is the union of six unit circles centered at &lt;math&gt;\text{cis}\, \frac{k\pi}{6}&lt;/math&gt;, &lt;math&gt;k = 0,1,2,3,4,5&lt;/math&gt;, as shown below. <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(&quot;4 4&quot;); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i &lt; 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(&quot;$1/\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8)); <br /> &lt;/asy&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;math&gt;\Longrightarrow&lt;/math&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;asy&gt;<br /> defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1));<br /> draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(&quot;4 4&quot;)); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(&quot;4 4&quot;)); draw(Circle((0,0),1),linetype(&quot;4 4&quot;)); label(&quot;$\sqrt{3}$&quot;,(0,-0.5),W,fontsize(8));<br /> add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> The area of the regular hexagon is &lt;math&gt;6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}&lt;/math&gt;. The total area of the six &lt;math&gt;120^{\circ}&lt;/math&gt; sectors is &lt;math&gt;6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}&lt;/math&gt;. Their sum is &lt;math&gt;2\pi + \sqrt{27}&lt;/math&gt;, and &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> - Th3Numb3rThr33<br /> <br /> == Solution 3 (Calculus) ==<br /> One can describe the line parallel to the imaginary axis &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; using polar coordinates as &lt;math&gt;r(\theta)=\dfrac{1}{2\cos{\theta}}&lt;/math&gt;<br /> <br /> so &lt;math&gt;z&lt;/math&gt; is equal to &lt;math&gt;z=(\dfrac{1}{2\cos{\theta}})(cis{\theta})<br /> \rightarrow \frac{1}{z}=2\cos{\theta}cis(-\theta)&lt;/math&gt;<br /> <br /> Dividing the hexagon to 12 equal parts we get that <br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta&lt;/math&gt;<br /> <br /> which is a routine computation:<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta&lt;/math&gt;<br /> <br /> &lt;math&gt;Area = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12[\frac{1}{2}\sin{2\theta}+\theta]_0^{\frac{\pi}{6}}=12(\frac{\sqrt{3}}{4}+\frac{\pi}{6})=2\pi+3\sqrt{3}=2\pi + \sqrt{27}&lt;/math&gt;<br /> <br /> &lt;math&gt;a+b = \boxed{029}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_12&diff=128770 2013 AIME II Problems/Problem 12 2020-07-20T23:52:20Z <p>Hi13: /* Solution */</p> <hr /> <div>==Problem 12==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of all polynomials of the form &lt;math&gt;z^3 + az^2 + bz + c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers. Find the number of polynomials in &lt;math&gt;S&lt;/math&gt; such that each of its roots &lt;math&gt;z&lt;/math&gt; satisfies either &lt;math&gt;|z| = 20&lt;/math&gt; or &lt;math&gt;|z| = 13&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from [[Vieta's formulas]]. <br /> <br /> *Case 1: &lt;math&gt;f(z)=(z-r)(z-\omega)(z-\omega^*)&lt;/math&gt;, where &lt;math&gt;r\in \mathbb{R}&lt;/math&gt;, &lt;math&gt;\omega&lt;/math&gt; is nonreal, and &lt;math&gt;\omega^*&lt;/math&gt; is the complex conjugate of omega (note that we may assume that &lt;math&gt;\Im(\omega)&gt;0&lt;/math&gt;).<br /> <br /> The real root &lt;math&gt;r&lt;/math&gt; must be one of &lt;math&gt;-20&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt;, &lt;math&gt;-13&lt;/math&gt;, or &lt;math&gt;13&lt;/math&gt;. By Viète's formulas, &lt;math&gt;a=-(r+\omega+\omega^*)&lt;/math&gt;, &lt;math&gt;b=|\omega|^2+r(\omega+\omega^*)&lt;/math&gt;, and &lt;math&gt;c=-r|\omega|^2&lt;/math&gt;. But &lt;math&gt;\omega+\omega^*=2\Re{(\omega)}&lt;/math&gt; (i.e., adding the conjugates cancels the imaginary part). Therefore, to make &lt;math&gt;a&lt;/math&gt; an integer, &lt;math&gt;2\Re{(\omega)}&lt;/math&gt; must be an integer. Conversely, if &lt;math&gt;\omega+\omega^*=2\Re{(\omega)}&lt;/math&gt; is an integer, then &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are clearly integers. Therefore &lt;math&gt;2\Re{(\omega)}\in \mathbb{Z}&lt;/math&gt; is equivalent to the desired property. Let &lt;math&gt;\omega=\alpha+i\beta&lt;/math&gt;.<br /> <br /> *Subcase 1.1: &lt;math&gt;|\omega|=20&lt;/math&gt;.<br /> In this case, &lt;math&gt;\omega&lt;/math&gt; lies on a circle of radius &lt;math&gt;20&lt;/math&gt; in the complex plane. As &lt;math&gt;\omega&lt;/math&gt; is nonreal, we see that &lt;math&gt;\beta\ne 0&lt;/math&gt;. Hence &lt;math&gt;-20&lt;\Re{(\omega)}&lt; 20&lt;/math&gt;, or rather &lt;math&gt;-40&lt;2\Re{(\omega)}&lt; 40&lt;/math&gt;. We count &lt;math&gt;79&lt;/math&gt; integers in this interval, each of which corresponds to a unique complex number on the circle of radius &lt;math&gt;20&lt;/math&gt; with positive imaginary part.<br /> <br /> *Subcase 1.2: &lt;math&gt;|\omega|=13&lt;/math&gt;.<br /> In this case, &lt;math&gt;\omega&lt;/math&gt; lies on a circle of radius &lt;math&gt;13&lt;/math&gt; in the complex plane. As &lt;math&gt;\omega&lt;/math&gt; is nonreal, we see that &lt;math&gt;\beta\ne 0&lt;/math&gt;. Hence &lt;math&gt;-13&lt;\Re{(\omega)}&lt; 13&lt;/math&gt;, or rather &lt;math&gt;-26&lt;2\Re{(\omega)}&lt; 26&lt;/math&gt;. We count &lt;math&gt;51&lt;/math&gt; integers in this interval, each of which corresponds to a unique complex number on the circle of radius &lt;math&gt;13&lt;/math&gt; with positive imaginary part.<br /> <br /> Therefore, there are &lt;math&gt;79+51=130&lt;/math&gt; choices for &lt;math&gt;\omega&lt;/math&gt;. We also have &lt;math&gt;4&lt;/math&gt; choices for &lt;math&gt;r&lt;/math&gt;, hence there are &lt;math&gt;4\cdot 130=520&lt;/math&gt; total polynomials in this case.<br /> <br /> *Case 2: &lt;math&gt;f(z)=(z-r_1)(z-r_2)(z-r_3)&lt;/math&gt;, where &lt;math&gt;r_1,r_2,r_3&lt;/math&gt; are all real.<br /> In this case, there are four possible real roots, namely &lt;math&gt;\pm 13, \pm20&lt;/math&gt;. Let &lt;math&gt;p&lt;/math&gt; be the number of times that &lt;math&gt;13&lt;/math&gt; appears among &lt;math&gt;r_1,r_2,r_3&lt;/math&gt;, and define &lt;math&gt;q,r,s&lt;/math&gt; similarly for &lt;math&gt;-13,20&lt;/math&gt;, and &lt;math&gt;-20&lt;/math&gt;, respectively. Then &lt;math&gt;p+q+r+s=3&lt;/math&gt; because there are three roots. We wish to find the number of ways to choose nonnegative integers &lt;math&gt;p,q,r,s&lt;/math&gt; that satisfy that equation. By balls and urns, these can be chosen in &lt;math&gt;\binom{6}{3}=20&lt;/math&gt; ways.<br /> <br /> Therefore, there are a total of &lt;math&gt;520+20=\boxed{540}&lt;/math&gt; polynomials with the desired property.<br /> <br /> ==Solution Systematics==<br /> This combinatorics problem involves counting, and casework is most appropriate.<br /> There are two cases: either all three roots are real, or one is real and there are two imaginary roots.<br /> <br /> Case 1: Three roots are of the set &lt;math&gt;{13, -13, 20, -20}&lt;/math&gt;. By stars and bars, there is &lt;math&gt;\binom{6}{3}=20&lt;/math&gt; ways (3 bars between all four possibilities, and then 3 stars that represent the roots themselves).<br /> <br /> Case 2: One real root: one of &lt;math&gt;13, -13, 20, -20&lt;/math&gt;. Then two imaginary roots left; it is well known that because coefficients of the polynomial are integral (and thus not imaginary), these roots are conjugates. Therefore, either both roots have a norm (also called magnitude) of &lt;math&gt;20&lt;/math&gt; or &lt;math&gt;13&lt;/math&gt;. Call the root &lt;math&gt;a+bi&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; is not the magnitude of the root; otherwise, it would be case 1. We need integral coefficients: expansion of &lt;math&gt;(x-(a+bi))(x-(a-bi))=-2ax+x^2+(a^2+b^2)&lt;/math&gt; tells us that we just need &lt;math&gt;2a&lt;/math&gt; to be integral, because &lt;math&gt;a^2+b^2&lt;/math&gt; IS the norm of the root! (Note that it is not necessary to multiply by the real root. That won't affect whether or not a coefficient is imaginary.) <br /> Therefore, when the norm is &lt;math&gt;20&lt;/math&gt;, the &lt;math&gt;a&lt;/math&gt; term can range from &lt;math&gt;-19.5, -19, ...., 0, 0.5, ..., 19.5&lt;/math&gt; or &lt;math&gt;79&lt;/math&gt; solutions. When the norm is &lt;math&gt;13&lt;/math&gt;, the &lt;math&gt;a&lt;/math&gt; term has &lt;math&gt;51&lt;/math&gt; possibilities from &lt;math&gt;-12.5, -12, ..., 12.5&lt;/math&gt;. In total that's 130 total ways to choose the imaginary root. Now, multiply by the ways to choose the real root, &lt;math&gt;4&lt;/math&gt;, and you get &lt;math&gt;520&lt;/math&gt; for this case.<br /> <br /> And &lt;math&gt;520+20=540&lt;/math&gt; and we are done.<br /> <br /> ==Comments==<br /> If the polynomial has one real root and two complex roots, then it can be factored as &lt;math&gt;(z-r)(z^2+pz+q), &lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is real with &lt;math&gt;|r|=13,20&lt;/math&gt; and &lt;math&gt;p,q&lt;/math&gt; are integers with &lt;math&gt;p^2 &lt;4q.&lt;/math&gt; The roots &lt;math&gt;z_1&lt;/math&gt; and &lt;math&gt;z_2&lt;/math&gt; are conjugates. We have &lt;math&gt;|z_1|^2=|z_2|^2=z_1z_2=q.&lt;/math&gt; So &lt;math&gt;q&lt;/math&gt; is either &lt;math&gt;20^2&lt;/math&gt; or &lt;math&gt;13^2&lt;/math&gt;. The only requirement for &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;p&lt;\sqrt{4q^2}=2\sqrt{q}.&lt;/math&gt; All such quadratic equations are listed as follows:<br /> <br /> &lt;math&gt;z^2+pz+20^2,&lt;/math&gt; where &lt;math&gt;p=0,\pm1,\pm2,\cdots,\pm 39,&lt;/math&gt;<br /> <br /> &lt;math&gt;z^2+pz+13^2,&lt;/math&gt; where &lt;math&gt;p=0,\pm1,\pm2,\cdots,\pm 25&lt;/math&gt;.<br /> <br /> Total of 130 equations, multiplied by 4 (the number of cases for real &lt;math&gt;r&lt;/math&gt;, we have 520 equations, as indicated in the solution.<br /> <br /> -JZ<br /> <br /> ==See Also==<br /> {{AIME box|year=2013|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_10&diff=128761 2016 AIME II Problems/Problem 10 2020-07-20T20:38:09Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\omega&lt;/math&gt;. Points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; are on side &lt;math&gt;\overline{AB}&lt;/math&gt; with &lt;math&gt;AP&lt;AQ&lt;/math&gt;. Rays &lt;math&gt;CP&lt;/math&gt; and &lt;math&gt;CQ&lt;/math&gt; meet &lt;math&gt;\omega&lt;/math&gt; again at &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (other than &lt;math&gt;C&lt;/math&gt;), respectively. If &lt;math&gt;AP=4,PQ=3,QB=6,BT=5,&lt;/math&gt; and &lt;math&gt;AS=7&lt;/math&gt;, then &lt;math&gt;ST=\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> &lt;asy&gt;<br /> import cse5;<br /> pathpen = black; pointpen = black;<br /> pointfontsize = 9;<br /> size(8cm);<br /> <br /> pair A = origin, B = (13,0), P = (4,0), Q = (7,0),<br /> T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)),<br /> S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10));<br /> <br /> Drawing(A--B--C--cycle);<br /> D(circumcircle(A,B,C),rgb(0,0.6,1));<br /> DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1));<br /> DrawPathArray(A--S^^B--T,rgb(0,0.4,0));<br /> D(S--T,rgb(1,0.2,0.4));<br /> <br /> D(&quot;A&quot;,A,dir(215));<br /> D(&quot;B&quot;,B,dir(330));<br /> D(&quot;P&quot;,P,dir(240));<br /> D(&quot;Q&quot;,Q,dir(240));<br /> D(&quot;T&quot;,T,dir(290));<br /> D(&quot;C&quot;,C,dir(120));<br /> D(&quot;S&quot;,S,dir(250));<br /> <br /> MP(&quot;4&quot;,(A+P)/2,dir(90));<br /> MP(&quot;3&quot;,(P+Q)/2,dir(90));<br /> MP(&quot;6&quot;,(Q+B)/2,dir(90));<br /> MP(&quot;5&quot;,(B+T)/2,dir(140));<br /> MP(&quot;7&quot;,(A+S)/2,dir(40));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;\angle ACP=\alpha&lt;/math&gt;, &lt;math&gt;\angle PCQ=\beta&lt;/math&gt;, and &lt;math&gt;\angle QCB=\gamma&lt;/math&gt;. Note that since &lt;math&gt;\triangle ACQ\sim\triangle TBQ&lt;/math&gt; we have &lt;math&gt;\tfrac{AC}{CQ}=\tfrac56&lt;/math&gt;, so by the Ratio Lemma &lt;cmath&gt;\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.&lt;/cmath&gt;Similarly, we can deduce &lt;math&gt;\tfrac{PC}{CB}=\tfrac47&lt;/math&gt; and hence &lt;math&gt;\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}&lt;/math&gt;.<br /> <br /> Now Law of Sines on &lt;math&gt;\triangle ACS&lt;/math&gt;, &lt;math&gt;\triangle SCT&lt;/math&gt;, and &lt;math&gt;\triangle TCB&lt;/math&gt; yields &lt;cmath&gt;\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.&lt;/cmath&gt;Hence &lt;cmath&gt;\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},&lt;/cmath&gt;so &lt;cmath&gt;TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.&lt;/cmath&gt;Hence &lt;math&gt;ST=\tfrac{35}8&lt;/math&gt; and the requested answer is &lt;math&gt;35+8=\boxed{43}&lt;/math&gt;.<br /> <br /> Edit: Note that the finish is much simpler. Once you get &lt;math&gt;\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}&lt;/math&gt;, you can solve quickly from there getting &lt;math&gt;ST=\dfrac{AS \sin(\beta)}{\sin(\alpha)}=7\cdot \dfrac{15}{24}=\dfrac{35}{8}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Projective Geometry)==<br /> Projecting through &lt;math&gt;C&lt;/math&gt; we have &lt;cmath&gt;\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}&lt;/cmath&gt; which easily gives &lt;math&gt;ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{43.}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> By Ptolemy's Theorem applied to quadrilateral &lt;math&gt;ASTB&lt;/math&gt;, we find<br /> &lt;cmath&gt;5\cdot 7+13\cdot ST=AT\cdot BS.&lt;/cmath&gt;<br /> Therefore, in order to find &lt;math&gt;ST&lt;/math&gt;, it suffices to find &lt;math&gt;AT\cdot BS&lt;/math&gt;. We do this using similar triangles, which can be found by using Power of a Point theorem.<br /> <br /> As &lt;math&gt;\triangle APS\sim \triangle CPB&lt;/math&gt;, we find<br /> &lt;cmath&gt;\frac{4}{PC}=\frac{7}{BC}.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\frac{BC}{PC}=\frac{7}{4}&lt;/math&gt;.<br /> <br /> As &lt;math&gt;\triangle BQT\sim\triangle CQA&lt;/math&gt;, we find<br /> &lt;cmath&gt;\frac{6}{CQ}=\frac{5}{AC}.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;\frac{AC}{CQ}=\frac{5}{6}&lt;/math&gt;.<br /> <br /> As &lt;math&gt;\triangle ATQ\sim\triangle CBQ&lt;/math&gt;, we find<br /> &lt;cmath&gt;\frac{AT}{BC}=\frac{7}{CQ}.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;AT=\frac{7\cdot BC}{CQ}&lt;/math&gt;.<br /> <br /> As &lt;math&gt;\triangle BPS\sim \triangle CPA&lt;/math&gt;, we find<br /> &lt;cmath&gt;\frac{9}{PC}=\frac{BS}{AC}.&lt;/cmath&gt;<br /> Therefore, &lt;math&gt;BS=\frac{9\cdot AC}{PC}&lt;/math&gt;. Thus we find<br /> &lt;cmath&gt;AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).&lt;/cmath&gt;<br /> But now we can substitute in our previously found values for &lt;math&gt;\frac{BC}{PC}&lt;/math&gt; and &lt;math&gt;\frac{AC}{CQ}&lt;/math&gt;, finding<br /> &lt;cmath&gt;AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.&lt;/cmath&gt;<br /> Substituting this into our original expression from Ptolemy's Theorem, we find<br /> &lt;cmath&gt;\begin{align*}35+13ST&amp;=\frac{21\cdot 35}{8}\\13ST&amp;=\frac{13\cdot 35}{8}\\ST&amp;=\frac{35}{8}.\end{align*}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;\boxed{43}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_12&diff=128734 2014 AIME II Problems/Problem 12 2020-07-20T04:36:52Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Suppose that the angles of &lt;math&gt;\triangle ABC&lt;/math&gt; satisfy &lt;math&gt;\cos(3A)+\cos(3B)+\cos(3C)=1.&lt;/math&gt; Two sides of the triangle have lengths 10 and 13. There is a positive integer &lt;math&gt;m&lt;/math&gt; so that the maximum possible length for the remaining side of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\sqrt{m}.&lt;/math&gt; Find &lt;math&gt;m.&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Note that &lt;math&gt;\cos{3C}=-\cos{(3A+3B)}&lt;/math&gt;. Thus, our expression is of the form &lt;math&gt;\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1&lt;/math&gt;. Let &lt;math&gt;\cos{3A}=x&lt;/math&gt; and &lt;math&gt;\cos{3B}=y&lt;/math&gt;.<br /> <br /> Using the fact that &lt;math&gt;\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}&lt;/math&gt;, we get &lt;math&gt;x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1&lt;/math&gt;, or &lt;math&gt;\sqrt{1-x^2}\sqrt{1-y^2}=xy-x-y+1=(x-1)(y-1)&lt;/math&gt;. <br /> <br /> Squaring both sides, we get &lt;math&gt;(1-x^2)(1-y^2) = [(x-1)(y-1)]^2&lt;/math&gt;. Cancelling factors, &lt;math&gt;(1+x)(1+y) = (1-x)(1-y)&lt;/math&gt;.<br /> <br /> *Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have <br /> <br /> &lt;math&gt;cos(3A)-1=0, cos(3A)=1&lt;/math&gt; <br /> <br /> For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side).<br /> <br /> Expanding, &lt;math&gt;1+x+y+xy=1-x-y+xy\rightarrow x+y=-x-y&lt;/math&gt;. <br /> <br /> Simplification leads to &lt;math&gt;x+y=0&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\cos(3C)=1&lt;/math&gt;. So &lt;math&gt;\angle C&lt;/math&gt; could be &lt;math&gt;0^\circ&lt;/math&gt; or &lt;math&gt;120^\circ&lt;/math&gt;. We eliminate &lt;math&gt;0^\circ&lt;/math&gt; and use law of cosines to get our answer: <br /> <br /> &lt;cmath&gt;m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C&lt;/cmath&gt;<br /> &lt;cmath&gt;\rightarrow m=269-260\cos 120^\circ=269-260\left(\text{-}\frac{1}{2}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;\rightarrow m=269+130=399&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\framebox{399}&lt;/math&gt;<br /> <br /> NOTE: This solution forgot the case of dividing by 0<br /> <br /> NOTE FROM DIFFERENT PERSON: Pretty sure they addressed that in the notice<br /> <br /> ==Solution 2==<br /> As above, we can see that &lt;math&gt;\cos3A+\cos3B-\cos(3A+3B)=1&lt;/math&gt;<br /> <br /> Expanding, we get<br /> <br /> &lt;math&gt;\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B&lt;/math&gt;<br /> <br /> &lt;math&gt;(\cos3A-1)(\cos3B-1)=\sin3A\sin3B&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\tan{x}=\frac{1}{\tan(90-x)}&lt;/math&gt;, or &lt;math&gt;\tan{x}\tan(90-x)=1&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{3A}{2}+\frac{3B}{2}=90&lt;/math&gt;, or &lt;math&gt;A+B=60&lt;/math&gt;.<br /> <br /> Now we know that &lt;math&gt;C=120&lt;/math&gt;, so we can just use the Law of Cosines to get &lt;math&gt;\boxed{399}&lt;/math&gt;<br /> <br /> Note: This solution also forgets that &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; might be 120 when dividing by &lt;math&gt;\sin 3A&lt;/math&gt; and &lt;math&gt;\sin 3B&lt;/math&gt;<br /> ==Solution 3==<br /> &lt;cmath&gt;\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)&lt;/cmath&gt;<br /> &lt;cmath&gt;2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)&lt;/cmath&gt;<br /> If &lt;math&gt;\cos\frac{3}{2}(A+B) = 0&lt;/math&gt;, then &lt;math&gt;\frac{3}{2}(A+B)=90&lt;/math&gt;, &lt;math&gt;A+B=60&lt;/math&gt;, so &lt;math&gt;C=120&lt;/math&gt;; otherwise,<br /> &lt;cmath&gt;2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin\frac{3}{2}A\sin\frac{3}{2}B=0&lt;/cmath&gt;<br /> so either &lt;math&gt;\sin\frac{3}{2}A=0&lt;/math&gt; or &lt;math&gt;\sin\frac{3}{2}B=0&lt;/math&gt;, i.e., either &lt;math&gt;A=120&lt;/math&gt; or &lt;math&gt;B=120&lt;/math&gt;. In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. &lt;math&gt;\boxed{399}&lt;/math&gt;<br /> <br /> -Mathdummy<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_12&diff=128733 2014 AIME II Problems/Problem 12 2020-07-20T04:34:16Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Suppose that the angles of &lt;math&gt;\triangle ABC&lt;/math&gt; satisfy &lt;math&gt;\cos(3A)+\cos(3B)+\cos(3C)=1.&lt;/math&gt; Two sides of the triangle have lengths 10 and 13. There is a positive integer &lt;math&gt;m&lt;/math&gt; so that the maximum possible length for the remaining side of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;\sqrt{m}.&lt;/math&gt; Find &lt;math&gt;m.&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Note that &lt;math&gt;\cos{3C}=-\cos{(3A+3B)}&lt;/math&gt;. Thus, our expression is of the form &lt;math&gt;\cos{3A}+\cos{3B}-\cos{(3A+3B)}=1&lt;/math&gt;. Let &lt;math&gt;\cos{3A}=x&lt;/math&gt; and &lt;math&gt;\cos{3B}=y&lt;/math&gt;.<br /> <br /> Using the fact that &lt;math&gt;\cos(3A+3B)=\cos 3A\cos 3B-\sin 3A\sin 3B=xy-\sqrt{1-x^2}\sqrt{1-y^2}&lt;/math&gt;, we get &lt;math&gt;x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}=1&lt;/math&gt;, or &lt;math&gt;\sqrt{1-x^2}\sqrt{1-y^2}=xy-x-y+1=(x-1)(y-1)&lt;/math&gt;. <br /> <br /> Squaring both sides, we get &lt;math&gt;(1-x^2)(1-y^2) = [(x-1)(y-1)]^2&lt;/math&gt;. Cancelling factors, &lt;math&gt;(1+x)(1+y) = (1-x)(1-y)&lt;/math&gt;.<br /> <br /> *Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have <br /> <br /> &lt;math&gt;cos(3A)-1=0, cos(3A)=1&lt;/math&gt; <br /> <br /> For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side).<br /> <br /> Expanding, &lt;math&gt;1+x+y+xy=1-x-y+xy\rightarrow x+y=-x-y&lt;/math&gt;. <br /> <br /> Simplification leads to &lt;math&gt;x+y=0&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;\cos(3C)=1&lt;/math&gt;. So &lt;math&gt;\angle C&lt;/math&gt; could be &lt;math&gt;0^\circ&lt;/math&gt; or &lt;math&gt;120^\circ&lt;/math&gt;. We eliminate &lt;math&gt;0^\circ&lt;/math&gt; and use law of cosines to get our answer: <br /> <br /> &lt;cmath&gt;m=10^2+13^2-2\cdot 10\cdot 13\cos\angle C&lt;/cmath&gt;<br /> &lt;cmath&gt;\rightarrow m=269-260\cos 120^\circ=269-260\left(\text{-}\frac{1}{2}\right)&lt;/cmath&gt;<br /> &lt;cmath&gt;\rightarrow m=269+130=399&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\framebox{399}&lt;/math&gt;<br /> <br /> NOTE: This solution forgot the case of dividing by 0<br /> NOTE FROM DIFFERENT PERSON: Pretty sure they addressed that in the notice<br /> <br /> ==Solution 2==<br /> As above, we can see that &lt;math&gt;\cos3A+\cos3B-\cos(3A+3B)=1&lt;/math&gt;<br /> <br /> Expanding, we get<br /> <br /> &lt;math&gt;\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1&lt;/math&gt;<br /> <br /> &lt;math&gt;\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B&lt;/math&gt;<br /> <br /> &lt;math&gt;(\cos3A-1)(\cos3B-1)=\sin3A\sin3B&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;\tan{x}=\frac{1}{\tan(90-x)}&lt;/math&gt;, or &lt;math&gt;\tan{x}\tan(90-x)=1&lt;/math&gt;<br /> <br /> Thus &lt;math&gt;\frac{3A}{2}+\frac{3B}{2}=90&lt;/math&gt;, or &lt;math&gt;A+B=60&lt;/math&gt;.<br /> <br /> Now we know that &lt;math&gt;C=120&lt;/math&gt;, so we can just use the Law of Cosines to get &lt;math&gt;\boxed{399}&lt;/math&gt;<br /> <br /> Note: This solution also forgets that &lt;math&gt;A&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; might be 120 when dividing by &lt;math&gt;\sin 3A&lt;/math&gt; and &lt;math&gt;\sin 3B&lt;/math&gt;<br /> ==Solution 3==<br /> &lt;cmath&gt;\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)&lt;/cmath&gt;<br /> &lt;cmath&gt;2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)&lt;/cmath&gt;<br /> If &lt;math&gt;\cos\frac{3}{2}(A+B) = 0&lt;/math&gt;, then &lt;math&gt;\frac{3}{2}(A+B)=90&lt;/math&gt;, &lt;math&gt;A+B=60&lt;/math&gt;, so &lt;math&gt;C=120&lt;/math&gt;; otherwise,<br /> &lt;cmath&gt;2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin\frac{3}{2}A\sin\frac{3}{2}B=0&lt;/cmath&gt;<br /> so either &lt;math&gt;\sin\frac{3}{2}A=0&lt;/math&gt; or &lt;math&gt;\sin\frac{3}{2}B=0&lt;/math&gt;, i.e., either &lt;math&gt;A=120&lt;/math&gt; or &lt;math&gt;B=120&lt;/math&gt;. In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. &lt;math&gt;\boxed{399}&lt;/math&gt;<br /> <br /> -Mathdummy<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=II|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_15&diff=128505 2015 AIME II Problems/Problem 15 2020-07-18T00:44:53Z <p>Hi13: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; have radii &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt;, respectively, and are externally tangent at point &lt;math&gt;A&lt;/math&gt;. Point &lt;math&gt;B&lt;/math&gt; is on &lt;math&gt;\mathcal{P}&lt;/math&gt; and point &lt;math&gt;C&lt;/math&gt; is on &lt;math&gt;\mathcal{Q}&lt;/math&gt; so that line &lt;math&gt;BC&lt;/math&gt; is a common external tangent of the two circles. A line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; again at &lt;math&gt;D&lt;/math&gt; and intersects &lt;math&gt;\mathcal{Q}&lt;/math&gt; again at &lt;math&gt;E&lt;/math&gt;. Points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; lie on the same side of &lt;math&gt;\ell&lt;/math&gt;, and the areas of &lt;math&gt;\triangle DBA&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equal. This common area is &lt;math&gt;\frac{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 /> &lt;asy&gt;<br /> import cse5;<br /> pathpen=black; pointpen=black;<br /> size(6cm);<br /> <br /> pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689);<br /> <br /> filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7));<br /> filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7));<br /> D(CR((0,1),1)); D(CR((4,4),4,150,390));<br /> D(L(MP(&quot;D&quot;,D(D),N),MP(&quot;A&quot;,D((0.8,1.6)),NE),1,5.5));<br /> D((-1.2,0)--MP(&quot;B&quot;,D((0,0)),S)--MP(&quot;C&quot;,D((4,0)),S)--(8,0));<br /> D(MP(&quot;E&quot;,E,N));<br /> &lt;/asy&gt;<br /> <br /> ==Hint==<br /> &lt;math&gt;[ABC] = \frac{1}{2}ab \sin C&lt;/math&gt; is your friend for a quick solve. If you know about homotheties, go ahead, but you'll still need to do quite a bit of computation. If you're completely lost and you have a lot of time left in your mocking of this AIME, go ahead and coordinate bash.<br /> <br /> ==Solution 1 (guys trig is fast)==<br /> Let &lt;math&gt;M&lt;/math&gt; be the intersection of &lt;math&gt;\overline{BC}&lt;/math&gt; and the common internal tangent of &lt;math&gt;\mathcal P&lt;/math&gt; and &lt;math&gt;\mathcal Q.&lt;/math&gt; We claim that &lt;math&gt;M&lt;/math&gt; is the circumcenter of right &lt;math&gt;\triangle{ABC}.&lt;/math&gt; Indeed, we have &lt;math&gt;AM = BM&lt;/math&gt; and &lt;math&gt;BM = CM&lt;/math&gt; by equal tangents to circles, and since &lt;math&gt;BM = CM, M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{BC},&lt;/math&gt; implying that &lt;math&gt;\angle{BAC} = 90.&lt;/math&gt; Now draw &lt;math&gt;\overline{PA}, \overline{PB}, \overline{PM},&lt;/math&gt; where &lt;math&gt;P&lt;/math&gt; is the center of circle &lt;math&gt;\mathcal P.&lt;/math&gt; Quadrilateral &lt;math&gt;PAMB&lt;/math&gt; is cyclic, and by Pythagorean Theorem &lt;math&gt;PM = \sqrt{5},&lt;/math&gt; so by Ptolemy on &lt;math&gt;PAMB&lt;/math&gt; we have &lt;cmath&gt;AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.&lt;/cmath&gt; Do the same thing on cyclic quadrilateral &lt;math&gt;QAMC&lt;/math&gt; (where &lt;math&gt;Q&lt;/math&gt; is the center of circle &lt;math&gt;\mathcal Q&lt;/math&gt; and get &lt;math&gt;AC = \frac{8 \sqrt{5}}{5}.&lt;/math&gt; <br /> <br /> Let &lt;math&gt;\angle A = \angle{DAB}.&lt;/math&gt; By Law of Sines, &lt;math&gt;BD = 2R \sin A = 2 \sin A.&lt;/math&gt; Note that &lt;math&gt;\angle{D} = \angle{ABC}&lt;/math&gt; from inscribed angles, so <br /> &lt;cmath&gt;\begin{align*}<br /> [ABD] &amp;= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\<br /> &amp;= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\<br /> &amp;= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\<br /> &amp;= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right)<br /> \end{align*}&lt;/cmath&gt; after angle addition identity.<br /> <br /> Similarly, &lt;math&gt;\angle{EAC} = 90 - \angle A,&lt;/math&gt; and by Law of Sines &lt;math&gt;CE = 8 \sin{\angle{EAC}} = 8 \cos A.&lt;/math&gt; Note that &lt;math&gt;\angle{E} = \angle{ACB}&lt;/math&gt; from inscribed angles, so<br /> &lt;cmath&gt;\begin{align*}<br /> [ACE] &amp;= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\<br /> &amp;= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\<br /> &amp;= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\<br /> &amp;= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\<br /> &amp;= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right)<br /> \end{align*}&lt;/cmath&gt; after angle addition identity.<br /> Setting the two areas equal, we get &lt;cmath&gt;\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}&lt;/cmath&gt; after Pythagorean Identity. Now plug back in and the common area is &lt;math&gt;\frac{64}{65} \iff \boxed{129}.&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> <br /> &lt;/asy&gt;<br /> <br /> Call &lt;math&gt;O_1&lt;/math&gt; and &lt;math&gt;O_2&lt;/math&gt; the centers of circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt;, respectively, and extend &lt;math&gt;CB&lt;/math&gt; and &lt;math&gt;O_2O_1&lt;/math&gt; to meet at point &lt;math&gt;N&lt;/math&gt;. Call &lt;math&gt;K&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;O_1N&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;O_2N&lt;/math&gt;, respectively. Using the fact that &lt;math&gt;\triangle{O_1BN} \sim \triangle{O_2CN}&lt;/math&gt; and setting &lt;math&gt;NO_1 = k&lt;/math&gt;, we have that &lt;math&gt;\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}&lt;/math&gt;. We can do some more length chasing using triangles similar to &lt;math&gt;O_1BN&lt;/math&gt; to get that &lt;math&gt;AK = AL = \frac{24}{15}&lt;/math&gt;, &lt;math&gt;BK = \frac{12}{15}&lt;/math&gt;, and &lt;math&gt;CL = \frac{48}{15}&lt;/math&gt;. Now, consider the circles &lt;math&gt;\mathcal{P}&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; on the coordinate plane, where &lt;math&gt;A&lt;/math&gt; is the origin. If the line &lt;math&gt;\ell&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt;\mathcal{P}&lt;/math&gt; at &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;\mathcal{Q}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt; then &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;. To verify this, notice that &lt;math&gt;\triangle{AO_1D} \sim \triangle{EO_2A}&lt;/math&gt; from the fact that both triangles are isosceles with &lt;math&gt;\angle{O_1AD} \cong \angle{O_2AE}&lt;/math&gt;, which are corresponding angles. Since &lt;math&gt;O_2A = 4\cdot O_1A&lt;/math&gt;, we can conclude that &lt;math&gt;4 \cdot DA = AE&lt;/math&gt;.<br /> <br /> <br /> Hence, we need to find the slope &lt;math&gt;m&lt;/math&gt; of line &lt;math&gt;\ell&lt;/math&gt; such that the perpendicular distance &lt;math&gt;n&lt;/math&gt; from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is four times the perpendicular distance &lt;math&gt;p&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;AE&lt;/math&gt;. This will mean that the product of the bases and heights of triangles &lt;math&gt;ACE&lt;/math&gt; and &lt;math&gt;DBA&lt;/math&gt; will be equal, which in turn means that their areas will be equal. Let the line &lt;math&gt;\ell&lt;/math&gt; have the equation &lt;math&gt;y = -mx \implies mx + y = 0&lt;/math&gt;, and let &lt;math&gt;m&lt;/math&gt; be a positive real number so that the negative slope of &lt;math&gt;\ell&lt;/math&gt; is preserved. Setting &lt;math&gt;A = (0,0)&lt;/math&gt;, the coordinates of &lt;math&gt;B&lt;/math&gt; are &lt;math&gt;(x_B, y_B) = \left(\frac{-24}{15}, \frac{-12}{15}\right)&lt;/math&gt;, and the coordinates of &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;(x_C, y_C) = \left(\frac{24}{15}, \frac{-48}{15}\right)&lt;/math&gt;. Using the point-to-line distance formula and the condition &lt;math&gt;n = 4p&lt;/math&gt;, we have &lt;cmath&gt;\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}&lt;/cmath&gt; &lt;cmath&gt;\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.&lt;/cmath&gt; If &lt;math&gt;m &gt; 2&lt;/math&gt;, then clearly &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; would not lie on the same side of &lt;math&gt;\ell&lt;/math&gt;. Thus since &lt;math&gt;m &gt; 0&lt;/math&gt;, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have &lt;cmath&gt;\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.&lt;/cmath&gt; Thus, the equation of &lt;math&gt;\ell&lt;/math&gt; is &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt;.<br /> <br /> <br /> Then we can find the coordinates of &lt;math&gt;D&lt;/math&gt; by finding the point &lt;math&gt;(x,y)&lt;/math&gt; other than &lt;math&gt;A = (0,0)&lt;/math&gt; where the circle &lt;math&gt;\mathcal{P}&lt;/math&gt; intersects &lt;math&gt;\ell&lt;/math&gt;. &lt;math&gt;\mathcal{P}&lt;/math&gt; can be represented with the equation &lt;math&gt;(x + 1)^2 + y^2 = 1&lt;/math&gt;, and substituting &lt;math&gt;y = -\frac{3}{2}x&lt;/math&gt; into this equation yields &lt;math&gt;x = 0, -\frac{8}{13}&lt;/math&gt; as solutions. Discarding &lt;math&gt;x = 0&lt;/math&gt;, the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}&lt;/math&gt;. The distance from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;A&lt;/math&gt; is then &lt;math&gt;\frac{4}{\sqrt{13}}.&lt;/math&gt; The perpendicular distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; or the height of &lt;math&gt;\triangle{DBA}&lt;/math&gt; is &lt;math&gt;\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.&lt;/math&gt; Finally, the common area is &lt;math&gt;\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}&lt;/math&gt;, and &lt;math&gt;m + n = 64 + 65 = \boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> By [[homothety]], we deduce that &lt;math&gt;AE = 4 AD&lt;/math&gt;. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt; is four times that from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;l&lt;/math&gt;. Let the distance from &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and the distance from &lt;math&gt;B&lt;/math&gt; be &lt;math&gt;4x&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; be the centers of their respective circles. Then dropping a perpendicular from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; creates a &lt;math&gt;3-4-5&lt;/math&gt; right triangle, from which &lt;math&gt;BC = 4&lt;/math&gt; and, if &lt;math&gt;\alpha = \angle{AQC}&lt;/math&gt;, that &lt;math&gt;\cos \alpha = \dfrac{3}{5}&lt;/math&gt;. Then &lt;math&gt;\angle{BPA} = 180^\circ - \alpha&lt;/math&gt;, and the Law of Cosines on triangles &lt;math&gt;APB&lt;/math&gt; and &lt;math&gt;AQC&lt;/math&gt; gives &lt;math&gt;AB = \dfrac{4}{\sqrt{5}}&lt;/math&gt; and &lt;math&gt;AC = \dfrac{8}{\sqrt{5}}.&lt;/math&gt;<br /> <br /> Now, using the Pythagorean Theorem to express the length of the projection of &lt;math&gt;BC&lt;/math&gt; onto line &lt;math&gt;l&lt;/math&gt; gives<br /> &lt;cmath&gt;\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.&lt;/cmath&gt;<br /> Squaring and simplifying gives<br /> &lt;cmath&gt;\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,&lt;/cmath&gt;<br /> and squaring and solving gives &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}.&lt;/math&gt;<br /> <br /> By the Law of Sines on triangle &lt;math&gt;ABD&lt;/math&gt;, we have<br /> &lt;cmath&gt;\frac{BD}{\sin A} = 2.&lt;/cmath&gt;<br /> But we know &lt;math&gt;\sin A = \dfrac{4x}{AB}&lt;/math&gt;, and so a small computation gives &lt;math&gt;BD = \dfrac{16}{\sqrt{65}}.&lt;/math&gt; The Pythagorean Theorem now gives<br /> &lt;cmath&gt;AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},&lt;/cmath&gt;<br /> and so the common area is &lt;math&gt;\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.&lt;/math&gt; The answer is &lt;math&gt;\boxed{129}.&lt;/math&gt;<br /> <br /> ==Alternate Path to x==<br /> Call the intersection of lines &lt;math&gt;l&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;E&lt;/math&gt;.You can use similar triangles to find that the distance from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt; is four times the distance from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. Then draw a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call the point &lt;math&gt;F&lt;/math&gt;. &lt;math&gt;AF = \frac{8}{5}&lt;/math&gt; and &lt;math&gt;FE = FC + CE = \frac{16}{5} + \frac{4}{3} = \frac{68}{15}&lt;/math&gt;, so by the Pythagorean Theorem, &lt;math&gt;AE = \dfrac{20\sqrt{13}}{5}&lt;/math&gt;. You can now use similar triangles to find that &lt;math&gt;x = \dfrac{8}{5\sqrt{13}}&lt;/math&gt; and continue on like in solution 2.<br /> <br /> ==Solution 4==<br /> &lt;math&gt;DE&lt;/math&gt; goes through &lt;math&gt;A&lt;/math&gt;, the point of tangency of both circles. So &lt;math&gt;DE&lt;/math&gt; intercepts equal arcs in circle &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;: [[homothety]]. Hence, &lt;math&gt;AE=4AD&lt;/math&gt;. We will use such similarity later.<br /> <br /> The diagonal distance between the centers of the circles is &lt;math&gt;4+1=5&lt;/math&gt;. The difference in heights is &lt;math&gt;4-1=3&lt;/math&gt;. So &lt;math&gt;BC=\sqrt{5^2-3^2}=4&lt;/math&gt;.<br /> <br /> The triangle connecting the centers with a side parallel to &lt;math&gt;BC&lt;/math&gt; is a &lt;math&gt;3-4-5&lt;/math&gt; right triangle. Since &lt;math&gt;O_PA=1&lt;/math&gt;, the height of &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;1+3/5=8/5&lt;/math&gt;. Drop an altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call it &lt;math&gt;I&lt;/math&gt;: &lt;math&gt;IB=4/5&lt;/math&gt; and &lt;math&gt;IC=4-4/5=32/5&lt;/math&gt;. Since right &lt;math&gt;\triangle AIB\sim\triangle CIB&lt;/math&gt;, &lt;math&gt;ABC&lt;/math&gt; is a right triangle also; &lt;math&gt;IB:IA:IC&lt;/math&gt; form a geometric progression &lt;math&gt;\times 2&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;BA&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; to a point &lt;math&gt;G&lt;/math&gt; on the other side of &lt;math&gt;\circ Q&lt;/math&gt;. By [[homothety]], &lt;math&gt;\triangle DAB\sim\triangle EAG&lt;/math&gt;. By angle chasing &lt;math&gt;\triangle DAB&lt;/math&gt; through right triangle &lt;math&gt;ABC&lt;/math&gt;, we deduce that &lt;math&gt;\angle CEG&lt;/math&gt; is a right angle. Since &lt;math&gt;ACEG&lt;/math&gt; is cyclic, &lt;math&gt;\angle GAC&lt;/math&gt; is also right. So &lt;math&gt;CG&lt;/math&gt; is a diameter of &lt;math&gt;\circ G&lt;/math&gt;. Because of this, &lt;math&gt;CG \perp BC&lt;/math&gt;, the tangent line. &lt;math&gt;\triangle BCG&lt;/math&gt; is right and &lt;math&gt;\triangle BCG\sim\triangle ABC\sim\triangle CAG&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;AC=\sqrt{(8/5)^2+(16/5)^2}=8\sqrt{5}/5&lt;/math&gt; so &lt;math&gt;AG=2AC=16\sqrt{5}/5&lt;/math&gt; and &lt;math&gt;[\triangle CAG]=64/5&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;[\triangle DAB]=[\triangle ACE]&lt;/math&gt;, the common area is &lt;math&gt;[ACEG]/17&lt;/math&gt;. &lt;math&gt;16[\triangle DAB]=[\triangle GAE]&lt;/math&gt; because the triangles are similar with a ratio of &lt;math&gt;1:4&lt;/math&gt;. So we only need to find &lt;math&gt;[\triangle CEG]&lt;/math&gt; now.<br /> <br /> <br /> Extend &lt;math&gt;DE&lt;/math&gt; through &lt;math&gt;E&lt;/math&gt; to intersect the tangent at &lt;math&gt;F&lt;/math&gt;. Because &lt;math&gt;4DA=AE&lt;/math&gt;, the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AD&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; times the height from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt;EA&lt;/math&gt;. So &lt;math&gt;BC=3/4BF&lt;/math&gt; and &lt;math&gt;BF=16/3&lt;/math&gt;. We look at right triangle &lt;math&gt;\triangle AIF&lt;/math&gt;. &lt;math&gt;IF=68/15&lt;/math&gt; and &lt;math&gt;AI=8/5&lt;/math&gt;. &lt;math&gt;\triangle AIF&lt;/math&gt; is a &lt;math&gt;17-6-5\sqrt{13}&lt;/math&gt; right triangle. Hypotenuse &lt;math&gt;AF&lt;/math&gt; intersects &lt;math&gt;CG&lt;/math&gt; at a point, we call it &lt;math&gt;H&lt;/math&gt;. &lt;math&gt;CH=4/3\div 68/15\cdot 8/5=8/17&lt;/math&gt;. So &lt;math&gt;HG=8-8/17=128/17&lt;/math&gt;.<br /> <br /> <br /> By [[Power of a Point Theorem|Power of a Point]], &lt;math&gt;CH\cdot HG=AH\cdot HE&lt;/math&gt;. &lt;math&gt;AH=16/5\cdot 5\sqrt{13}/17=16\sqrt{13}/17.&lt;/math&gt; So &lt;math&gt;HE=1024/289\cdot 17/(16\sqrt{13})=64/(17\sqrt{13})&lt;/math&gt;. The height from &lt;math&gt;E&lt;/math&gt; to &lt;math&gt;CG&lt;/math&gt; is &lt;math&gt;17/(5\sqrt{13})\cdot 64/(17\sqrt{13})=64/65&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;[\triangle CEG]=64/65\cdot 8\div 2=256/65&lt;/math&gt;. The area of the whole cyclic quadrilateral is &lt;math&gt;64/5+256/65=(832+256)/65=1088/65&lt;/math&gt;. Lastly, the common area is &lt;math&gt;1/17&lt;/math&gt; the area of the quadrilateral, or &lt;math&gt;64/65&lt;/math&gt;. So &lt;math&gt;64+65=\boxed{129}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Easy computation)==<br /> &lt;asy&gt;<br /> unitsize(35);<br /> draw(Circle((-1,0),1));<br /> draw(Circle((4,0),4));<br /> pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;<br /> A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);<br /> label(&quot;$A$&quot;,A,NE);label(&quot;$O_1$&quot;,O_1,NE);label(&quot;$O_2$&quot;,O_2,NE);label(&quot;$B$&quot;,B,SW);label(&quot;$C$&quot;,C,SW);label(&quot;$D$&quot;,D,NE);label(&quot;$E$&quot;,E,NE);label(&quot;$N$&quot;,N,W);label(&quot;$K$&quot;,(-24/15,0.2));label(&quot;$L$&quot;,(24/15,0.2));label(&quot;$n$&quot;,(-0.8,-0.12));label(&quot;$p$&quot;,((29/15,-48/15)));label(&quot;$\mathcal{P}$&quot;,(-1.6,1.1));label(&quot;$\mathcal{Q}$&quot;,(6,4));<br /> draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);<br /> dot(O_1);dot(O_2);<br /> draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));<br /> //draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);<br /> path circle2 = Circle((4,0),4);<br /> N = (-8/3,0);<br /> pair X =rotate(180,O_2)*E;<br /> pair Y = (8,0);<br /> draw(X--Y,dashed); draw(E--Y,dashed);draw(E--X,dashed); draw(Y--C,dashed); draw(C--X,dashed); draw(O_2--Y);<br /> dot(&quot;$X$&quot;, X, NE);dot(&quot;$Y$&quot;, Y, NE);<br /> &lt;/asy&gt;<br /> <br /> Consider the homothety that takes triangle BDA onto CXY on the big circle, as plotted. Some hidden congruence angles are revealed which help reduce computation complexity. Just some angle chasing and straight forward trigs. Because &lt;math&gt;AE=XY&lt;/math&gt; and &lt;math&gt;AE \parallel XY&lt;/math&gt;, &lt;math&gt;XYE&lt;/math&gt; is right angle. <br /> <br /> First, &lt;math&gt;\frac{NO_1}{NO_1+5} = \frac{1}{4}&lt;/math&gt;, so &lt;math&gt;NO_1=\frac{5}{3}&lt;/math&gt;. And,<br /> &lt;cmath&gt;\cos{\angle{AO_2C}}=\cos{2\angle{AYC}} = \frac{O_2C}{NO_1+5} = \frac{3}{5}&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin{\angle{AYC}} = \sqrt{\dfrac{1-\cos{2\angle{AYC}}}{2}}=\frac{1}{\sqrt{5}}&lt;/cmath&gt;<br /> &lt;cmath&gt; \cos{\angle{AYC}} = \frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> Then,<br /> &lt;cmath&gt;[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}&lt;/cmath&gt;<br /> &lt;cmath&gt;[CXY] = \frac{1}{2}CX*XY*\sin{\angle{CXY}}=\frac{1}{2}*8\sin{\angle{XYC}}*XY*\sin{\angle{CAY}}&lt;/cmath&gt;<br /> Since &lt;math&gt;\angle{CAY} = 90 - \angle{AYC}&lt;/math&gt;, &lt;math&gt;\angle{XYC} = 90 - \angle{CYE}&lt;/math&gt;, &lt;math&gt;XY = AE&lt;/math&gt;, we have<br /> &lt;cmath&gt;[CXY] = \frac{1}{2}*8AE\cos{\angle{CYE}}*\cos{\angle{AYC}}=\frac{1}{2}*8AE\cos{\angle{CYE}}*\frac{2}{\sqrt{5}}&lt;/cmath&gt;<br /> Since &lt;math&gt;\triangle{CXY}&lt;/math&gt; is four times in scale to &lt;math&gt;\triangle{AEC}&lt;/math&gt;, their area ratio is 16. Divide the two equations for the two areas, we have<br /> &lt;cmath&gt;\tan{\angle{CYE}} = \frac{1}{8}&lt;/cmath&gt;<br /> With this angle found, everything else just follows. <br /> &lt;cmath&gt;\sin{\angle{CYE}} = \dfrac{\tan{\angle{CYE}}}{\sqrt{1+\tan^2{\angle{CYE}}}}=\dfrac{1}{\sqrt{65}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\cos{\angle{CYE}} = \dfrac{8}{\sqrt{65}}&lt;/cmath&gt;<br /> &lt;cmath&gt;\sin{\angle{AYE}} = \sin({\angle{AYC}+\angle{CYE}} )= \frac{1}{\sqrt{5}}*\frac{8}{\sqrt{65}} + \frac{2}{\sqrt{5}}*\frac{1}{\sqrt{65}} = \frac{2}{\sqrt{13}}&lt;/cmath&gt;<br /> &lt;cmath&gt;AE = 8\sin{\angle{AYE}} = \frac{16}{\sqrt{13}}&lt;/cmath&gt;<br /> &lt;cmath&gt;[AEC] = \frac{1}{2}*8*\frac{16}{\sqrt{13}}*\dfrac{1}{\sqrt{65}}*\frac{1}{\sqrt{5}}=\frac{64}{65}&lt;/cmath&gt;<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_11&diff=128496 2015 AIME II Problems/Problem 11 2020-07-18T00:06:26Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> The circumcircle of acute &lt;math&gt;\triangle ABC&lt;/math&gt; has center &lt;math&gt;O&lt;/math&gt;. The line passing through point &lt;math&gt;O&lt;/math&gt; perpendicular to &lt;math&gt;\overline{OB}&lt;/math&gt; intersects lines &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. Also &lt;math&gt;AB=5&lt;/math&gt;, &lt;math&gt;BC=4&lt;/math&gt;, &lt;math&gt;BQ=4.5&lt;/math&gt;, and &lt;math&gt;BP=\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> unitsize(30);<br /> draw(Circle((0,0),3));<br /> pair A,B,C,O, Q, P, M, N;<br /> A=(2.5, -sqrt(11/4));<br /> B=(-2.5, -sqrt(11/4));<br /> C=(-1.96, 2.28);<br /> Q=(-1.89, 2.81);<br /> P=(1.13, -1.68);<br /> O=origin;<br /> M=foot(O,C,B);<br /> N=foot(O,A,B);<br /> draw(A--B--C--cycle);<br /> label(&quot;$A$&quot;,A,SE);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,NW);<br /> label(&quot;$Q$&quot;,Q,NW);<br /> dot(O);<br /> label(&quot;$O$&quot;,O,NE);<br /> label(&quot;$M$&quot;,M,W);<br /> label(&quot;$N$&quot;,N,S);<br /> label(&quot;$P$&quot;,P,S);<br /> draw(B--O);<br /> draw(C--Q);<br /> draw(Q--O);<br /> draw(O--C);<br /> draw(O--A);<br /> draw(O--P);<br /> draw(O--M, dashed);<br /> draw(O--N, dashed);<br /> draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5));<br /> draw(rightanglemark(O,N,B,5));<br /> draw(rightanglemark(B,O,P,5));<br /> draw(rightanglemark(O,M,C,5));<br /> &lt;/asy&gt;<br /> ===Solution 1===<br /> Call &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; the feet of the altitudes from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Let &lt;math&gt;OB = r&lt;/math&gt; . Notice that &lt;math&gt;\triangle{OMB} \sim \triangle{QOB}&lt;/math&gt; because both are right triangles, and &lt;math&gt;\angle{OBQ} \cong \angle{OBM}&lt;/math&gt;. By &lt;math&gt;\frac{MB}{BO}=\frac{BO}{BQ}&lt;/math&gt;, &lt;math&gt;MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}&lt;/math&gt;. However, since &lt;math&gt;O&lt;/math&gt; is the circumcenter of triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;OM&lt;/math&gt; is a perpendicular bisector by the definition of a circumcenter. Hence, &lt;math&gt;\frac{r^2}{4.5} = 2 \implies r = 3&lt;/math&gt;. Since we know &lt;math&gt;BN=\frac{5}{2}&lt;/math&gt; and &lt;math&gt;\triangle BOP \sim \triangle NBO&lt;/math&gt;, we have &lt;math&gt;\frac{BP}{3} = \frac{3}{\frac{5}{2}}&lt;/math&gt;. Thus, &lt;math&gt;BP = \frac{18}{5}&lt;/math&gt;. &lt;math&gt;m + n=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Notice that &lt;math&gt;\angle{CBO}=90-A&lt;/math&gt;, so &lt;math&gt;\angle{BQO}=A&lt;/math&gt;. From this we get that &lt;math&gt;\triangle{BPQ}\sim \triangle{BCA}&lt;/math&gt;. So &lt;math&gt;\dfrac{BP}{BC}=\dfrac{BQ}{BA}&lt;/math&gt;, plugging in the given values we get &lt;math&gt;\dfrac{BP}{4}=\dfrac{4.5}{5}&lt;/math&gt;, so &lt;math&gt;BP=\dfrac{18}{5}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Let &lt;math&gt;r=BO&lt;/math&gt;. Drawing perpendiculars, &lt;math&gt;BM=MC=2&lt;/math&gt; and &lt;math&gt;BN=NA=2.5&lt;/math&gt;. From there, &lt;math&gt;OM=\sqrt{r^2-4}&lt;/math&gt;. Thus, &lt;math&gt;OQ=\frac{\sqrt{4r^2+9}}{2}&lt;/math&gt;. Using &lt;math&gt;\triangle{BOQ}&lt;/math&gt;, we get &lt;math&gt;r=3&lt;/math&gt;. Now let's find &lt;math&gt;NP&lt;/math&gt;. After some calculations with &lt;math&gt;\triangle{BON}&lt;/math&gt; ~ &lt;math&gt;\triangle{OPN}&lt;/math&gt;, &lt;math&gt;{NP=11/10}&lt;/math&gt;. Therefore, &lt;math&gt;BP=\frac{5}{2}+\frac{11}{10}=18/5&lt;/math&gt;. &lt;math&gt;18+5=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Let &lt;math&gt;\angle{BQO}=\alpha&lt;/math&gt;. Extend &lt;math&gt;OB&lt;/math&gt; to touch the circumcircle at a point &lt;math&gt;K&lt;/math&gt;. Then, note that &lt;math&gt;\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha&lt;/math&gt;. But since &lt;math&gt;BK&lt;/math&gt; is a diameter, &lt;math&gt;\angle{KAB}=90^\circ&lt;/math&gt;, implying &lt;math&gt;\angle{CAB}=\alpha&lt;/math&gt;. It follows that &lt;math&gt;APCQ&lt;/math&gt; is a cyclic quadrilateral.<br /> <br /> Let &lt;math&gt;BP=x&lt;/math&gt;. By Power of a Point, &lt;cmath&gt;5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.&lt;/cmath&gt;The answer is &lt;math&gt;18+5=\boxed{023}&lt;/math&gt;.<br /> <br /> ===Solution 5===<br /> Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.<br /> <br /> Denote the circumradius of &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;R&lt;/math&gt;, the circumcircle of &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;O&lt;/math&gt;, and the shortest distance from &lt;math&gt;Q&lt;/math&gt; to circle &lt;math&gt;O&lt;/math&gt; to be &lt;math&gt;x&lt;/math&gt;. <br /> <br /> Using Power of a Point on &lt;math&gt;Q&lt;/math&gt; relative to circle &lt;math&gt;O&lt;/math&gt;, we get that &lt;math&gt;x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}&lt;/math&gt;. Using Pythagorean Theorem on triangle &lt;math&gt;QOB&lt;/math&gt; to get &lt;math&gt;(x + r)^2 + r^2 = \frac{81}{4}&lt;/math&gt;. Subtracting the first equation from the second, we get that &lt;math&gt;2r^2 = 18&lt;/math&gt; and therefore &lt;math&gt;r = 3&lt;/math&gt;. Now, set &lt;math&gt;\cos{ABC} = y&lt;/math&gt;. Using law of cosines on &lt;math&gt;ABC&lt;/math&gt; to find &lt;math&gt;AC&lt;/math&gt; in terms of &lt;math&gt;y&lt;/math&gt; and plugging that into the extended law of sines, we get &lt;math&gt;\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6&lt;/math&gt;. Squaring both sides and cross multiplying, we get &lt;math&gt;36x^2 - 40x + 5 = 0&lt;/math&gt;. Now, we get &lt;math&gt;x = \frac{10 \pm \sqrt{55}}{18}&lt;/math&gt; using quadratic formula. If you drew a decent diagram, &lt;math&gt;B&lt;/math&gt; is acute and therefore &lt;math&gt;x = \frac{10 + \sqrt{55}}{18}&lt;/math&gt;(You can also try plugging in both in the end and seeing which gives a rational solution). Note that &lt;math&gt;BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.&lt;/math&gt; Using the cosine addition formula and then plugging in what we know about &lt;math&gt;QBO&lt;/math&gt;, we get that &lt;math&gt;BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}&lt;/math&gt;. Now, the hard part is to find what &lt;math&gt;\sin{B}&lt;/math&gt; is. We therefore want &lt;math&gt;\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}&lt;/math&gt;. For the numerator, by inspection &lt;math&gt;(a + b\sqrt{55})^2&lt;/math&gt; will not work for integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. The other case is if there is &lt;math&gt;(a\sqrt{5} + b\sqrt{11})^2&lt;/math&gt;. By inspection, &lt;math&gt;5\sqrt{5} - 2\sqrt{11}&lt;/math&gt; works. Therefore, plugging all this in yields the answer, &lt;math&gt;\frac{18}{5} \rightarrow \boxed{23}&lt;/math&gt;. Solution by hyxue<br /> <br /> ===Solution 6===<br /> &lt;asy&gt; <br /> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br /> import graph; size(15cm); <br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br /> pen dotstyle = black; /* point style */ <br /> real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */<br /> pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br /> /* draw figures */<br /> draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); <br /> draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); <br /> draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */<br /> draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */<br /> draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); <br /> /* dots and labels */<br /> dot((-1.82,0.87),dotstyle); <br /> label(&quot;$A$&quot;, (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); <br /> dot((3.178984115621537,0.7692140299269852),dotstyle); <br /> label(&quot;$B$&quot;, (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); <br /> dot((2.6857306099246263,4.738685150758791),dotstyle); <br /> label(&quot;$C$&quot;, (2.7238749148597092,4.831703985774336), NE * labelscalefactor); <br /> dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); <br /> label(&quot;$O$&quot;, (0.7539479965810783,2.556577122410283), NE * labelscalefactor); <br /> dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); <br /> label(&quot;$P$&quot;, (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); <br /> dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); <br /> label(&quot;$Q$&quot;, (2.6591355325688624,5.312625111363486), NE * labelscalefactor); <br /> dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); <br /> label(&quot;$A'$&quot;, (1.3643478867519216,5.488346291867214), NE * labelscalefactor); <br /> dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); <br /> label(&quot;$P'$&quot;, (1.8822629450786978,4.184310162865866), NE * labelscalefactor); <br /> dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); <br /> label(&quot;$D$&quot;, (2.603644633462422,5.802794720137042), NE * labelscalefactor); <br /> clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br /> &lt;/asy&gt;<br /> Reflect &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;P&lt;/math&gt; across &lt;math&gt;OB&lt;/math&gt; to points &lt;math&gt;A'&lt;/math&gt; and &lt;math&gt;P'&lt;/math&gt;, respectively with &lt;math&gt;A'&lt;/math&gt; on the circle and &lt;math&gt;P, O, P'&lt;/math&gt; collinear. Now, &lt;math&gt;\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB&lt;/math&gt; by parallel lines. From here, &lt;math&gt;\angle P'PB = \angle PP'B = \angle A'P'Q&lt;/math&gt; as &lt;math&gt;P, P', Q&lt;/math&gt; collinear. From here, &lt;math&gt;A'P'QC&lt;/math&gt; is cyclic, and by power of a point we obtain &lt;math&gt;\frac{18}{5} \implies \boxed{023}&lt;/math&gt;.<br /> ~awang11's sol<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=128190 2019 AIME I Problems/Problem 11 2020-07-13T04:36:39Z <p>Hi13: /* Solution 2 (Lots of Pythagorean Theorem) */</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> &lt;asy&gt;<br /> <br /> unitsize(1cm);<br /> <br /> <br /> var x = 9;<br /> <br /> pair A = (0,sqrt(x^2-1));<br /> pair B = (-1,0);<br /> pair C = (1,0);<br /> <br /> dot(Label(&quot;$A$&quot;,A,NE),A);<br /> dot(Label(&quot;$B$&quot;,B,SW),B);<br /> dot(Label(&quot;$C$&quot;,C,SE),C);<br /> <br /> draw(A--B--C--cycle);<br /> <br /> <br /> var r = sqrt((x-1)/(x+1));<br /> <br /> pair I = (0,r);<br /> dot(Label(&quot;$I$&quot;,I,SE),I);<br /> draw(circle(I,r));<br /> draw(Label(&quot;$r$&quot;),I--I+r*SSW,dashed);<br /> <br /> <br /> pair M = intersectionpoint(A--B,circle(I,r));<br /> pair N = (0,0);<br /> pair O = intersectionpoint(A--C,circle(I,r));<br /> <br /> dot(Label(&quot;$M$&quot;,M,W),M);<br /> dot(Label(&quot;$N$&quot;,N,S),N);<br /> dot(Label(&quot;$O$&quot;,O,E),O);<br /> <br /> var rN = sqrt((x+1)/(x-1));<br /> <br /> pair EN = (0,-rN);<br /> dot(Label(&quot;$E_N$&quot;,EN,SE),EN);<br /> draw(circle(EN,rN));<br /> draw(Label(&quot;$r_N$&quot;),EN--EN+rN*SSW,dashed);<br /> <br /> <br /> pair AB = (-1-2/(x-1),-2rN);<br /> pair AC = (1+2/(x-1),-2rN);<br /> <br /> draw(B--AB,EndArrow);<br /> draw(C--AC,EndArrow);<br /> <br /> pair H = intersectionpoint(B--AB,circle(EN,rN));<br /> dot(Label(&quot;$H$&quot;,H,W),H);<br /> <br /> <br /> var rM = sqrt(x^2-1);<br /> <br /> pair EM = (-x,rM);<br /> dot(Label(&quot;$E_M$&quot;,EM,SW),EM);<br /> draw(Label(&quot;$r_M$&quot;),EM--EM+rM*SSE,dashed);<br /> <br /> <br /> pair CB = (-x-1,0);<br /> pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));<br /> <br /> draw(B--CB,EndArrow);<br /> draw(A--CA,EndArrow);<br /> <br /> <br /> pair J = intersectionpoint(A--B,circle(EM,rM));<br /> pair K = intersectionpoint(B--CB,circle(EM,rM));<br /> <br /> dot(Label(&quot;$J$&quot;,J,W),J);<br /> dot(Label(&quot;$K$&quot;,K,S),K);<br /> <br /> <br /> draw(arc(EM,rM,-100,15),Arrows);<br /> <br /> &lt;/asy&gt;<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;AC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;BC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4,&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than &lt;math&gt;1&lt;/math&gt;, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution (WAY Too Long, Really)==<br /> <br /> https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==Video Solution 2 (I would recommend this one because it's more concise)==<br /> <br /> https://www.youtube.com/watch?v=ldr4yi3t6hQ<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=128189 2019 AIME I Problems/Problem 11 2020-07-13T04:33:59Z <p>Hi13: /* Solution 2 (Lots of Pythagorean Theorem) */</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> &lt;asy&gt;<br /> <br /> unitsize(1cm);<br /> <br /> <br /> var x = 9;<br /> <br /> pair A = (0,sqrt(x^2-1));<br /> pair B = (-1,0);<br /> pair C = (1,0);<br /> <br /> dot(Label(&quot;$A$&quot;,A,NE),A);<br /> dot(Label(&quot;$B$&quot;,B,SW),B);<br /> dot(Label(&quot;$C$&quot;,C,SE),C);<br /> <br /> draw(A--B--C--cycle);<br /> <br /> <br /> var r = sqrt((x-1)/(x+1));<br /> <br /> pair I = (0,r);<br /> dot(Label(&quot;$I$&quot;,I,SE),I);<br /> draw(circle(I,r));<br /> draw(Label(&quot;$r$&quot;),I--I+r*SSW,dashed);<br /> <br /> <br /> pair M = intersectionpoint(A--B,circle(I,r));<br /> pair N = (0,0);<br /> pair O = intersectionpoint(A--C,circle(I,r));<br /> <br /> dot(Label(&quot;$M$&quot;,M,W),M);<br /> dot(Label(&quot;$N$&quot;,N,S),N);<br /> dot(Label(&quot;$O$&quot;,O,E),O);<br /> <br /> var rN = sqrt((x+1)/(x-1));<br /> <br /> pair EN = (0,-rN);<br /> dot(Label(&quot;$E_N$&quot;,EN,SE),EN);<br /> draw(circle(EN,rN));<br /> draw(Label(&quot;$r_N$&quot;),EN--EN+rN*SSW,dashed);<br /> <br /> <br /> pair AB = (-1-2/(x-1),-2rN);<br /> pair AC = (1+2/(x-1),-2rN);<br /> <br /> draw(B--AB,EndArrow);<br /> draw(C--AC,EndArrow);<br /> <br /> pair H = intersectionpoint(B--AB,circle(EN,rN));<br /> dot(Label(&quot;$H$&quot;,H,W),H);<br /> <br /> <br /> var rM = sqrt(x^2-1);<br /> <br /> pair EM = (-x,rM);<br /> dot(Label(&quot;$E_M$&quot;,EM,SW),EM);<br /> draw(Label(&quot;$r_M$&quot;),EM--EM+rM*SSE,dashed);<br /> <br /> <br /> pair CB = (-x-1,0);<br /> pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));<br /> <br /> draw(B--CB,EndArrow);<br /> draw(A--CA,EndArrow);<br /> <br /> <br /> pair J = intersectionpoint(A--B,circle(EM,rM));<br /> pair K = intersectionpoint(B--CB,circle(EM,rM));<br /> <br /> dot(Label(&quot;$J$&quot;,J,W),J);<br /> dot(Label(&quot;$K$&quot;,K,S),K);<br /> <br /> <br /> draw(arc(EM,rM,-100,15),Arrows);<br /> <br /> &lt;/asy&gt;<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;AC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4,&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than &lt;math&gt;1&lt;/math&gt;, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution (WAY Too Long, Really)==<br /> <br /> https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==Video Solution 2 (I would recommend this one because it's more concise)==<br /> <br /> https://www.youtube.com/watch?v=ldr4yi3t6hQ<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_7&diff=128188 2019 AIME I Problems/Problem 7 2020-07-13T04:18:28Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>==Problem 7==<br /> There are positive integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; that satisfy the system of equations &lt;cmath&gt;<br /> \begin{align*}<br /> \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &amp;= 60\\<br /> \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &amp;= 570.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Let &lt;math&gt;m&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;x&lt;/math&gt;, and let &lt;math&gt;n&lt;/math&gt; be the number of (not necessarily distinct) prime factors in the prime factorization of &lt;math&gt;y&lt;/math&gt;. Find &lt;math&gt;3m+2n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Add the two equations to get that &lt;math&gt;\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Then, we use the theorem &lt;math&gt;\log a+\log b=\log ab&lt;/math&gt; to get the equation, &lt;math&gt;\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630&lt;/math&gt;.<br /> Using the theorem that &lt;math&gt;\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y&lt;/math&gt;, along with the previously mentioned theorem, we can get the equation &lt;math&gt;3\log(xy)=630&lt;/math&gt;.<br /> This can easily be simplified to &lt;math&gt;\log(xy)=210&lt;/math&gt;, or &lt;math&gt;xy = 10^{210}&lt;/math&gt;.<br /> <br /> &lt;math&gt;10^{210}&lt;/math&gt; can be factored into &lt;math&gt;2^{210} \cdot 5^{210}&lt;/math&gt;, and &lt;math&gt;m+n&lt;/math&gt; equals to the sum of the exponents of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, which is &lt;math&gt;210+210 = 420&lt;/math&gt;.<br /> Multiply by two to get &lt;math&gt;2m +2n&lt;/math&gt;, which is &lt;math&gt;840&lt;/math&gt;.<br /> Then, use the first equation (&lt;math&gt;\log x + 2\log(\gcd(x,y)) = 60&lt;/math&gt;) to show that &lt;math&gt;x&lt;/math&gt; has to have lower degrees of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; than &lt;math&gt;y&lt;/math&gt; (you can also test when &lt;math&gt;x&gt;y&lt;/math&gt;, which is a contradiction to the restrains you set before). Therefore, &lt;math&gt;\gcd(x,y)=x&lt;/math&gt;. Then, turn the equation into &lt;math&gt;3\log x = 60&lt;/math&gt;, which yields &lt;math&gt;\log x = 20&lt;/math&gt;, or &lt;math&gt;x = 10^{20}&lt;/math&gt;.<br /> Factor this into &lt;math&gt;2^{20} \cdot 5^{20}&lt;/math&gt;, and add the two 20's, resulting in &lt;math&gt;m&lt;/math&gt;, which is &lt;math&gt;40&lt;/math&gt;.<br /> Add &lt;math&gt;m&lt;/math&gt; to &lt;math&gt;2m + 2n&lt;/math&gt; (which is &lt;math&gt;840&lt;/math&gt;) to get &lt;math&gt;40+840 = \boxed{880}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Bashier Solution)==<br /> <br /> First simplifying the first and second equations, we get that <br /> <br /> &lt;cmath&gt;\log_{10}(x\cdot\text{gcd}(x,y)^2)=60&lt;/cmath&gt;<br /> &lt;cmath&gt;\log_{10}(y\cdot\text{lcm}(x,y)^2)=570&lt;/cmath&gt;<br /> <br /> <br /> Thus, when the two equations are added, we have that<br /> &lt;cmath&gt;\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630&lt;/cmath&gt;<br /> When simplified, this equals <br /> &lt;cmath&gt;\log_{10}(x^3y^3)=630&lt;/cmath&gt;<br /> so this means that<br /> &lt;cmath&gt;x^3y^3=10^{630}&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt;xy=10^{210}.&lt;/cmath&gt;<br /> <br /> Now, the following cannot be done on a proof contest but let's (intuitively) assume that &lt;math&gt;x&lt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both powers of &lt;math&gt;10&lt;/math&gt;. This means the first equation would simplify to &lt;cmath&gt;x^3=10^{60}&lt;/cmath&gt; and &lt;cmath&gt;y^3=10^{570}.&lt;/cmath&gt; Therefore, &lt;math&gt;x=10^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}&lt;/math&gt; and if we plug these values back, it works! &lt;math&gt;10^{20}&lt;/math&gt; has &lt;math&gt;20\cdot2=40&lt;/math&gt; total factors and &lt;math&gt;10^{190}&lt;/math&gt; has &lt;math&gt;190\cdot2=380&lt;/math&gt; so &lt;cmath&gt;3\cdot 40 + 2\cdot 380 = \boxed{880}.&lt;/cmath&gt;<br /> <br /> Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.<br /> <br /> ==Solution 3 (Easy Solution)==<br /> Let &lt;math&gt;x=10^a&lt;/math&gt; and &lt;math&gt;y=10^b&lt;/math&gt; and &lt;math&gt;a&lt;b&lt;/math&gt;. Then the given equations become &lt;math&gt;3a=60&lt;/math&gt; and &lt;math&gt;3b=570&lt;/math&gt;. Therefore, &lt;math&gt;x=10^{20}=2^{20}\cdot5^{20}&lt;/math&gt; and &lt;math&gt;y=10^{190}=2^{190}\cdot5^{190}&lt;/math&gt;. Our answer is &lt;math&gt;3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> ==Solution 4 ==<br /> We will use the notation &lt;math&gt;(a, b)&lt;/math&gt; for &lt;math&gt;\gcd(a, b)&lt;/math&gt; and &lt;math&gt;[a, b]&lt;/math&gt; as &lt;math&gt;\text{lcm}(a, b)&lt;/math&gt;. <br /> We can start with a similar way to Solution 1. We have, by logarithm properties, &lt;math&gt;\log_{10}{x}+\log_{10}{(x, y)^2}=60&lt;/math&gt; or &lt;math&gt;x(x, y)^2=10^{60}&lt;/math&gt;. We can do something similar to the second equation and our two equations become &lt;cmath&gt;x(x, y)^2=10^{60}&lt;/cmath&gt; &lt;cmath&gt;y[x, y]^2=10^{570}&lt;/cmath&gt;Adding the two equations gives us &lt;math&gt;xy(x, y)^2[x, y]^2=10^{630}&lt;/math&gt;. Since we know that &lt;math&gt;(a, b)\cdot[a, b]=ab&lt;/math&gt;, &lt;math&gt;x^3y^3=10^{630}&lt;/math&gt;, or &lt;math&gt;xy=10^{210}&lt;/math&gt;. We can express &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;2^a5^b&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;2^c5^d&lt;/math&gt;. Another way to express &lt;math&gt;(x, y)&lt;/math&gt; is now &lt;math&gt;2^{min(a, c)}5^{min(b, d)}&lt;/math&gt;, and &lt;math&gt;[x, y]&lt;/math&gt; is now &lt;math&gt;2^{max(a, c)}5^{max(b, d)}&lt;/math&gt;. We know that &lt;math&gt;x&lt;y&lt;/math&gt;, and thus, &lt;math&gt;a&lt;c&lt;/math&gt;, and &lt;math&gt;b&lt;d&lt;/math&gt;. Our equations for &lt;math&gt;lcm&lt;/math&gt; and &lt;math&gt;gcd&lt;/math&gt; now become &lt;cmath&gt;2^a5^b(2^a5^a)^2=10^{60}&lt;/cmath&gt; or &lt;math&gt;a=b=20&lt;/math&gt;. Doing the same for the &lt;math&gt;lcm&lt;/math&gt; equation, we have &lt;math&gt;c=d=190&lt;/math&gt;, and &lt;math&gt;190+20=210&lt;/math&gt;, which satisfies &lt;math&gt;xy=210&lt;/math&gt;. Thus, &lt;math&gt;3m+2n=3(20+20)+2(190+190)=\boxed{880}&lt;/math&gt;.<br /> ~awsomek<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_13&diff=127752 2014 AIME I Problems/Problem 13 2020-07-07T23:38:43Z <p>Hi13: /* Solution */</p> <hr /> <div>==Problem 13==<br /> On square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E,F,G&lt;/math&gt;, and &lt;math&gt;H&lt;/math&gt; lie on sides &lt;math&gt;\overline{AB},\overline{BC},\overline{CD},&lt;/math&gt; and &lt;math&gt;\overline{DA},&lt;/math&gt; respectively, so that &lt;math&gt;\overline{EG} \perp \overline{FH}&lt;/math&gt; and &lt;math&gt;EG=FH = 34&lt;/math&gt;. Segments &lt;math&gt;\overline{EG}&lt;/math&gt; and &lt;math&gt;\overline{FH}&lt;/math&gt; intersect at a point &lt;math&gt;P&lt;/math&gt;, and the areas of the quadrilaterals &lt;math&gt;AEPH, BFPE, CGPF,&lt;/math&gt; and &lt;math&gt;DHPG&lt;/math&gt; are in the ratio &lt;math&gt;269:275:405:411.&lt;/math&gt; Find the area of square &lt;math&gt;ABCD&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> pair A = (0,sqrt(850));<br /> pair B = (0,0);<br /> pair C = (sqrt(850),0);<br /> pair D = (sqrt(850),sqrt(850));<br /> draw(A--B--C--D--cycle);<br /> dotfactor = 3;<br /> dot(&quot;$A$&quot;,A,dir(135));<br /> dot(&quot;$B$&quot;,B,dir(215));<br /> dot(&quot;$C$&quot;,C,dir(305));<br /> dot(&quot;$D$&quot;,D,dir(45));<br /> pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850));<br /> pair F = ((2sqrt(850)+sqrt(306)+7)/6,0);<br /> dot(&quot;$H$&quot;,H,dir(90));<br /> dot(&quot;$F$&quot;,F,dir(270));<br /> draw(H--F);<br /> pair E = (0,(sqrt(850)-6)/2);<br /> pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2);<br /> dot(&quot;$E$&quot;,E,dir(180));<br /> dot(&quot;$G$&quot;,G,dir(0));<br /> draw(E--G);<br /> pair P = extension(H,F,E,G);<br /> dot(&quot;$P$&quot;,P,dir(60));<br /> label(&quot;$w$&quot;, intersectionpoint( A--P, E--H ));<br /> label(&quot;$x$&quot;, intersectionpoint( B--P, E--F ));<br /> label(&quot;$y$&quot;, intersectionpoint( C--P, G--F ));<br /> label(&quot;$z$&quot;, intersectionpoint( D--P, G--H ));&lt;/asy&gt;<br /> <br /> == Solution ==<br /> <br /> Notice that &lt;math&gt;269+411=275+405&lt;/math&gt;. This means &lt;math&gt;\overline{EG}&lt;/math&gt; passes through the center of the square. <br /> <br /> Draw &lt;math&gt;\overline{IJ} \parallel \overline{HF}&lt;/math&gt; with &lt;math&gt;I&lt;/math&gt; on &lt;math&gt;\overline{AD}&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt; such that &lt;math&gt;\overline{IJ}&lt;/math&gt; and &lt;math&gt;\overline{EG}&lt;/math&gt; intersects at the center of the square which I'll label as &lt;math&gt;O&lt;/math&gt;.<br /> <br /> Let the area of the square be &lt;math&gt;1360a&lt;/math&gt;. Then the area of &lt;math&gt;HPOI=71a&lt;/math&gt; and the area of &lt;math&gt;FPOJ=65a&lt;/math&gt;. This is because &lt;math&gt;\overline{HF}&lt;/math&gt; is perpendicular to &lt;math&gt;\overline{EG}&lt;/math&gt; (given in the problem), so &lt;math&gt;\overline{IJ}&lt;/math&gt; is also perpendicular to &lt;math&gt;\overline{EG}&lt;/math&gt;. These two orthogonal lines also pass through the center of the square, so they split it into 4 congruent quadrilaterals.<br /> <br /> Let the side length of the square be &lt;math&gt;d=\sqrt{1360a}&lt;/math&gt;. <br /> <br /> Draw &lt;math&gt;\overline{OK}\parallel \overline{HI}&lt;/math&gt; and intersects &lt;math&gt;\overline{HF}&lt;/math&gt; at &lt;math&gt;K&lt;/math&gt;. &lt;math&gt;OK=d\cdot\frac{[HFJI]}{[ABCD]}=\frac{d}{10}&lt;/math&gt;. <br /> <br /> The area of &lt;math&gt;HKOI=\frac12\cdot HFJI=68a&lt;/math&gt;, so the area of &lt;math&gt;POK=3a&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;\overline{PO}=h&lt;/math&gt;. Then &lt;math&gt;KP=\frac{6a}{h}&lt;/math&gt;<br /> <br /> Consider the area of &lt;math&gt;PFJO&lt;/math&gt;. <br /> &lt;cmath&gt;\frac12(PF+OJ)(PO)=65a&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(17-\frac{3a}{h}\right)h=65a&lt;/cmath&gt;<br /> &lt;cmath&gt;h=4a&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;KP=1.5&lt;/math&gt;.<br /> <br /> Solving &lt;math&gt;(4a)^2+1.5^2=\left(\frac{d}{10}\right)^2=13.6a&lt;/math&gt;, we get &lt;math&gt;a=\frac58&lt;/math&gt;.<br /> <br /> Therefore, the area of &lt;math&gt;ABCD=1360a=\boxed{850}&lt;/math&gt;<br /> <br /> ==Lazy Solution==<br /> &lt;math&gt;269+275+405+411=1360&lt;/math&gt;, a multiple of &lt;math&gt;17&lt;/math&gt;. In addition, &lt;math&gt;EG=FH=34&lt;/math&gt;, which is &lt;math&gt;17\cdot 2&lt;/math&gt;.<br /> Therefore, we suspect the square of the &quot;hypotenuse&quot; of a right triangle, corresponding to &lt;math&gt;EG&lt;/math&gt; and &lt;math&gt;FH&lt;/math&gt; must be a multiple of &lt;math&gt;17&lt;/math&gt;. All of these triples are primitive:<br /> <br /> &lt;cmath&gt;17=1^2+4^2&lt;/cmath&gt;<br /> &lt;cmath&gt;34=3^2+5^2&lt;/cmath&gt;<br /> &lt;cmath&gt;51=\emptyset&lt;/cmath&gt;<br /> &lt;cmath&gt;68=\emptyset\text{ others}&lt;/cmath&gt;<br /> &lt;cmath&gt;85=2^2+9^2=6^2+7^2&lt;/cmath&gt;<br /> &lt;cmath&gt;102=\emptyset&lt;/cmath&gt;<br /> &lt;cmath&gt;119=\emptyset \dots&lt;/cmath&gt;<br /> <br /> The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting &lt;math&gt;EG=FH=34&lt;/math&gt;:<br /> &lt;cmath&gt;\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}&lt;/cmath&gt;<br /> <br /> Thus, &lt;math&gt;\boxed{850}&lt;/math&gt; is the only valid answer.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_I_Problems/Problem_10&diff=127750 2014 AIME I Problems/Problem 10 2020-07-07T23:31:38Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem 10 ==<br /> A disk with radius &lt;math&gt;1&lt;/math&gt; is externally tangent to a disk with radius &lt;math&gt;5&lt;/math&gt;. Let &lt;math&gt;A&lt;/math&gt; be the point where the disks are tangent, &lt;math&gt;C&lt;/math&gt; be the center of the smaller disk, and &lt;math&gt;E&lt;/math&gt; be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of &lt;math&gt;360^\circ&lt;/math&gt;. That is, if the center of the smaller disk has moved to the point &lt;math&gt;D&lt;/math&gt;, and the point on the smaller disk that began at &lt;math&gt;A&lt;/math&gt; has now moved to point &lt;math&gt;B&lt;/math&gt;, then &lt;math&gt;\overline{AC}&lt;/math&gt; is parallel to &lt;math&gt;\overline{BD}&lt;/math&gt;. Then &lt;math&gt;\sin^2(\angle BEA)=\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> <br /> <br /> &lt;asy&gt;<br /> size(150);<br /> pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0);<br /> draw(circle(e,5));<br /> draw(circle(c,1));<br /> draw(circle(d,1));<br /> dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2));<br /> label(&quot;$A$&quot;,a,W,fontsize(9));<br /> label(&quot;$B$&quot;,b,NW,fontsize(9));<br /> label(&quot;$C$&quot;,c,E,fontsize(9));<br /> label(&quot;$D$&quot;,d,E,fontsize(9));<br /> label(&quot;$E$&quot;,e,SW,fontsize(9));<br /> label(&quot;$F$&quot;,(5/2,5*sqrt(3)/2),SSW,fontsize(9));<br /> &lt;/asy&gt;<br /> <br /> Let &lt;math&gt;F&lt;/math&gt; be the new tangency point of the two disks. The smaller disk rolled along minor arc &lt;math&gt;\overarc{AF}&lt;/math&gt; on the larger disk. Let &lt;math&gt;\alpha = \angle AEF&lt;/math&gt;, in radians. The smaller disk must then have rolled along an arc of length &lt;math&gt;5\alpha&lt;/math&gt;, since the larger disk has a radius of &lt;math&gt;5&lt;/math&gt;. Since all of the points on major arc &lt;math&gt;\overarc{BF}&lt;/math&gt; on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did, the length of major arc &lt;math&gt;\overarc{BF}&lt;/math&gt; equals the length of minor arc &lt;math&gt;\overarc{AF}&lt;/math&gt;, or &lt;math&gt;5\alpha&lt;/math&gt;. Since &lt;math&gt;\overline{AC} || \overline{BD}&lt;/math&gt;, &lt;math&gt;\angle BDF \cong \angle FEA&lt;/math&gt;, so the angles of minor arc &lt;math&gt;\overarc{BF}&lt;/math&gt; and minor arc &lt;math&gt;\overarc{AF}&lt;/math&gt; are equal, so minor arc &lt;math&gt;\overarc{BF}&lt;/math&gt; has an angle of &lt;math&gt;\alpha&lt;/math&gt;. Since the smaller disk has a radius of &lt;math&gt;1&lt;/math&gt;, the length of minor arc &lt;math&gt;\overarc{BF}&lt;/math&gt; is &lt;math&gt;\alpha&lt;/math&gt;. This means that &lt;math&gt;5\alpha + \alpha&lt;/math&gt; equals the circumference of the smaller disk, so &lt;math&gt;6\alpha = 2\pi&lt;/math&gt;, or &lt;math&gt;\alpha = \frac{\pi}{3}&lt;/math&gt;.<br /> <br /> Now, to find &lt;math&gt;\sin^2{\angle BEA}&lt;/math&gt;, we construct &lt;math&gt;\triangle BDE&lt;/math&gt;. Also, drop a perpendicular from &lt;math&gt;D&lt;/math&gt; to &lt;math&gt;\overline{EA}&lt;/math&gt;, and call this point &lt;math&gt;X&lt;/math&gt;. Since &lt;math&gt;\alpha = \frac{\pi}{3}&lt;/math&gt; and &lt;math&gt;\angle DXE&lt;/math&gt; is right, &lt;math&gt;DE = 6&lt;/math&gt;, &lt;math&gt;EX = 3&lt;/math&gt; and &lt;math&gt;DX = 3\sqrt{3}&lt;/math&gt;. Now drop a perpendicular from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{EA}&lt;/math&gt;, and call this point &lt;math&gt;Y&lt;/math&gt;. Since &lt;math&gt;\overline{BD} || \overline{EA}&lt;/math&gt;, &lt;math&gt;XY = BD = 1&lt;/math&gt;, and &lt;math&gt;BY = DX = 3\sqrt{3}&lt;/math&gt;. Thus, we know that &lt;math&gt;EY = EX - XY = 3 - 1 = 2&lt;/math&gt;, and by using the Pythagorean Theorem on &lt;math&gt;\triangle BEY&lt;/math&gt;, we get that &lt;math&gt;BE = \sqrt{31}&lt;/math&gt;. Thus, &lt;math&gt;\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}&lt;/math&gt;, so &lt;math&gt;\sin^2{\angle BEA} = \frac{27}{31}&lt;/math&gt;, and our answer is &lt;math&gt;27 + 31 = \boxed{058}&lt;/math&gt;.<br /> <br /> == Solution 2 (Trigonometry) ==<br /> <br /> First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate &lt;math&gt;5&lt;/math&gt; times (the circumference of the small circle is &lt;math&gt;2\pi&lt;/math&gt; while the larger one has a circumference of &lt;math&gt;10\pi&lt;/math&gt;) plus the extra rotation the circle gets for rotating around the circle, for a total of &lt;math&gt;6&lt;/math&gt; times. Therefore, one rotation will bring point &lt;math&gt;D&lt;/math&gt; &lt;math&gt;60^\circ&lt;/math&gt; from &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Now, draw &lt;math&gt;\triangle DBE&lt;/math&gt;, and call &lt;math&gt;\angle BED&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;, in degrees. We know that &lt;math&gt;\overline{ED}&lt;/math&gt; is 6, and &lt;math&gt;\overline{BD}&lt;/math&gt; is 1. Since &lt;math&gt;EC ||<br /> BD&lt;/math&gt;, &lt;math&gt;\angle BDE = 60^\circ&lt;/math&gt;. By the Law of Cosines, &lt;math&gt;\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31&lt;/math&gt;, and since lengths are positive, &lt;math&gt;\overline{BE}=\sqrt{31}&lt;/math&gt;. <br /> <br /> By the Law of Sines, we know that &lt;math&gt;\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}&lt;/math&gt;, so &lt;math&gt;\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}&lt;/math&gt;. As &lt;math&gt;x&lt;/math&gt; is clearly between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;90^\circ&lt;/math&gt;, &lt;math&gt;\cos{x}&lt;/math&gt; is positive. As &lt;math&gt;\cos{x}=\sqrt{1-\sin^2{x}}&lt;/math&gt;, &lt;math&gt;\cos{x} = \frac{11\sqrt{31}}{62}&lt;/math&gt;. <br /> <br /> Now we use the angle sum formula to find the sine of &lt;math&gt;\angle BEA&lt;/math&gt;: &lt;math&gt;\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62} = \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}&lt;/math&gt;. <br /> <br /> Finally, we square this to get &lt;math&gt;\frac{9\times 3}{31}=\frac{27}{31}&lt;/math&gt;, so our answer is &lt;math&gt;27+31=\boxed{058}&lt;/math&gt;.<br /> <br /> ==Condensed Solution==<br /> Convince yourself that the small circle actually only rotates &lt;math&gt;\frac{1}{1+5}=\frac{1}{6}&lt;/math&gt; of the way by looking at tangents in various locations with respect to the small circle's center. Then, use a coordinate system where &lt;math&gt;E=(0, 0), A=(5, 0)&lt;/math&gt;, and the circle rotates up in Quadrant I, and plot the other points. You see that &lt;math&gt;BD=1, DE=6, BE=\sqrt{2^2+(3\sqrt{3})^2}&lt;/math&gt;. From the Law of Cosines for an angle, see that angle &lt;math&gt;BED&lt;/math&gt; is &lt;math&gt;\arccos(\frac{11}{2\sqrt{31}}) = \arcsin(\frac{\sqrt{3}}{2\sqrt{31}})&lt;/math&gt;. Using the angle addition for &lt;math&gt;\sin&lt;/math&gt;, you get that &lt;math&gt;\sin(BEA)=\frac{3\sqrt{3}}{\sqrt{31}}&lt;/math&gt;, meaning your answer is &lt;math&gt;27 + 31 = \boxed{058}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2014|n=I|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=127467 2011 AIME I Problems/Problem 14 2020-07-04T21:16:25Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> <br /> <br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n-k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\cos\theta-\sin\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> == Solution 4 ==<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> O2=(A+E)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> label(&quot;$O$&quot;,O2,dir(152.5));<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> draw(O2--(W+X)/2,red);<br /> draw(O2--N,red);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_2$&quot;,(W+X)/2,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&diff=127466 2011 AIME I Problems/Problem 14 2020-07-04T21:14:36Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; be a regular octagon. Let &lt;math&gt;M_1&lt;/math&gt;, &lt;math&gt;M_3&lt;/math&gt;, &lt;math&gt;M_5&lt;/math&gt;, and &lt;math&gt;M_7&lt;/math&gt; be the midpoints of sides &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt;, &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt;, &lt;math&gt;\overline{A_5 A_6}&lt;/math&gt;, and &lt;math&gt;\overline{A_7 A_8}&lt;/math&gt;, respectively. For &lt;math&gt;i = 1, 3, 5, 7&lt;/math&gt;, ray &lt;math&gt;R_i&lt;/math&gt; is constructed from &lt;math&gt;M_i&lt;/math&gt; towards the interior of the octagon such that &lt;math&gt;R_1 \perp R_3&lt;/math&gt;, &lt;math&gt;R_3 \perp R_5&lt;/math&gt;, &lt;math&gt;R_5 \perp R_7&lt;/math&gt;, and &lt;math&gt;R_7 \perp R_1&lt;/math&gt;. Pairs of rays &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_3&lt;/math&gt;, &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_5&lt;/math&gt;, &lt;math&gt;R_5&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt;, and &lt;math&gt;R_7&lt;/math&gt; and &lt;math&gt;R_1&lt;/math&gt; meet at &lt;math&gt;B_1&lt;/math&gt;, &lt;math&gt;B_3&lt;/math&gt;, &lt;math&gt;B_5&lt;/math&gt;, &lt;math&gt;B_7&lt;/math&gt; respectively. If &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, then &lt;math&gt;\cos 2 \angle A_3 M_3 B_1&lt;/math&gt; can be written in the form &lt;math&gt;m - \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution 1==<br /> &lt;asy&gt;<br /> size(200);<br /> defaultpen(linewidth(0.8));<br /> real dif = 45;<br /> pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15;<br /> pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2;<br /> pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1));<br /> draw(M1--B1^^M3--B3^^M5--B5^^M7--B7);<br /> draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle);<br /> &lt;/asy&gt;<br /> <br /> <br /> Let &lt;math&gt;\theta=\angle M_1 M_3 B_1&lt;/math&gt;. Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8&lt;/math&gt; is a regular octagon and &lt;math&gt;B_1 B_3 = A_1 A_2&lt;/math&gt;, let &lt;math&gt;k=A_1 A_2 = A_2 A_3 = B_1 B_3&lt;/math&gt;.<br /> <br /> <br /> Extend &lt;math&gt;\overline{A_1 A_2}&lt;/math&gt; and &lt;math&gt;\overline{A_3 A_4}&lt;/math&gt; until they intersect. Denote their intersection as &lt;math&gt;I_1&lt;/math&gt;. Through similar triangles &amp; the &lt;math&gt;45-45-90&lt;/math&gt; triangles formed, we find that &lt;math&gt;M_1 M_3=\frac{k}{2}(2+\sqrt2)&lt;/math&gt;.<br /> <br /> We also have that&lt;math&gt;\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3&lt;/math&gt; through ASA congruence (&lt;math&gt;\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3&lt;/math&gt;, &lt;math&gt;M_7 M_1 = M_1 M_3&lt;/math&gt;, &lt;math&gt;\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1&lt;/math&gt;). Therefore, we may let &lt;math&gt;n=M_1 B_7 = M_3 B_1&lt;/math&gt;.<br /> <br /> Thus, we have that &lt;math&gt;\sin\theta=\frac{n-k}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt; and that &lt;math&gt;\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}&lt;/math&gt;. Therefore &lt;math&gt;\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2&lt;/math&gt;.<br /> <br /> Squaring gives that &lt;math&gt;\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2&lt;/math&gt; and consequently that &lt;math&gt;-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta&lt;/math&gt; through the identities &lt;math&gt;\sin^2\theta + \cos^2\theta = 1&lt;/math&gt; and &lt;math&gt;\sin2\theta = 2\sin\theta\cos\theta&lt;/math&gt;.<br /> <br /> Thus we have that &lt;math&gt;\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;. Therefore &lt;math&gt;m+n=5+32=\boxed{037}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;A_1A_2 = 2&lt;/math&gt;. Then &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; are the projections of &lt;math&gt;M_1&lt;/math&gt; and &lt;math&gt;M_5&lt;/math&gt; onto the line &lt;math&gt;B_1B_3&lt;/math&gt;, so &lt;math&gt;2=B_1B_3=-M_1M_5\cos x&lt;/math&gt;, where &lt;math&gt;x = \angle A_3M_3B_1&lt;/math&gt;. Then since &lt;math&gt;M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}&lt;/math&gt;,&lt;math&gt;\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> == Solution 3 ==<br /> Notice that &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; are parallel (&lt;math&gt;B_1B_3B_5B_7&lt;/math&gt; is a square by symmetry and since the rays are perpendicular) and &lt;math&gt;B_1B_3=B_3B_5=s=&lt;/math&gt; the distance between the parallel rays. If the regular hexagon as a side length of &lt;math&gt;s&lt;/math&gt;, then &lt;math&gt;M_3M_7&lt;/math&gt; has a length of &lt;math&gt;s+s\sqrt{2}&lt;/math&gt;. Let &lt;math&gt;X&lt;/math&gt; be on &lt;math&gt;R_3&lt;/math&gt; such that &lt;math&gt;M_7X&lt;/math&gt; is perpendicular to &lt;math&gt;M_3X&lt;/math&gt;, and &lt;math&gt;\phi=\angle M_7M_3X&lt;/math&gt;. The distance between &lt;math&gt;R_3&lt;/math&gt; and &lt;math&gt;R_7&lt;/math&gt; is &lt;math&gt;s=M_7X&lt;/math&gt;, so &lt;math&gt;\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}&lt;/math&gt;. <br /> <br /> Since we are considering a regular hexagon, &lt;math&gt;M_3&lt;/math&gt; is directly opposite to &lt;math&gt;M_7&lt;/math&gt; and &lt;math&gt;\angle A_3M_3B_1=90 ^\circ +\phi&lt;/math&gt;. All that's left is to calculate &lt;math&gt;\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi&lt;/math&gt;. By drawing a right triangle or using the Pythagorean identity, &lt;math&gt;\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}&lt;/math&gt; and &lt;math&gt;\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}&lt;/math&gt;, so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> <br /> == Solution 4 ==<br /> Assume that &lt;math&gt;A_1A_2=1.&lt;/math&gt;<br /> Denote the center &lt;math&gt;O&lt;/math&gt;, and the midpoint of &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;B_3&lt;/math&gt; as &lt;math&gt;B_2&lt;/math&gt;. Then we have that&lt;cmath&gt;\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.&lt;/cmath&gt;Thus, by the cosine double-angle theorem,&lt;cmath&gt;\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},&lt;/cmath&gt;so &lt;math&gt;m+n=\boxed{037}&lt;/math&gt;.<br /> <br /> ==Diagram==<br /> &lt;asy&gt;<br /> size(250);<br /> pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z;<br /> A=(-76.537,184.776);<br /> B=(76.537,184.776);<br /> C=(184.776,76.537);<br /> D=(184.776,-76.537);<br /> E=(76.537,-184.776);<br /> F=(-76.537,-184.776);<br /> G=(-184.776,-76.537);<br /> H=(-184.776,76.537);<br /> M=(A+B)/2;<br /> N=(C+D)/2;<br /> O=(E+F)/2;<br /> O2=(A+E)/2;<br /> P=(G+H)/2;<br /> W=(100,-41.421);<br /> X=(-41.421,-100);<br /> Y=(-100,41.421);<br /> Z=(41.421,100);<br /> draw(A--B--C--D--E--F--G--H--A);<br /> label(&quot;$A_1$&quot;,A,dir(112.5));<br /> label(&quot;$A_2$&quot;,B,dir(67.5));<br /> label(&quot;$\textcolor{blue}{A_3}$&quot;,C,dir(22.5));<br /> label(&quot;$A_4$&quot;,D,dir(337.5));<br /> label(&quot;$A_5$&quot;,E,dir(292.5));<br /> label(&quot;$A_6$&quot;,F,dir(247.5));<br /> label(&quot;$A_7$&quot;,G,dir(202.5));<br /> label(&quot;$A_8$&quot;,H,dir(152.5));<br /> label(&quot;$M_1$&quot;,M,dir(90));<br /> label(&quot;$\textcolor{blue}{M_3}$&quot;,N,dir(0));<br /> label(&quot;$M_5$&quot;,O,dir(270));<br /> label(&quot;$M_7$&quot;,P,dir(180));<br /> label(&quot;$O$&quot;,O2,dir(152.5));<br /> draw(M--W,red);<br /> draw(N--X,red);<br /> draw(O--Y,red);<br /> draw(P--Z,red);<br /> draw(O2--(W+X)/2,red);<br /> draw(O2--N,red);<br /> label(&quot;$\textcolor{blue}{B_1}$&quot;,W,dir(292.5));<br /> label(&quot;$B_2$&quot;,(W+X)/2,dir(292.5));<br /> label(&quot;$B_3$&quot;,X,dir(202.5));<br /> label(&quot;$B_5$&quot;,Y,dir(112.5));<br /> label(&quot;$B_7$&quot;,Z,dir(22.5));<br /> &lt;/asy&gt;<br /> All distances are to scale.<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=127462 2011 AIME I Problems/Problem 7 2020-07-04T19:14:57Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem 7 ==<br /> Find the number of positive integers &lt;math&gt;m&lt;/math&gt; for which there exist nonnegative integers &lt;math&gt;x_0&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; , &lt;math&gt;\dots&lt;/math&gt; , &lt;math&gt;x_{2011}&lt;/math&gt; such that<br /> &lt;cmath&gt;m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt; m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}&lt;/math&gt;. Now, divide by &lt;math&gt;m^{x_0}&lt;/math&gt; to get &lt;math&gt;1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}&lt;/math&gt;. Notice that since we can choose all nonnegative &lt;math&gt;x_0,...,x_{2011}&lt;/math&gt;, we can make &lt;math&gt;x_n-x_0&lt;/math&gt; whatever we desire. WLOG, let &lt;math&gt;x_0\geq...\geq x_{2011}&lt;/math&gt; and let &lt;math&gt;a_n=x_n-x_0&lt;/math&gt;. Notice that, also, &lt;math&gt;m^{a_{2011}}&lt;/math&gt; doesn't matter if we are able to make &lt;math&gt; m^{a_1} +m^{a_2} + .... + m^{a_{2010}}&lt;/math&gt; equal to &lt;math&gt;1-\left(\frac{1}{m}\right)^x&lt;/math&gt; for any power of &lt;math&gt;x&lt;/math&gt;. Consider &lt;math&gt;m=2&lt;/math&gt;. We can achieve a sum of &lt;math&gt;1-\left(\frac{1}{2}\right)^x&lt;/math&gt; by doing &lt;math&gt;\frac{1}{2}+\frac{1}{4}+...&lt;/math&gt; (the &quot;simplest&quot; sequence). If we don't have &lt;math&gt;\frac{1}{2}&lt;/math&gt;, to compensate, we need &lt;math&gt;2\cdot 1\frac{1}{4}&lt;/math&gt;'s. Now, let's try to generalize. The &quot;simplest&quot; sequence is having &lt;math&gt;\frac{1}{m}&lt;/math&gt; &lt;math&gt;m-1&lt;/math&gt; times, &lt;math&gt;\frac{1}{m^2}&lt;/math&gt; &lt;math&gt;m-1&lt;/math&gt; times, &lt;math&gt;\ldots&lt;/math&gt;. To make other sequences, we can split &lt;math&gt;m-1&lt;/math&gt; &lt;math&gt;\frac{1}{m^i}&lt;/math&gt;s into &lt;math&gt;m(m-1)&lt;/math&gt; &lt;math&gt;\frac{1}{m^{i+1}}&lt;/math&gt;s since &lt;math&gt;m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}&lt;/math&gt;. Since we want &lt;math&gt;2010&lt;/math&gt; terms, we have &lt;math&gt;\sum&lt;/math&gt; &lt;math&gt;(m-1)\cdot m^x=2010&lt;/math&gt;. However, since we can set &lt;math&gt;x&lt;/math&gt; to be anything we want (including 0), all we care about is that &lt;math&gt;m-1 | 2010&lt;/math&gt; which happens &lt;math&gt;\boxed{016}&lt;/math&gt; times.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}&lt;/math&gt;. The problem then becomes finding the number of positive integer roots &lt;math&gt;m&lt;/math&gt; for which &lt;math&gt;P(m) = 0&lt;/math&gt; and &lt;math&gt;x_0, x_1, ..., x_{2011}&lt;/math&gt; are nonnegative integers. We plug in &lt;math&gt;m = 1&lt;/math&gt; and see that &lt;math&gt;P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010&lt;/math&gt;. Now, we can say that &lt;math&gt;P(m) = (m-1)Q(m) - 2010&lt;/math&gt; for some polynomial &lt;math&gt;Q(m)&lt;/math&gt; with integer coefficients. Then if &lt;math&gt;P(m) = 0&lt;/math&gt;, &lt;math&gt;(m-1)Q(m) = 2010&lt;/math&gt;. Thus, if &lt;math&gt;P(m) = 0&lt;/math&gt;, then &lt;math&gt;m-1 | 2010&lt;/math&gt; .<br /> Now, we need to show that for all &lt;math&gt;m-1 | 2010&lt;/math&gt;, &lt;math&gt; m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. &lt;/math&gt;. We try with the first few &lt;math&gt;m&lt;/math&gt; that satisfy this.<br /> For &lt;math&gt;m = 2&lt;/math&gt;, we see we can satisfy this if &lt;math&gt;x_0 = 2010&lt;/math&gt;, &lt;math&gt;x_1 = 2009&lt;/math&gt;, &lt;math&gt;x_2 = 2008&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt; , &lt;math&gt;x_{2008} = 2&lt;/math&gt;, &lt;math&gt;x_{2009} = 1&lt;/math&gt;, &lt;math&gt; x_{2010} = 0&lt;/math&gt;, &lt;math&gt;x_{2011} = 0&lt;/math&gt;, because &lt;math&gt;2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots&lt;/math&gt; (based on the idea &lt;math&gt;2^n + 2^n = 2^{n+1}&lt;/math&gt;, leading to a chain of substitutions of this kind) &lt;math&gt;= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}&lt;/math&gt;. Thus &lt;math&gt;2&lt;/math&gt; is a possible value of &lt;math&gt;m&lt;/math&gt;. For other values, for example &lt;math&gt;m = 3&lt;/math&gt;, we can use the same strategy, with &lt;math&gt;x_{2011} = x_{2010} = x_{2009} = 0&lt;/math&gt;, &lt;math&gt;x_{2008} = x_{2007} = 1&lt;/math&gt;, &lt;math&gt;x_{2006} = x_{2005} = 2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;x_2 = x_1 = 1004&lt;/math&gt; and &lt;math&gt;x_0 = 1005&lt;/math&gt;, because <br /> &lt;math&gt;3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004} = \cdots&lt;/math&gt;<br /> &lt;math&gt;=3^{1004} +3^{1004}+3^{1004} = 3^{1005}&lt;/math&gt;.<br /> It's clearly seen we can use the same strategy for all &lt;math&gt;m-1 |2010&lt;/math&gt;. We count all positive &lt;math&gt;m&lt;/math&gt; satisfying &lt;math&gt;m-1 |2010&lt;/math&gt;, and see there are &lt;math&gt;\boxed{016}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> One notices that &lt;math&gt;m-1 \mid 2010&lt;/math&gt; if and only if there exist non-negative integers &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt; such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;.<br /> <br /> To prove the forward case, we proceed by directly finding &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt;. Suppose &lt;math&gt;m&lt;/math&gt; is an integer such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;. We will count how many &lt;math&gt;x_k = 0&lt;/math&gt;, how many &lt;math&gt;x_k = 1&lt;/math&gt;, etc. Suppose the number of &lt;math&gt;x_k = 0&lt;/math&gt; is non-zero. Then, there must be at least &lt;math&gt;m&lt;/math&gt; such &lt;math&gt;x_k&lt;/math&gt; since &lt;math&gt;m&lt;/math&gt; divides all the remaining terms, so &lt;math&gt;m&lt;/math&gt; must also divide the sum of all the &lt;math&gt;m^0&lt;/math&gt; terms. Thus, if we let &lt;math&gt;x_k = 0&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,m&lt;/math&gt;, we have,<br /> &lt;cmath&gt;m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> Well clearly, &lt;math&gt;m^{x_0}&lt;/math&gt; is greater than &lt;math&gt;m&lt;/math&gt;, so &lt;math&gt;m^2 \mid m^{x_0}&lt;/math&gt;. &lt;math&gt;m^2&lt;/math&gt; will also divide every term, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k \geq 2&lt;/math&gt;. So, all the terms, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k &lt; 2&lt;/math&gt; must sum to a multiple of &lt;math&gt;m^2&lt;/math&gt;. If there are exactly &lt;math&gt;m&lt;/math&gt; terms where &lt;math&gt;x_k = 0&lt;/math&gt;, then we must have at least &lt;math&gt;m-1&lt;/math&gt; terms where &lt;math&gt;x_k = 1&lt;/math&gt;. Suppose there are exactly &lt;math&gt;m-1&lt;/math&gt; such terms and &lt;math&gt;x_k = 1&lt;/math&gt; for &lt;math&gt;k = m+1,m+2,2m-1&lt;/math&gt;. Now, we have,<br /> &lt;cmath&gt;m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> One can repeat this process for successive powers of &lt;math&gt;m&lt;/math&gt; until the number of terms reaches 2011. Since there are &lt;math&gt;m + j(m-1)&lt;/math&gt; terms after the &lt;math&gt;j&lt;/math&gt;th power, we will only hit exactly 2011 terms if &lt;math&gt;m-1&lt;/math&gt; is a factor of 2010. To see this, <br /> &lt;cmath&gt;m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.&lt;/cmath&gt;<br /> Thus, when &lt;math&gt;j = 2010/(m-1) - 1&lt;/math&gt; (which is an integer since &lt;math&gt;m-1 \mid 2010&lt;/math&gt; by assumption, there are exactly 2011 terms. To see that these terms sum to a power of &lt;math&gt;m&lt;/math&gt;, we realize that the sum is a geometric series:<br /> &lt;cmath&gt;1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.&lt;/cmath&gt;<br /> Thus, we have found a solution for the case &lt;math&gt;m-1 \mid 2010&lt;/math&gt;.<br /> <br /> Now, for the reverse case, we use the formula &lt;cmath&gt;x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).&lt;/cmath&gt; Suppose &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x^k}&lt;/math&gt; has a solution. Subtract 2011 from both sides to get &lt;cmath&gt;m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).&lt;/cmath&gt; Now apply the formula to get &lt;cmath&gt;(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],&lt;/cmath&gt; where &lt;math&gt;a_k&lt;/math&gt; are some integers. Rearranging this equation, we find &lt;cmath&gt;(m-1)A = 2010,&lt;/cmath&gt; where &lt;math&gt;A = a_0 - \sum_{k=1}^{2011}a_k&lt;/math&gt;. Thus, if &lt;math&gt;m&lt;/math&gt; is a solution, then &lt;math&gt;m-1 \mid 2010&lt;/math&gt;. <br /> <br /> So, there is one positive integer solution corresponding to each factor of 2010. Since &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;, the number of solutions is &lt;math&gt;2^4 = \boxed{016}&lt;/math&gt;.<br /> <br /> ==Solution 4 (for noobs like me)==<br /> The problem is basically asking how many integers &lt;math&gt;m&lt;/math&gt; have a power that can be expressed as the sum of 2011 other powers of &lt;math&gt;m&lt;/math&gt; (not necessarily distinct). Notice that &lt;math&gt;2+2+4+8+16=32&lt;/math&gt;, &lt;math&gt;3+3+3+9+9+27+27+81+81=243&lt;/math&gt;, and &lt;math&gt;4+4+4+4+16+16+16+64+64+64+256+256+256=1024&lt;/math&gt;. Thus, we can safely assume that the equation &lt;math&gt;2011 = (m-1)x + m&lt;/math&gt; must have an integer solution &lt;math&gt;x&lt;/math&gt;. To find the number of &lt;math&gt;m&lt;/math&gt;-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant &lt;math&gt;m&lt;/math&gt; to make the equation equal a friendlier number, &lt;math&gt;2010&lt;/math&gt;, instead of the ugly prime number &lt;math&gt;2011&lt;/math&gt;: &lt;math&gt;2010 = (m-1)x+(m-1)&lt;/math&gt;. Factor the equation and we get &lt;math&gt;2010 = (m-1)(x+1)&lt;/math&gt;. The number of values of &lt;math&gt;m-1&lt;/math&gt; that allow &lt;math&gt;x+1&lt;/math&gt; to be an integer is quite obviously the number of factors of &lt;math&gt;2010&lt;/math&gt;. Factoring &lt;math&gt;2010&lt;/math&gt;, we obtain &lt;math&gt;2010 = 2 \times 3 \times 5 \times 67&lt;/math&gt;, so the number of positive integers &lt;math&gt;m&lt;/math&gt; that satisfy the required condition is &lt;math&gt;2^4 = \boxed{016}&lt;/math&gt;.<br /> <br /> -fidgetboss_4000<br /> <br /> ==Solution 5==<br /> <br /> First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of &lt;math&gt;m&lt;/math&gt; we can multiply to get to non-negative powers.<br /> Now we see that our problem is just a matter of &lt;math&gt;&quot;m&quot;&lt;/math&gt;-chopping blocks.<br /> What is meant by &lt;math&gt;m&lt;/math&gt;-chopping is taking an existing block of say &lt;math&gt;m^{k}&lt;/math&gt; and turning it into &lt;math&gt;m&lt;/math&gt; blocks of &lt;math&gt;m^{k-1}&lt;/math&gt;. This process increases the total # of blocks by &lt;math&gt;m-1&lt;/math&gt; per chop.<br /> The problem wants us to find the number of positive integers &lt;math&gt;m&lt;/math&gt; where some number of chops will turn &lt;math&gt;1&lt;/math&gt; block into &lt;math&gt;2011&lt;/math&gt; such blocks, thus increasing the total amount by &lt;math&gt;2010= 2 \cdot 3 \cdot 5 \cdot 67&lt;/math&gt;. Thus &lt;math&gt;m-1 | 2010&lt;/math&gt;, and a cursory check on extreme cases will confirm that there are indeed &lt;math&gt;\boxed{016}&lt;/math&gt; possible &lt;math&gt;m&lt;/math&gt;s.<br /> <br /> ==An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)==<br /> https://artofproblemsolving.com/community/c6h84155p486903<br /> 2001 Austrian-Polish Math Individual Competition #1<br /> ~MSC<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=6|num-a=8}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=127461 2011 AIME I Problems/Problem 7 2020-07-04T19:13:44Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem 7 ==<br /> Find the number of positive integers &lt;math&gt;m&lt;/math&gt; for which there exist nonnegative integers &lt;math&gt;x_0&lt;/math&gt;, &lt;math&gt;x_1&lt;/math&gt; , &lt;math&gt;\dots&lt;/math&gt; , &lt;math&gt;x_{2011}&lt;/math&gt; such that<br /> &lt;cmath&gt;m^{x_0} = \sum_{k = 1}^{2011} m^{x_k}.&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> &lt;math&gt; m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}&lt;/math&gt;. Now, divide by &lt;math&gt;m^{x_0}&lt;/math&gt; to get &lt;math&gt;1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}&lt;/math&gt;. Notice that since we can choose all nonnegative &lt;math&gt;x_0,...,x_{2010}&lt;/math&gt;, we can make &lt;math&gt;x_n-x_0&lt;/math&gt; whatever we desire. WLOG, let &lt;math&gt;x_0\geq...\geq x_{2011}&lt;/math&gt; and let &lt;math&gt;a_n=x_n-x_0&lt;/math&gt;. Notice that, also, &lt;math&gt;m^{a_{2011}}&lt;/math&gt; doesn't matter if we are able to make &lt;math&gt; m^{a_1} +m^{a_2} + .... + m^{a_{2011}}&lt;/math&gt; equal to &lt;math&gt;1-\left(\frac{1}{m}\right)^x&lt;/math&gt; for any power of &lt;math&gt;x&lt;/math&gt;. Consider &lt;math&gt;m=2&lt;/math&gt;. We can achieve a sum of &lt;math&gt;1-\left(\frac{1}{2}\right)^x&lt;/math&gt; by doing &lt;math&gt;\frac{1}{2}+\frac{1}{4}+...&lt;/math&gt; (the &quot;simplest&quot; sequence). If we don't have &lt;math&gt;\frac{1}{2}&lt;/math&gt;, to compensate, we need &lt;math&gt;2\cdot 1\frac{1}{4}&lt;/math&gt;'s. Now, let's try to generalize. The &quot;simplest&quot; sequence is having &lt;math&gt;\frac{1}{m}&lt;/math&gt; &lt;math&gt;m-1&lt;/math&gt; times, &lt;math&gt;\frac{1}{m^2}&lt;/math&gt; &lt;math&gt;m-1&lt;/math&gt; times, &lt;math&gt;\ldots&lt;/math&gt;. To make other sequences, we can split &lt;math&gt;m-1&lt;/math&gt; &lt;math&gt;\frac{1}{m^i}&lt;/math&gt;s into &lt;math&gt;m(m-1)&lt;/math&gt; &lt;math&gt;\frac{1}{m^{i+1}}&lt;/math&gt;s since &lt;math&gt;m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}&lt;/math&gt;. Since we want &lt;math&gt;2010&lt;/math&gt; terms, we have &lt;math&gt;\sum&lt;/math&gt; &lt;math&gt;(m-1)\cdot m^x=2010&lt;/math&gt;. However, since we can set &lt;math&gt;x&lt;/math&gt; to be anything we want (including 0), all we care about is that &lt;math&gt;m-1 | 2010&lt;/math&gt; which happens &lt;math&gt;\boxed{016}&lt;/math&gt; times.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}&lt;/math&gt;. The problem then becomes finding the number of positive integer roots &lt;math&gt;m&lt;/math&gt; for which &lt;math&gt;P(m) = 0&lt;/math&gt; and &lt;math&gt;x_0, x_1, ..., x_{2011}&lt;/math&gt; are nonnegative integers. We plug in &lt;math&gt;m = 1&lt;/math&gt; and see that &lt;math&gt;P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010&lt;/math&gt;. Now, we can say that &lt;math&gt;P(m) = (m-1)Q(m) - 2010&lt;/math&gt; for some polynomial &lt;math&gt;Q(m)&lt;/math&gt; with integer coefficients. Then if &lt;math&gt;P(m) = 0&lt;/math&gt;, &lt;math&gt;(m-1)Q(m) = 2010&lt;/math&gt;. Thus, if &lt;math&gt;P(m) = 0&lt;/math&gt;, then &lt;math&gt;m-1 | 2010&lt;/math&gt; .<br /> Now, we need to show that for all &lt;math&gt;m-1 | 2010&lt;/math&gt;, &lt;math&gt; m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}. &lt;/math&gt;. We try with the first few &lt;math&gt;m&lt;/math&gt; that satisfy this.<br /> For &lt;math&gt;m = 2&lt;/math&gt;, we see we can satisfy this if &lt;math&gt;x_0 = 2010&lt;/math&gt;, &lt;math&gt;x_1 = 2009&lt;/math&gt;, &lt;math&gt;x_2 = 2008&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt; , &lt;math&gt;x_{2008} = 2&lt;/math&gt;, &lt;math&gt;x_{2009} = 1&lt;/math&gt;, &lt;math&gt; x_{2010} = 0&lt;/math&gt;, &lt;math&gt;x_{2011} = 0&lt;/math&gt;, because &lt;math&gt;2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots&lt;/math&gt; (based on the idea &lt;math&gt;2^n + 2^n = 2^{n+1}&lt;/math&gt;, leading to a chain of substitutions of this kind) &lt;math&gt;= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}&lt;/math&gt;. Thus &lt;math&gt;2&lt;/math&gt; is a possible value of &lt;math&gt;m&lt;/math&gt;. For other values, for example &lt;math&gt;m = 3&lt;/math&gt;, we can use the same strategy, with &lt;math&gt;x_{2011} = x_{2010} = x_{2009} = 0&lt;/math&gt;, &lt;math&gt;x_{2008} = x_{2007} = 1&lt;/math&gt;, &lt;math&gt;x_{2006} = x_{2005} = 2&lt;/math&gt;, &lt;math&gt;\cdots&lt;/math&gt;, &lt;math&gt;x_2 = x_1 = 1004&lt;/math&gt; and &lt;math&gt;x_0 = 1005&lt;/math&gt;, because <br /> &lt;math&gt;3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004} = \cdots&lt;/math&gt;<br /> &lt;math&gt;=3^{1004} +3^{1004}+3^{1004} = 3^{1005}&lt;/math&gt;.<br /> It's clearly seen we can use the same strategy for all &lt;math&gt;m-1 |2010&lt;/math&gt;. We count all positive &lt;math&gt;m&lt;/math&gt; satisfying &lt;math&gt;m-1 |2010&lt;/math&gt;, and see there are &lt;math&gt;\boxed{016}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> One notices that &lt;math&gt;m-1 \mid 2010&lt;/math&gt; if and only if there exist non-negative integers &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt; such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;.<br /> <br /> To prove the forward case, we proceed by directly finding &lt;math&gt;x_0,x_1,\ldots,x_{2011}&lt;/math&gt;. Suppose &lt;math&gt;m&lt;/math&gt; is an integer such that &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x_k}&lt;/math&gt;. We will count how many &lt;math&gt;x_k = 0&lt;/math&gt;, how many &lt;math&gt;x_k = 1&lt;/math&gt;, etc. Suppose the number of &lt;math&gt;x_k = 0&lt;/math&gt; is non-zero. Then, there must be at least &lt;math&gt;m&lt;/math&gt; such &lt;math&gt;x_k&lt;/math&gt; since &lt;math&gt;m&lt;/math&gt; divides all the remaining terms, so &lt;math&gt;m&lt;/math&gt; must also divide the sum of all the &lt;math&gt;m^0&lt;/math&gt; terms. Thus, if we let &lt;math&gt;x_k = 0&lt;/math&gt; for &lt;math&gt;k = 1,2,\ldots,m&lt;/math&gt;, we have,<br /> &lt;cmath&gt;m^{x_0} = m + \sum_{k=m+1}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> Well clearly, &lt;math&gt;m^{x_0}&lt;/math&gt; is greater than &lt;math&gt;m&lt;/math&gt;, so &lt;math&gt;m^2 \mid m^{x_0}&lt;/math&gt;. &lt;math&gt;m^2&lt;/math&gt; will also divide every term, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k \geq 2&lt;/math&gt;. So, all the terms, &lt;math&gt;m^{x_k}&lt;/math&gt;, where &lt;math&gt;x_k &lt; 2&lt;/math&gt; must sum to a multiple of &lt;math&gt;m^2&lt;/math&gt;. If there are exactly &lt;math&gt;m&lt;/math&gt; terms where &lt;math&gt;x_k = 0&lt;/math&gt;, then we must have at least &lt;math&gt;m-1&lt;/math&gt; terms where &lt;math&gt;x_k = 1&lt;/math&gt;. Suppose there are exactly &lt;math&gt;m-1&lt;/math&gt; such terms and &lt;math&gt;x_k = 1&lt;/math&gt; for &lt;math&gt;k = m+1,m+2,2m-1&lt;/math&gt;. Now, we have,<br /> &lt;cmath&gt;m^{x_0} = m^2 + \sum_{k=2m}^{2011}m^{x_k}.&lt;/cmath&gt;<br /> One can repeat this process for successive powers of &lt;math&gt;m&lt;/math&gt; until the number of terms reaches 2011. Since there are &lt;math&gt;m + j(m-1)&lt;/math&gt; terms after the &lt;math&gt;j&lt;/math&gt;th power, we will only hit exactly 2011 terms if &lt;math&gt;m-1&lt;/math&gt; is a factor of 2010. To see this, <br /> &lt;cmath&gt;m+j(m-1) = 2011 \Rightarrow m-1+j(m-1) = 2010 \Rightarrow (m-1)(j+1) = 2010.&lt;/cmath&gt;<br /> Thus, when &lt;math&gt;j = 2010/(m-1) - 1&lt;/math&gt; (which is an integer since &lt;math&gt;m-1 \mid 2010&lt;/math&gt; by assumption, there are exactly 2011 terms. To see that these terms sum to a power of &lt;math&gt;m&lt;/math&gt;, we realize that the sum is a geometric series:<br /> &lt;cmath&gt;1 + (m-1) + (m-1)m+(m-1)m^2 + \cdots + (m-1)m^j = 1+(m-1)\frac{m^{j+1}-1}{m-1} = m^{j+1}.&lt;/cmath&gt;<br /> Thus, we have found a solution for the case &lt;math&gt;m-1 \mid 2010&lt;/math&gt;.<br /> <br /> Now, for the reverse case, we use the formula &lt;cmath&gt;x^k-1 = (x-1)(x^{k-1}+x^{k-2}+\cdots+1).&lt;/cmath&gt; Suppose &lt;math&gt;m^{x_0} = \sum_{k=1}^{2011}m^{x^k}&lt;/math&gt; has a solution. Subtract 2011 from both sides to get &lt;cmath&gt;m^{x_0}-1-2010 = \sum_{k=1}^{2011}(m^{x^k}-1).&lt;/cmath&gt; Now apply the formula to get &lt;cmath&gt;(m-1)a_0-2010 = \sum_{k=1}^{2011}[(m-1)a_k],&lt;/cmath&gt; where &lt;math&gt;a_k&lt;/math&gt; are some integers. Rearranging this equation, we find &lt;cmath&gt;(m-1)A = 2010,&lt;/cmath&gt; where &lt;math&gt;A = a_0 - \sum_{k=1}^{2011}a_k&lt;/math&gt;. Thus, if &lt;math&gt;m&lt;/math&gt; is a solution, then &lt;math&gt;m-1 \mid 2010&lt;/math&gt;. <br /> <br /> So, there is one positive integer solution corresponding to each factor of 2010. Since &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;, the number of solutions is &lt;math&gt;2^4 = \boxed{016}&lt;/math&gt;.<br /> <br /> ==Solution 4 (for noobs like me)==<br /> The problem is basically asking how many integers &lt;math&gt;m&lt;/math&gt; have a power that can be expressed as the sum of 2011 other powers of &lt;math&gt;m&lt;/math&gt; (not necessarily distinct). Notice that &lt;math&gt;2+2+4+8+16=32&lt;/math&gt;, &lt;math&gt;3+3+3+9+9+27+27+81+81=243&lt;/math&gt;, and &lt;math&gt;4+4+4+4+16+16+16+64+64+64+256+256+256=1024&lt;/math&gt;. Thus, we can safely assume that the equation &lt;math&gt;2011 = (m-1)x + m&lt;/math&gt; must have an integer solution &lt;math&gt;x&lt;/math&gt;. To find the number of &lt;math&gt;m&lt;/math&gt;-values that allow the aforementioned equation to have an integer solution, we can subtract 1 from the constant &lt;math&gt;m&lt;/math&gt; to make the equation equal a friendlier number, &lt;math&gt;2010&lt;/math&gt;, instead of the ugly prime number &lt;math&gt;2011&lt;/math&gt;: &lt;math&gt;2010 = (m-1)x+(m-1)&lt;/math&gt;. Factor the equation and we get &lt;math&gt;2010 = (m-1)(x+1)&lt;/math&gt;. The number of values of &lt;math&gt;m-1&lt;/math&gt; that allow &lt;math&gt;x+1&lt;/math&gt; to be an integer is quite obviously the number of factors of &lt;math&gt;2010&lt;/math&gt;. Factoring &lt;math&gt;2010&lt;/math&gt;, we obtain &lt;math&gt;2010 = 2 \times 3 \times 5 \times 67&lt;/math&gt;, so the number of positive integers &lt;math&gt;m&lt;/math&gt; that satisfy the required condition is &lt;math&gt;2^4 = \boxed{016}&lt;/math&gt;.<br /> <br /> -fidgetboss_4000<br /> <br /> ==Solution 5==<br /> <br /> First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of &lt;math&gt;m&lt;/math&gt; we can multiply to get to non-negative powers.<br /> Now we see that our problem is just a matter of &lt;math&gt;&quot;m&quot;&lt;/math&gt;-chopping blocks.<br /> What is meant by &lt;math&gt;m&lt;/math&gt;-chopping is taking an existing block of say &lt;math&gt;m^{k}&lt;/math&gt; and turning it into &lt;math&gt;m&lt;/math&gt; blocks of &lt;math&gt;m^{k-1}&lt;/math&gt;. This process increases the total # of blocks by &lt;math&gt;m-1&lt;/math&gt; per chop.<br /> The problem wants us to find the number of positive integers &lt;math&gt;m&lt;/math&gt; where some number of chops will turn &lt;math&gt;1&lt;/math&gt; block into &lt;math&gt;2011&lt;/math&gt; such blocks, thus increasing the total amount by &lt;math&gt;2010= 2 \cdot 3 \cdot 5 \cdot 67&lt;/math&gt;. Thus &lt;math&gt;m-1 | 2010&lt;/math&gt;, and a cursory check on extreme cases will confirm that there are indeed &lt;math&gt;\boxed{016}&lt;/math&gt; possible &lt;math&gt;m&lt;/math&gt;s.<br /> <br /> ==An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)==<br /> https://artofproblemsolving.com/community/c6h84155p486903<br /> 2001 Austrian-Polish Math Individual Competition #1<br /> ~MSC<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=6|num-a=8}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_4&diff=127460 2011 AIME I Problems/Problem 4 2020-07-04T19:08:26Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=125&lt;/math&gt;, &lt;math&gt;AC=117&lt;/math&gt; and &lt;math&gt;BC=120&lt;/math&gt;. The angle bisector of angle &lt;math&gt;A&lt;/math&gt; intersects &lt;math&gt; \overline{BC} &lt;/math&gt; at point &lt;math&gt;L&lt;/math&gt;, and the angle bisector of angle &lt;math&gt;B&lt;/math&gt; intersects &lt;math&gt; \overline{AC} &lt;/math&gt; at point &lt;math&gt;K&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;C&lt;/math&gt; to &lt;math&gt; \overline{BK}&lt;/math&gt; and &lt;math&gt; \overline{AL}&lt;/math&gt;, respectively. Find &lt;math&gt;MN&lt;/math&gt;.<br /> <br /> == Solution 1 == <br /> Extend &lt;math&gt;{CM}&lt;/math&gt; and &lt;math&gt;{CN}&lt;/math&gt; such that they intersect line &lt;math&gt;{AB}&lt;/math&gt; at points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. <br /> Since &lt;math&gt;{BM}&lt;/math&gt; is the angle bisector of angle &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;{CM}&lt;/math&gt; is perpendicular to &lt;math&gt;{BM}&lt;/math&gt;, so &lt;math&gt;BP=BC=120&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;{CP}&lt;/math&gt;. For the same reason, &lt;math&gt;AQ=AC=117&lt;/math&gt;, and &lt;math&gt;N&lt;/math&gt; is the midpoint of &lt;math&gt;{CQ}&lt;/math&gt;.<br /> Hence &lt;math&gt;MN=\frac{PQ}{2}&lt;/math&gt;. &lt;math&gt;PQ=BP+AQ-AB=120+117-125=112&lt;/math&gt;, so &lt;math&gt;MN=\boxed{056}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;ABC&lt;/math&gt;. Now, since &lt;math&gt;IM \perp MC&lt;/math&gt; and &lt;math&gt;IN \perp NC&lt;/math&gt;, we have &lt;math&gt;CMIN&lt;/math&gt; is a cyclic quadrilateral. Consequently, &lt;math&gt;\frac{MN}{\sin \angle MIN} = 2R = CI&lt;/math&gt;. Since &lt;math&gt;\sin \angle MIN = \sin (90 - \frac{\angle BAC}{2}) = \cos \angle IAK&lt;/math&gt;, we have that &lt;math&gt;MN = AI \cdot \cos \angle IAK&lt;/math&gt;. Letting &lt;math&gt;X&lt;/math&gt; be the point of contact of the incircle of &lt;math&gt;ABC&lt;/math&gt; with side &lt;math&gt;AC&lt;/math&gt;, we have &lt;math&gt;AX=MN&lt;/math&gt;. Thus, &lt;math&gt;MN=\frac{117+120-125}{2}=\boxed{056}&lt;/math&gt;<br /> <br /> == Solution 3 (Bash) == <br /> Project &lt;math&gt;I&lt;/math&gt; onto &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; as &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;. &lt;math&gt;ID&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; are both in-radii of &lt;math&gt;\triangle ABC&lt;/math&gt; so we get right triangles with legs &lt;math&gt;r&lt;/math&gt; (the in-radius length) and &lt;math&gt;s - c = 56&lt;/math&gt;. Since &lt;math&gt;IC&lt;/math&gt; is the hypotenuse for the 4 triangles (&lt;math&gt;\triangle INC, \triangle IMC, \triangle IDC,&lt;/math&gt; and &lt;math&gt;\triangle IEC&lt;/math&gt;), &lt;math&gt;C, D, M, I, N, E&lt;/math&gt; are con-cyclic on a circle we shall denote as &lt;math&gt;\omega&lt;/math&gt; which is also the circumcircle of &lt;math&gt;\triangle CMN&lt;/math&gt; and &lt;math&gt;\triangle CDE&lt;/math&gt;. To find &lt;math&gt;MN&lt;/math&gt;, we can use the Law of Cosines on &lt;math&gt;\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})&lt;/math&gt; where &lt;math&gt;O&lt;/math&gt; is the center of &lt;math&gt;\omega&lt;/math&gt;. Now, the circumradius &lt;math&gt;R&lt;/math&gt; can be found with Pythagorean Theorem with &lt;math&gt;\triangle CDI&lt;/math&gt; or &lt;math&gt;\triangle CEI&lt;/math&gt;: &lt;math&gt;r^2 + 56^2 = (2R)^2&lt;/math&gt;. To find &lt;math&gt;r&lt;/math&gt;, we can use the formula &lt;math&gt;rs = [ABC]&lt;/math&gt; and by Heron's, &lt;math&gt;[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}&lt;/math&gt;. To find &lt;math&gt;\angle MCN&lt;/math&gt;, we can find &lt;math&gt;\angle MIN&lt;/math&gt; since &lt;math&gt;\angle MCN = 180 - \angle MIN&lt;/math&gt;. &lt;math&gt;\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}&lt;/math&gt;. Thus, &lt;math&gt;\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}&lt;/math&gt; and since &lt;math&gt;\angle A + \angle B + \angle C = 180&lt;/math&gt;, we have &lt;math&gt;\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}&lt;/math&gt;. Plugging this into our Law of Cosines (LoC) formula gives &lt;math&gt;MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})&lt;/math&gt;. To find &lt;math&gt;\cos{\angle C}&lt;/math&gt;, we use LoC on &lt;math&gt;\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}&lt;/math&gt;. Our formula now becomes &lt;math&gt;MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}&lt;/math&gt;. After simplifying, we get &lt;math&gt;MN^2 = 3136 \implies MN = \boxed{056}&lt;/math&gt;.<br /> <br /> --lucasxia01<br /> <br /> == Solution 4==<br /> <br /> Because &lt;math&gt;\angle CMI = \angle CNI = 90&lt;/math&gt;, &lt;math&gt;CMIN&lt;/math&gt; is cyclic. <br /> <br /> Ptolemy on CMIN:<br /> <br /> &lt;math&gt;CN*MI+CM*IN=CI*MN&lt;/math&gt;<br /> <br /> &lt;math&gt;CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN&lt;/math&gt;<br /> <br /> &lt;math&gt;MN = CI \sin \angle MCN&lt;/math&gt; by angle addition formula.<br /> <br /> &lt;math&gt;\angle MCN = 180 - \angle MIN = 90 - \angle BCI&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;H&lt;/math&gt; be where the incircle touches &lt;math&gt;BC&lt;/math&gt;, then &lt;math&gt;CI \cos \angle BCI = CH = \frac{a+b-c}{2}&lt;/math&gt;.<br /> &lt;math&gt;a=120, b=117, c=125&lt;/math&gt;, for a final answer of &lt;math&gt;\boxed{056}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=vkniYGN45F4<br /> <br /> ~Shreyas S<br /> <br /> == See also ==<br /> {{AIME box|year=2011|n=I|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_12&diff=127260 2008 AIME I Problems/Problem 12 2020-07-02T04:04:17Z <p>Hi13: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let &lt;math&gt;M&lt;/math&gt; be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when &lt;math&gt;M&lt;/math&gt; is divided by &lt;math&gt;10&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the number of car lengths that separates each car. Then their speed is at most &lt;math&gt;15n&lt;/math&gt;. Let a ''unit'' be the distance between the cars (front to front). Then the length of each unit is &lt;math&gt;4(n + 1)&lt;/math&gt;. To maximize, in a unit, the CAR comes first, THEN the empty space. So at time zero, the car is right at the eye.<br /> <br /> Hence, we count the number of units that pass the eye in an hour: &lt;math&gt;\frac {15,000n\frac{\text{meters}}{\text{hour}}}{4(n + 1)\frac{\text{meters}}{\text{unit}}} = \frac {15,000n}{4(n + 1)}\frac{\text{units}}{\text{hour}}&lt;/math&gt;. We wish to maximize this.<br /> <br /> Observe that as &lt;math&gt;n&lt;/math&gt; gets larger, the &lt;math&gt; + 1&lt;/math&gt; gets less and less significant, so we take the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity<br /> &lt;center&gt;&lt;math&gt;\lim_{n\rightarrow \infty}\frac {15,000n}{4(n + 1)} = \lim_{n\rightarrow \infty}\frac {15,000}{4} = 3750&lt;/math&gt;&lt;/center&gt;<br /> Now, as the speeds are clearly finite, we can never actually reach &lt;math&gt;3750&lt;/math&gt; full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the &lt;math&gt;3750&lt;/math&gt;th unit has passed, but not all of the space behind it. Hence, &lt;math&gt;3750&lt;/math&gt; cars is possible, and the answer is &lt;math&gt;\boxed {375}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Disclaimer: This is for the people who may not understand calculus, and is also how I did it.<br /> First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we start with one car in front of the photoelectric eye.<br /> We first set the speed of the cars as &lt;math&gt;15k&lt;/math&gt;. Then, the distance between them is &lt;math&gt;\frac{4}{1000} \times k~km&lt;/math&gt;. Therefore, it takes the car closest to the eye not on the eye &lt;math&gt;\frac{\frac{k}{250}}{15k}&lt;/math&gt; hours to get to the eye. There is one hour, so the amount of cars that can pass is &lt;math&gt;\frac{1}{\frac{\frac{k}{250}}{15k}}&lt;/math&gt;, or &lt;math&gt;3750&lt;/math&gt; cars. When divided by ten, you get the quotient of &lt;math&gt;\boxed{375}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=11|num-a=13}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_15&diff=126544 2005 AIME I Problems/Problem 15 2020-06-26T00:20:16Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt; ABC &lt;/math&gt; has &lt;math&gt; BC=20. &lt;/math&gt; The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] &lt;math&gt; AD. &lt;/math&gt; If the area of the triangle is &lt;math&gt; m \sqrt{n} &lt;/math&gt; where &lt;math&gt; m &lt;/math&gt; and &lt;math&gt; n &lt;/math&gt; are integers and &lt;math&gt; n &lt;/math&gt; is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find &lt;math&gt; m+n. &lt;/math&gt;<br /> <br /> == Solution 1==<br /> &lt;center&gt;&lt;asy&gt;<br /> size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10);<br /> pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C);<br /> path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir);<br /> D(MP(&quot;A&quot;,A,s)--MP(&quot;B&quot;,B,s)--MP(&quot;C&quot;,C,N,s)--cycle); D(cir); <br /> D(A--MP(&quot;D&quot;,D,NE,s)); D(MP(&quot;E&quot;,E1,NE,s)); D(MP(&quot;F&quot;,F,NW,s)); D(MP(&quot;G&quot;,G,s)); D(MP(&quot;P&quot;,P,SW,s)); D(MP(&quot;Q&quot;,Q,SE,s));<br /> MP(&quot;10&quot;,(B+D)/2,NE); MP(&quot;10&quot;,(C+D)/2,NE);<br /> &lt;/asy&gt;&lt;/center&gt;&lt;!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps --&gt;<br /> <br /> Let &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; be the points of tangency of the incircle with &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Without loss of generality, let &lt;math&gt;AC &lt; AB&lt;/math&gt;, so that &lt;math&gt;E&lt;/math&gt; is between &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. Let the length of the median be &lt;math&gt;3m&lt;/math&gt;. Then by two applications of the [[Power of a Point Theorem]], &lt;math&gt;DE^2 = 2m \cdot m = AF^2&lt;/math&gt;, so &lt;math&gt;DE = AF&lt;/math&gt;. Now, &lt;math&gt;CE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; are two tangents to a circle from the same point, so by the Two Tangent Theorem &lt;math&gt;CE = CF = c&lt;/math&gt; and thus &lt;math&gt;AC = AF + CF = DE + CE = CD = 10&lt;/math&gt;. Then &lt;math&gt;DE = AF = AG = 10 - c&lt;/math&gt; so &lt;math&gt;BG = BE = BD + DE = 20 - c&lt;/math&gt; and thus &lt;math&gt;AB = AG + BG = 30 - 2c&lt;/math&gt;.<br /> <br /> Now, by [[Stewart's Theorem]] in triangle &lt;math&gt;\triangle ABC&lt;/math&gt; with [[cevian]] &lt;math&gt;\overline{AD}&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.&lt;/cmath&gt;<br /> <br /> Our earlier result from Power of a Point was that &lt;math&gt;2m^2 = (10 - c)^2&lt;/math&gt;, so we combine these two results to solve for &lt;math&gt;c&lt;/math&gt; and we get<br /> <br /> &lt;cmath&gt;9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;c = 2&lt;/math&gt; or &lt;math&gt; = 10&lt;/math&gt;. We discard the value &lt;math&gt;c = 10&lt;/math&gt; as extraneous (it gives us a line) and are left with &lt;math&gt;c = 2&lt;/math&gt;, so our triangle has area &lt;math&gt;\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}&lt;/math&gt; and so the answer is &lt;math&gt;24 + 14 = \boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Use Power of a point similar to the first solution to find that &lt;math&gt;AB = 30&lt;/math&gt; and that the side &lt;math&gt;AC = 2 \cdot x \sqrt{2}&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is one third of the median's length. Then use systems of law of cosines, creating two triangles, with &lt;math&gt;10-10-3x&lt;/math&gt; with angle &lt;math&gt;\theta&lt;/math&gt;, and &lt;math&gt;10-20-2 \cdot x \sqrt{2}&lt;/math&gt; with the same angle. Solving the system yields &lt;math&gt;x = 4 \sqrt{2}&lt;/math&gt;. Solving using Heron's Formula gets the answer &lt;math&gt;24 \sqrt{14}&lt;/math&gt;, or &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> WLOG let E be be between C &amp; D (as in solution 1). Assume &lt;math&gt;AD = 3m&lt;/math&gt;. We use power of a point to get that <br /> &lt;math&gt;AG = DE = \sqrt{2}m &lt;/math&gt; and &lt;math&gt;AB = AG + GB = AG + BE = 10+2\sqrt{2} m&lt;/math&gt; <br /> <br /> Since now we have &lt;math&gt;AC = 10&lt;/math&gt;, &lt;math&gt;BC = 20, AB = 10+2\sqrt{2} m &lt;/math&gt; in triangle &lt;math&gt;\triangle ABC&lt;/math&gt; and [[cevian]] &lt;math&gt;AD = 3m&lt;/math&gt;. Now, we can apply [[Stewart's Theorem]].<br /> <br /> &lt;cmath&gt;2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000&lt;/cmath&gt;<br /> &lt;cmath&gt;1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;100 m^2 = 400\sqrt{2}m&lt;/cmath&gt;<br /> <br /> &lt;math&gt;m = 4\sqrt{2}&lt;/math&gt; or &lt;math&gt;m = 0&lt;/math&gt; if &lt;math&gt;m = 0&lt;/math&gt;, we get a degenerate triangle, so &lt;math&gt;m = 4\sqrt{2}&lt;/math&gt;, and thus AB = 26. You can now use [[Heron's Formula]] to finish. The answer is &lt;math&gt;24 \sqrt{14}&lt;/math&gt;, or &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> -Alexlikemath<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=I|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_24&diff=124897 2010 AMC 12B Problems/Problem 24 2020-06-10T21:46:22Z <p>Hi13: /* Solution 3 */</p> <hr /> <div>== Problem 24 ==<br /> The set of real numbers &lt;math&gt;x&lt;/math&gt; for which <br /> <br /> &lt;cmath&gt;\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1&lt;/cmath&gt;<br /> <br /> is the union of intervals of the form &lt;math&gt;a&lt;x\le b&lt;/math&gt;. What is the sum of the lengths of these intervals?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}&lt;/math&gt;<br /> <br /> == Solution ==<br /> Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1&lt;/cmath&gt;<br /> We shall say that &lt;math&gt;f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}&lt;/math&gt;. &lt;math&gt;f(x)&lt;/math&gt; has three vertical asymptotes at &lt;math&gt;x=\{-1,0,1\}&lt;/math&gt;. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at &lt;math&gt;y=0&lt;/math&gt;. The function intersects &lt;math&gt;1&lt;/math&gt; at some point from &lt;math&gt;x=-1&lt;/math&gt; to &lt;math&gt;x=0&lt;/math&gt;, and at some point from &lt;math&gt;x=0&lt;/math&gt; to &lt;math&gt;x=1&lt;/math&gt;, and at some point to the right of &lt;math&gt;x=1&lt;/math&gt;. The intervals where the function is greater than &lt;math&gt;1&lt;/math&gt; are between the points where the function equals &lt;math&gt;1&lt;/math&gt; and the vertical asymptotes.<br /> <br /> If &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are values of x where &lt;math&gt;f(x)=1&lt;/math&gt;, then the sum of the lengths of the intervals is &lt;math&gt;(p-(-1))+(q-0)+(r-1)=p+q+r&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies x^3-3x^2-x+1=0&lt;/cmath&gt;<br /> <br /> And now our job is simply to find the sum of the roots of &lt;math&gt;x^3-3x^2-x+1&lt;/math&gt;.<br /> Using [[Vieta's formulas]], we find this to be &lt;math&gt;3&lt;/math&gt; &lt;math&gt;\Rightarrow\boxed{C}&lt;/math&gt;.<br /> <br /> ''NOTE''': For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose &lt;math&gt;C&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> As in the first solution, note that the expression can be translated into &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1&lt;/cmath&gt; without affecting the interval lengths.<br /> <br /> This simplifies into &lt;cmath&gt;\frac{-x^3+3x^2+x-1}{(x)(x+1)(x-1)}\ge0,&lt;/cmath&gt; so &lt;cmath&gt;-x^3+3x^2+x-1\ge0.&lt;/cmath&gt;<br /> Each interval is &lt;math&gt;(-1, a), (0, b), (1, c)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are the roots of &lt;math&gt;-x^3+3x^2+x-1=0&lt;/math&gt;<br /> so the total length is &lt;math&gt;a+b+c&lt;/math&gt;, which is the sum of the roots, or &lt;math&gt;\boxed3&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Note that all of the answers but &lt;math&gt;C&lt;/math&gt; have weird factors of 2010, but 2010 is a random number (set &lt;math&gt;y=x-2010&lt;/math&gt;). So therefore the answer is &lt;math&gt;\fbox{C(3)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=23|num-a=25|ab=B}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_24&diff=124896 2010 AMC 12B Problems/Problem 24 2020-06-10T21:44:15Z <p>Hi13: /* Solution 3 */</p> <hr /> <div>== Problem 24 ==<br /> The set of real numbers &lt;math&gt;x&lt;/math&gt; for which <br /> <br /> &lt;cmath&gt;\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge1&lt;/cmath&gt;<br /> <br /> is the union of intervals of the form &lt;math&gt;a&lt;x\le b&lt;/math&gt;. What is the sum of the lengths of these intervals?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \dfrac{1003}{335} \qquad \textbf{(B)}\ \dfrac{1004}{335} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \dfrac{403}{134} \qquad \textbf{(E)}\ \dfrac{202}{67}&lt;/math&gt;<br /> <br /> == Solution ==<br /> Because the right side of the inequality is a horizontal line, the left side can be translated horizontally by any value and the intervals will remain the same. For simplicity of calculation, we will find the intervals where &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1&lt;/cmath&gt;<br /> We shall say that &lt;math&gt;f(x)=\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}&lt;/math&gt;. &lt;math&gt;f(x)&lt;/math&gt; has three vertical asymptotes at &lt;math&gt;x=\{-1,0,1\}&lt;/math&gt;. As the sum of decreasing hyperbolas, the function is decreasing at all intervals. Values immediately to the left of each asymptote approach negative infinity, and values immediately to the right of each asymptote approach positive infinity. In addition, the function has a horizontal asymptote at &lt;math&gt;y=0&lt;/math&gt;. The function intersects &lt;math&gt;1&lt;/math&gt; at some point from &lt;math&gt;x=-1&lt;/math&gt; to &lt;math&gt;x=0&lt;/math&gt;, and at some point from &lt;math&gt;x=0&lt;/math&gt; to &lt;math&gt;x=1&lt;/math&gt;, and at some point to the right of &lt;math&gt;x=1&lt;/math&gt;. The intervals where the function is greater than &lt;math&gt;1&lt;/math&gt; are between the points where the function equals &lt;math&gt;1&lt;/math&gt; and the vertical asymptotes.<br /> <br /> If &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; are values of x where &lt;math&gt;f(x)=1&lt;/math&gt;, then the sum of the lengths of the intervals is &lt;math&gt;(p-(-1))+(q-0)+(r-1)=p+q+r&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}=1&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies x(x-1)+(x-1)(x+1)+x(x+1)=x(x-1)(x+1)&lt;/cmath&gt;<br /> &lt;cmath&gt;\implies x^3-3x^2-x+1=0&lt;/cmath&gt;<br /> <br /> And now our job is simply to find the sum of the roots of &lt;math&gt;x^3-3x^2-x+1&lt;/math&gt;.<br /> Using [[Vieta's formulas]], we find this to be &lt;math&gt;3&lt;/math&gt; &lt;math&gt;\Rightarrow\boxed{C}&lt;/math&gt;.<br /> <br /> ''NOTE''': For the AMC, one may note that the transformed inequality should not yield solutions that involve big numbers like 67 or 134, and immediately choose &lt;math&gt;C&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> As in the first solution, note that the expression can be translated into &lt;cmath&gt;\frac{1}{x+1}+\frac{1}{x}+\frac{1}{x-1}\ge1&lt;/cmath&gt; without affecting the interval lengths.<br /> <br /> This simplifies into &lt;cmath&gt;\frac{-x^3+3x^2+x-1}{(x)(x+1)(x-1)}\ge0,&lt;/cmath&gt; so &lt;cmath&gt;-x^3+3x^2+x-1\ge0.&lt;/cmath&gt;<br /> Each interval is &lt;math&gt;(-1, a), (0, b), (1, c)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are the roots of &lt;math&gt;-x^3+3x^2+x-1=0&lt;/math&gt;<br /> so the total length is &lt;math&gt;a+b+c&lt;/math&gt;, which is the sum of the roots, or &lt;math&gt;\boxed3&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Note that all of the answers but &lt;math&gt;C&lt;/math&gt; have weird factors of 2010, but 2010 is a random number (set &lt;math&gt;y=x-2010&lt;/math&gt;). So therefore the answer is &lt;math&gt;\fbox[C(3)]&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=23|num-a=25|ab=B}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_9&diff=122557 2020 AMC 12A Problems/Problem 9 2020-05-17T19:59:21Z <p>Hi13: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> How many solutions does the equation &lt;math&gt;\tan(2x)=\cos(\tfrac{x}{2})&lt;/math&gt; have on the interval &lt;math&gt;[0,2\pi]?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Draw a graph of &lt;math&gt;\tan(2x)&lt;/math&gt; and &lt;math&gt;\cos(\tfrac{x}{2})&lt;/math&gt;<br /> <br /> &lt;math&gt;\tan(2x)&lt;/math&gt; has a period of &lt;math&gt;\tfrac{\pi}{2},&lt;/math&gt; asymptotes at &lt;math&gt;x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},&lt;/math&gt; and zeroes at &lt;math&gt;\tfrac{k\pi}{2}&lt;/math&gt;. It is positive from &lt;math&gt;(0,\tfrac{\pi}{4}) \cup (\tfrac{\pi}{2},\tfrac{3\pi}{4}) \cup (\pi,\tfrac{5\pi}{4}) \cup (\tfrac{3\pi}{2},\tfrac{7\pi}{4})&lt;/math&gt; and negative elsewhere.<br /> <br /> cos&lt;math&gt;(\tfrac{x}{2})&lt;/math&gt; has a period of &lt;math&gt;4\pi&lt;/math&gt; and zeroes at &lt;math&gt;\pi&lt;/math&gt;. It is positive from &lt;math&gt;[0,\pi)&lt;/math&gt; and negative elsewhere.<br /> <br /> Drawing such a graph would get &lt;math&gt;\boxed{\textbf{E) }5}&lt;/math&gt; ~lopkiloinm<br /> <br /> Or you could see the points at which both graphs are positive or both are negative, again yielding 5 such areas. -hi13<br /> <br /> edited by - annabelle0913<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_20&diff=121976 2013 AMC 12A Problems/Problem 20 2020-05-04T01:58:14Z <p>Hi13: /* Solution 1 */</p> <hr /> <div>== Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Imagine that the 19 numbers are just 19 persons sitting evenly around a circle &lt;math&gt;C&lt;/math&gt;; each of them is facing to the center. <br /> <br /> One may check that &lt;math&gt;x \succ y&lt;/math&gt; if and only if &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the left of &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;y \succ x&lt;/math&gt; if and only if &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the right of &lt;math&gt;x&lt;/math&gt;. Therefore, &quot;&lt;math&gt;x \succ y&lt;/math&gt; and &lt;math&gt;y \succ z&lt;/math&gt; and &lt;math&gt;z \succ x&lt;/math&gt;&quot; implies that &lt;math&gt;x, y, z&lt;/math&gt; cuts the circumference of &lt;math&gt;C&lt;/math&gt; into three arcs, each of which has no more than &lt;math&gt;10&lt;/math&gt; numbers sitting on it (inclusive).<br /> <br /> We count the complement: where the cut generated by &lt;math&gt;(x,y,z)&lt;/math&gt; has ONE arc that has more than &lt;math&gt;10&lt;/math&gt; persons sitting on. Note that there can only be one such arc because there are only &lt;math&gt;19&lt;/math&gt; persons in total.<br /> <br /> Suppose the number of persons on the longest arc is &lt;math&gt;k&gt;10&lt;/math&gt;. Then two places of &lt;math&gt;x,y,z&lt;/math&gt; are just chosen from the two end-points of the arc, and there are &lt;math&gt;19-k&lt;/math&gt; possible places for the third person. Once the three places of &lt;math&gt;x,y,z&lt;/math&gt; are chosen, there are three possible ways to put &lt;math&gt;x,y,z&lt;/math&gt; into them clockwise. Also, note that for any &lt;math&gt;k&gt;10&lt;/math&gt;, there are &lt;math&gt;19&lt;/math&gt; ways to choose an arc of length &lt;math&gt;k&lt;/math&gt;. Therefore the total number of ways (of the complement) is <br /> <br /> &lt;cmath&gt;\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36&lt;/cmath&gt;<br /> <br /> So the answer is<br /> <br /> &lt;cmath&gt; 3\cdot \binom{19}{3} - 3\cdot 19\cdot 36 = 3\cdot 19 \cdot (51 - 36) = 855 &lt;/cmath&gt;<br /> <br /> NOTE: this multiple-choice problem can be done even faster -- after we realized the fact that each choice of the three places of &lt;math&gt;x,y,z&lt;/math&gt; corresponds to &lt;math&gt;3&lt;/math&gt; possible ways to put them in, and that each arc of length &lt;math&gt;k&gt;10&lt;/math&gt; has &lt;math&gt;19&lt;/math&gt; equitable positions, it is evident that the answer should be divisible by &lt;math&gt;3\cdot 19&lt;/math&gt;, which can only be &lt;math&gt;855&lt;/math&gt; from the five choices.<br /> <br /> ==Solution 2==<br /> <br /> First, we can find out that the only &lt;math&gt;x, y, z&lt;/math&gt; that satisfy the conditions in the problem are &lt;math&gt;(1 \leq z-y \leq 9 , 1 \leq y-x \leq 9 , z-x \geq 10)&lt;/math&gt; , &lt;math&gt;(1 \leq y-x \leq 9 , 1 \leq x-z \leq 9 , y-z \geq 10)&lt;/math&gt; , and &lt;math&gt;(1 \leq x-z \leq 9 , 1 \leq z-y \leq 9 , x-y \geq 10)&lt;/math&gt;. <br /> <br /> Consider the 1st set of conditions for &lt;math&gt;x, y, z&lt;/math&gt;. We get that there are<br /> <br /> &lt;cmath&gt;45\cdot 10 - \sum_{k=1}^{9} \binom{k+1}{2} = 285&lt;/cmath&gt;<br /> <br /> cases for the first set of conditions.<br /> <br /> Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is<br /> <br /> &lt;cmath&gt;3\cdot 285 = \fbox{(B)855} &lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> <br /> [Category: Introductory Set Theory Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_20&diff=121975 2013 AMC 12A Problems/Problem 20 2020-05-04T01:53:59Z <p>Hi13: /* Solution */</p> <hr /> <div>== Problem 20 ==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set &lt;math&gt;\{1,2,3,...,19\}&lt;/math&gt;. For &lt;math&gt;a,b \in S&lt;/math&gt;, define &lt;math&gt;a \succ b&lt;/math&gt; to mean that either &lt;math&gt;0 &lt; a - b \le 9&lt;/math&gt; or &lt;math&gt;b - a &gt; 9&lt;/math&gt;. How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of elements of &lt;math&gt;S&lt;/math&gt; have the property that &lt;math&gt;x \succ y&lt;/math&gt;, &lt;math&gt;y \succ z&lt;/math&gt;, and &lt;math&gt;z \succ x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)} \ 810 \qquad \textbf{(B)} \ 855 \qquad \textbf{(C)} \ 900 \qquad \textbf{(D)} \ 950 \qquad \textbf{(E)} \ 988 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Imagine 19 numbers are just 19 persons sitting evenly around a circle &lt;math&gt;C&lt;/math&gt;; each of them is facing to the center. <br /> <br /> One may check that &lt;math&gt;x \succ y&lt;/math&gt; if and only if &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the left of &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;y \succ x&lt;/math&gt; if and only if &lt;math&gt;y&lt;/math&gt; is one of the 9 persons on the right of &lt;math&gt;x&lt;/math&gt;. Therefore, &quot;&lt;math&gt;x \succ y&lt;/math&gt; and &lt;math&gt;y \succ z&lt;/math&gt; and &lt;math&gt;z \succ x&lt;/math&gt;&quot; implies that &lt;math&gt;x, y, z&lt;/math&gt; cuts the circumference of &lt;math&gt;C&lt;/math&gt; into three arcs, each of which has no more than &lt;math&gt;10&lt;/math&gt; numbers sitting on it (inclusive).<br /> <br /> We count the complement: where the cut generated by &lt;math&gt;(x,y,z)&lt;/math&gt; has ONE arc that has more than &lt;math&gt;10&lt;/math&gt; persons sitting on. Note that there can only be one such arc because there are only &lt;math&gt;19&lt;/math&gt; persons in total.<br /> <br /> Suppose the number of persons on the longest arc is &lt;math&gt;k&gt;10&lt;/math&gt;. Then two places of &lt;math&gt;x,y,z&lt;/math&gt; are just chosen from the two end-points of the arc, and there are &lt;math&gt;19-k&lt;/math&gt; possible places for the third person. Once the three places of &lt;math&gt;x,y,z&lt;/math&gt; are chosen, there are three possible ways to put &lt;math&gt;x,y,z&lt;/math&gt; into them clockwise. Also, note that for any &lt;math&gt;k&gt;10&lt;/math&gt;, there are &lt;math&gt;19&lt;/math&gt; ways to choose an arc of length &lt;math&gt;k&lt;/math&gt;. Therefore the total number of ways (of the complement) is <br /> <br /> &lt;cmath&gt;\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36&lt;/cmath&gt;<br /> <br /> So the answer is<br /> <br /> &lt;cmath&gt; 3\cdot \binom{19}{3} - 3\cdot 19\cdot 36 = 3\cdot 19 \cdot (51 - 36) = 855 &lt;/cmath&gt;<br /> <br /> NOTE: this multiple-choice problem can be done even faster -- after we realized the fact that each choice of the three places of &lt;math&gt;x,y,z&lt;/math&gt; corresponds to &lt;math&gt;3&lt;/math&gt; possible ways to put them in, and that each arc of length &lt;math&gt;k&gt;10&lt;/math&gt; has &lt;math&gt;19&lt;/math&gt; equitable positions, it is evident that the answer should be divisible by &lt;math&gt;3\cdot 19&lt;/math&gt;, which can only be &lt;math&gt;855&lt;/math&gt; from the five choices.<br /> <br /> ==Solution 2==<br /> <br /> First, we can find out that the only &lt;math&gt;x, y, z&lt;/math&gt; that satisfy the conditions in the problem are &lt;math&gt;(1 \leq z-y \leq 9 , 1 \leq y-x \leq 9 , z-x \geq 10)&lt;/math&gt; , &lt;math&gt;(1 \leq y-x \leq 9 , 1 \leq x-z \leq 9 , y-z \geq 10)&lt;/math&gt; , and &lt;math&gt;(1 \leq x-z \leq 9 , 1 \leq z-y \leq 9 , x-y \geq 10)&lt;/math&gt;. <br /> <br /> Consider the 1st set of conditions for &lt;math&gt;x, y, z&lt;/math&gt;. We get that there are<br /> <br /> &lt;cmath&gt;45\cdot 10 - \sum_{k=1}^{9} \binom{k+1}{2} = 285&lt;/cmath&gt;<br /> <br /> cases for the first set of conditions.<br /> <br /> Since the 2nd and 3rd set of conditions are simply rotations of the 1st set, the total number of cases is<br /> <br /> &lt;cmath&gt;3\cdot 285 = \fbox{(B)855} &lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2013|ab=A|num-b=19|num-a=21}}<br /> <br /> [Category: Introductory Set Theory Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_24&diff=121723 2010 AMC 12A Problems/Problem 24 2020-04-27T01:23:03Z <p>Hi13: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)&lt;/math&gt;. The intersection of the domain of &lt;math&gt;f(x)&lt;/math&gt; with the interval &lt;math&gt;[0,1]&lt;/math&gt; is a union of &lt;math&gt;n&lt;/math&gt; disjoint open intervals. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> == Solution 1==<br /> The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.<br /> <br /> We note that since all of the &lt;math&gt;\sin&lt;/math&gt; factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to &lt;math&gt;0&lt;/math&gt;.<br /> <br /> First, let us find the number of zeros of the inside of the logarithm.<br /> <br /> &lt;cmath&gt;\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &amp;= 0\\<br /> \sin(\pi x) &amp;= 0\\<br /> x &amp;= 0, 1\\<br /> \sin(2 \pi x) &amp;= 0\\<br /> x &amp;= 0, \frac{1}{2}, 1\\<br /> \sin(3 \pi x) &amp;= 0\\<br /> x &amp;= 0, \frac{1}{3}, \frac{2}{3}, 1\\<br /> \sin(4 \pi x) &amp;= 0\\<br /> x &amp;= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\<br /> &amp;\cdots\end{align*}&lt;/cmath&gt;<br /> <br /> After counting up the number of zeros for each factor and eliminating the excess cases we get &lt;math&gt;23&lt;/math&gt; zeros and &lt;math&gt;22&lt;/math&gt; intervals.<br /> <br /> In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.<br /> <br /> The first interval &lt;math&gt;(0, \frac{1}{8})&lt;/math&gt; is obviously positive. This means the next interval &lt;math&gt;(\frac{1}{8}, \frac{1}{7})&lt;/math&gt; is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are &lt;math&gt;5&lt;/math&gt; negative intervals from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Since the function is symmetric, we know that there are also &lt;math&gt;5&lt;/math&gt; negative intervals from &lt;math&gt;\frac{1}{2}&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;.<br /> <br /> And so, the total number of disjoint open intervals is &lt;math&gt;22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2 (cheap)==<br /> Note that the expression &lt;math&gt;\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)&lt;/math&gt; must be greater than zero, since logarithm functions are undefined for &lt;math&gt;0&lt;/math&gt; and negative numbers. Let &lt;math&gt;x_1, x_2, x_3, ..., x_8&lt;/math&gt; temporarily be the dependent variables of the functions &lt;math&gt;y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)&lt;/math&gt;. It is easy to see that for &lt;math&gt;y_i&lt;/math&gt; to be positive for &lt;math&gt;1\leq i\leq8&lt;/math&gt;, &lt;math&gt;\lfloor i x_i \rfloor&lt;/math&gt; must be even for &lt;math&gt;1 \leq i\leq8&lt;/math&gt;. Since an even number of positives times an even number of negatives equals a positive, there can be &lt;math&gt;2, 4, 6, &lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; positive values of &lt;math&gt;y_i&lt;/math&gt; for &lt;math&gt;1 \leq i\leq 8&lt;/math&gt; for a given value of &lt;math&gt;x&lt;/math&gt;. (since &lt;math&gt;y_1&lt;/math&gt; is always positive on the range &lt;math&gt;[0, 1]&lt;/math&gt;)<br /> Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;. (I recommend increments of at most &lt;math&gt;0.1&lt;/math&gt;.) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for &lt;math&gt;y_1, y_2, ..., y_8&lt;/math&gt;, respectively. Draw a large bar (label it with &lt;math&gt;+&lt;/math&gt; so you know it's positive) for all values of &lt;math&gt;x_i&lt;/math&gt; such that &lt;math&gt;\lfloor i x_i \rfloor&lt;/math&gt; is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at &lt;math&gt;0&lt;/math&gt;, and slowly work your way to &lt;math&gt;1&lt;/math&gt;, marking all disjoint intervals in which your straightedge touches &lt;math&gt;2, 4, 6, &lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. )<br /> If done correctly, you should obtain &lt;math&gt;\boxed{12\ \textbf{(B)}}&lt;/math&gt; as your answer.<br /> <br /> -fidgetboss_4000<br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=23|num-a=25|ab=A}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_23&diff=121720 2010 AMC 12A Problems/Problem 23 2020-04-27T01:14:35Z <p>Hi13: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The number obtained from the last two nonzero digits of &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;n&lt;/math&gt;. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68&lt;/math&gt;<br /> <br /> == Hints and Method of Attack ==<br /> Let &lt;math&gt;P&lt;/math&gt; be the result of dividing &lt;math&gt;90!&lt;/math&gt; by tens such that &lt;math&gt;P&lt;/math&gt; is not divisible by 10. We want to consider &lt;math&gt;P \mod 100&lt;/math&gt;. But because 100 is not prime, and because &lt;math&gt;P&lt;/math&gt; is obviously divisible by 4 (if in doubt, look at the answer choices), we only need to consider &lt;math&gt;P \mod 25&lt;/math&gt;.<br /> <br /> However, 25 is a very particular number. &lt;math&gt;1 * 2 * 3 * 4 \equiv -1 \mod 25&lt;/math&gt;, and so is &lt;math&gt;6 * 7 * 8 * 9&lt;/math&gt;. How can we group terms to take advantage of this fact?<br /> <br /> There might be a problem when you cancel out the 10s from &lt;math&gt;90!&lt;/math&gt;. One method is to cancel out a factor of 2 from an existing number along with a factor of 5. But this might prove cumbersome, as the grouping method will not be as effective. Instead, take advantage of ''inverses'' in modular arithmetic. Just leave the negative powers of 2 in a &quot;storage base,&quot; and take care of the other terms first. Then, use Fermat's Little Theorem to solve for the power of 2.<br /> <br /> == Solution 1==<br /> <br /> We will use the fact that for any integer &lt;math&gt;n&lt;/math&gt;,<br /> &lt;cmath&gt;\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&amp;=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\<br /> &amp;=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\<br /> &amp;=24\pmod{25}\equiv -1\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> First, we find that the number of factors of &lt;math&gt;10&lt;/math&gt; in &lt;math&gt;90!&lt;/math&gt; is equal to &lt;math&gt;\left\lfloor \frac{90}5\right\rfloor+\left\lfloor\frac{90}{25}\right\rfloor=18+3=21&lt;/math&gt;. Let &lt;math&gt;N=\frac{90!}{10^{21}}&lt;/math&gt;. The &lt;math&gt;n&lt;/math&gt; we want is therefore the last two digits of &lt;math&gt;N&lt;/math&gt;, or &lt;math&gt;N\pmod{100}&lt;/math&gt;. If instead we find &lt;math&gt;N\pmod{25}&lt;/math&gt;, we know that &lt;math&gt;N\pmod{100}&lt;/math&gt;, what we are looking for, could be &lt;math&gt;N\pmod{25}&lt;/math&gt;, &lt;math&gt;N\pmod{25}+25&lt;/math&gt;, &lt;math&gt;N\pmod{25}+50&lt;/math&gt;, or &lt;math&gt;N\pmod{25}+75&lt;/math&gt;. Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, because &lt;math&gt;N\pmod{100}&lt;/math&gt; has to be a multiple of 4.<br /> <br /> If we divide &lt;math&gt;N&lt;/math&gt; by &lt;math&gt;5^{21}&lt;/math&gt; by taking out all the factors of &lt;math&gt;5&lt;/math&gt; in &lt;math&gt;N&lt;/math&gt;, we can write &lt;math&gt;N&lt;/math&gt; as &lt;math&gt;\frac M{2^{21}}&lt;/math&gt; where<br /> &lt;cmath&gt;M=1\cdot 2\cdot 3\cdot 4\cdot 1\cdot 6\cdot 7\cdot 8\cdot 9\cdot 2\cdots 89\cdot 18,&lt;/cmath&gt;<br /> where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form &lt;math&gt;5n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;, and every number in the form &lt;math&gt;25n&lt;/math&gt; is replaced by &lt;math&gt;n&lt;/math&gt;.<br /> <br /> The number &lt;math&gt;M&lt;/math&gt; can be grouped as follows:<br /> <br /> &lt;cmath&gt;\begin{align*}M= &amp;(1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots(86\cdot 87\cdot 88\cdot 89)\\<br /> &amp;\cdot (1\cdot 2\cdot 3\cdot 4)(6\cdot 7\cdot 8\cdot 9)\cdots (16\cdot 17\cdot 18) \\<br /> &amp;\cdot (1\cdot 2\cdot 3).\end{align*}&lt;/cmath&gt;<br /> <br /> Where the first line is composed of the numbers in &lt;math&gt;90!&lt;/math&gt; that aren't multiples of five, the second line is the multiples of five '''and not 25''' after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.<br /> <br /> Using the identity at the beginning of the solution, we can reduce &lt;math&gt;M&lt;/math&gt; to<br /> <br /> &lt;cmath&gt;\begin{align*}M&amp;\equiv(-1)^{18} \cdot (-1)^3(16\cdot 17\cdot 18) \cdot (1\cdot 2\cdot 3) \\<br /> &amp;= 1\cdot -21\cdot 6\\<br /> &amp;= -1\pmod{25} =24\pmod{25}.\end{align*}&lt;/cmath&gt;<br /> <br /> Using the fact that &lt;math&gt;2^{10}=1024\equiv -1\pmod{25}&lt;/math&gt; (or simply the fact that &lt;math&gt;2^{21}=2097152&lt;/math&gt; if you have your powers of 2 memorized), we can deduce that &lt;math&gt;2^{21}\equiv 2\pmod{25}&lt;/math&gt;. Therefore &lt;math&gt;N=\frac M{2^{21}}\equiv \frac {24}2\pmod{25}=12\pmod{25}&lt;/math&gt;.<br /> <br /> Finally, combining with the fact that &lt;math&gt;N\equiv 0\pmod 4&lt;/math&gt; yields &lt;math&gt;n=\boxed{\textbf{(A)}\ 12}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;90!&lt;/math&gt; after we truncate its zeros. Notice that &lt;math&gt;90!&lt;/math&gt; has exactly (floored) &lt;math&gt;\left\lfloor\frac{90}{5}\right\rfloor + \left\lfloor\frac{90}{25}\right\rfloor = 21&lt;/math&gt; factors of 5; thus, &lt;cmath&gt;P = 2^{-21}*5^{-21}*90!.&lt;/cmath&gt; We shall consider &lt;math&gt;P&lt;/math&gt; modulo 4 and 25, to determine its residue modulo 100. It is easy to prove that &lt;math&gt;P&lt;/math&gt; is divisible by 4 (consider the number of 2s dividing &lt;math&gt;90!&lt;/math&gt; minus the number of 5s dividing &lt;math&gt;90!&lt;/math&gt;), and so we only need to consider &lt;math&gt;P&lt;/math&gt; modulo 25.<br /> <br /> Now, notice that for integers &lt;math&gt;a, n&lt;/math&gt; we have&lt;cmath&gt;(5n + a)(5n - a) \equiv -a^2 \mod 25.&lt;/cmath&gt;<br /> <br /> Thus, for integral a: &lt;cmath&gt;(10a + 1)(10a + 2)(10a + 3)(10a + 4)(10a + 6)(10a + 7)(10a + 8)(10a + 9) \equiv (-1)(-4)(-9)(-16) \equiv 576 \equiv 1 \mod 25.&lt;/cmath&gt; Using this process, we can essentially remove all the numbers which had not formerly been a multiple of 5 in &lt;math&gt;90!&lt;/math&gt; from consideration.<br /> <br /> Now, we consider the remnants of the 5, 10, 15, 20, ..., 90 not yet eliminated. The 10, 20, 30, ..., 90 becomes 1, 2, 3, 4, 1, 6, 7, 8, 9, whose product is 1 mod 25. Also, the 5, 5, 15, 25, ..., 85 becomes 1, 1, 3, 1, 7, 9, 11, 13, 3, 17 and &lt;math&gt;2^{-12}&lt;/math&gt;. We deduce that from multiplying out the 1, 1, 3, 1, 7, ..., 17 is equivalent to 2 modulo 25, and so we need to compute &lt;math&gt;2^{-11}&lt;/math&gt;. But this is simply by Fermat's Little Theorem &lt;math&gt;2^9 = 512 \equiv 12 \mod 25&lt;/math&gt;. Because 12 is also a multiple of 4, we can utilize the Chinese Remainder Theorem to show that &lt;math&gt;P = 12 \mod 100&lt;/math&gt; and so the answer is &lt;math&gt;\boxed{12}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=22|num-a=24|ab=A}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Hi13 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_25&diff=120806 2019 AMC 12B Problems/Problem 25 2020-04-11T01:56:40Z <p>Hi13: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a convex quadrilateral with &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;CD=6.&lt;/math&gt; Suppose that the centroids of &lt;math&gt;\triangle ABC,\triangle BCD,&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt; form the vertices of an equilateral triangle. What is the maximum possible value of the area of &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30&lt;/math&gt;<br /> <br /> ==Solution 1 (vectors)==<br /> Place an origin at &lt;math&gt;A&lt;/math&gt;, and assign position vectors of &lt;math&gt;B = \vec{p}&lt;/math&gt; and &lt;math&gt;D = \vec{q}&lt;/math&gt;. Since &lt;math&gt;AB&lt;/math&gt; is not parallel to &lt;math&gt;AD&lt;/math&gt;, vectors &lt;math&gt;\vec{p}&lt;/math&gt; and &lt;math&gt;\vec{q}&lt;/math&gt; are linearly independent, so we can write &lt;math&gt;C = m\vec{p} + n\vec{q}&lt;/math&gt; for some constants &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. Now, recall that the centroid of a triangle &lt;math&gt;\triangle XYZ&lt;/math&gt; has position vector &lt;math&gt;\frac{1}{3}\left(\vec{x}+\vec{y}+\vec{z}\right)&lt;/math&gt;. <br /> <br /> Thus the centroid of &lt;math&gt;\triangle ABC&lt;/math&gt; is &lt;math&gt;g_1 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}n\vec{q}&lt;/math&gt;; the centroid of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;g_2 = \frac{1}{3}(m+1)\vec{p} + \frac{1}{3}(n+1)\vec{q}&lt;/math&gt;; and the centroid of &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;g_3 = \frac{1}{3}m\vec{p} + \frac{1}{3}(n+1)\vec{q}&lt;/math&gt;. <br /> <br /> Hence &lt;math&gt;\overrightarrow{G_{1}G_{2}} = \frac{1}{3}\vec{q}&lt;/math&gt;, &lt;math&gt;\overrightarrow{G_{2}G_{3}} = -\frac{1}{3}\vec{p}&lt;/math&gt;, and &lt;math&gt;\overrightarrow{G_{3}G_{1}} = \frac{1}{3}\vec{p} - \frac{1}{3}\vec{q}&lt;/math&gt;. For &lt;math&gt;\triangle G_{1}G_{2}G_{3}&lt;/math&gt; to be equilateral, we need &lt;math&gt;\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{2}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{q}\right| \Rightarrow AB = AD&lt;/math&gt;. Further, &lt;math&gt;\left|\overrightarrow{G_{1}G_{2}}\right| = \left|\overrightarrow{G_{1}G_{3}}\right| \Rightarrow \left|\vec{p}\right| = \left|\vec{p} - \vec{q}\right| = BD&lt;/math&gt;. Hence we have &lt;math&gt;AB = AD = BD&lt;/math&gt;, so &lt;math&gt;\triangle ABD&lt;/math&gt; is equilateral.<br /> <br /> Now let the side length of &lt;math&gt;\triangle ABD&lt;/math&gt; be &lt;math&gt;k&lt;/math&gt;, and let &lt;math&gt;\angle BCD = \theta&lt;/math&gt;. By the Law of Cosines in &lt;math&gt;\triangle BCD&lt;/math&gt;, we have &lt;math&gt;k^2 = 2^2 + 6^2 - 2 \cdot 2 \cdot 6 \cdot \cos{\theta} = 40 - 24\cos{\theta}&lt;/math&gt;. Since &lt;math&gt;\triangle ABD&lt;/math&gt; is equilateral, its area is &lt;math&gt;\frac{\sqrt{3}}{4}k^2 = 10\sqrt{3} - 6\sqrt{3}\cos{\theta}&lt;/math&gt;, while the area of &lt;math&gt;\triangle BCD&lt;/math&gt; is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 6 \cdot \sin{\theta} = 6 \sin{\theta}&lt;/math&gt;. Thus the total area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;10\sqrt{3} + 6\left(\sin{\theta} - \sqrt{3}\cos{\theta}\right) = 10\sqrt{3} + 12\left(\sin{\theta} \frac{1}{2} - \frac{\sqrt{3}}{2}\cos{\theta}\right) = 10\sqrt{3}+12\sin{\left(\theta-60^{\circ}\right)}&lt;/math&gt;, where in the last step we used the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. Alternatively, we can use calculus to find the local maximum. Observe that &lt;math&gt;\sin{\left(\theta-60^{\circ}\right)}&lt;/math&gt; has maximum value &lt;math&gt;1&lt;/math&gt; when e.g. &lt;math&gt;\theta = 150^{\circ}&lt;/math&gt;, which is a valid configuration, so the maximum area is &lt;math&gt;10\sqrt{3} + 12(1) = \boxed{\textbf{(C) } 12+10\sqrt3}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Let &lt;math&gt;G_1&lt;/math&gt;, &lt;math&gt;G_2&lt;/math&gt;, &lt;math&gt;G_3&lt;/math&gt; be the centroids of &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;BCD&lt;/math&gt;, and &lt;math&gt;CDA&lt;/math&gt; respectively, and let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;G_1&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; are collinear due to well-known properties of the centroid. Likewise, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;G_2&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt; are collinear as well. Because (as is also well-known) &lt;math&gt;AG_1 = 3AM&lt;/math&gt; and &lt;math&gt;DG_2 = 3DM&lt;/math&gt;, we have &lt;math&gt;\triangle MG_1G_2\sim\triangle MAD&lt;/math&gt;. This implies that &lt;math&gt;AD&lt;/math&gt; is parallel to &lt;math&gt;G_1G_2&lt;/math&gt;, and in terms of lengths, &lt;math&gt;AD = 3G_1G_2&lt;/math&gt;.<br /> <br /> We can apply the same argument to the pair of triangles &lt;math&gt;\triangle BCD&lt;/math&gt; and &lt;math&gt;\triangle ACD&lt;/math&gt;, concluding that &lt;math&gt;AB&lt;/math&gt; is parallel to &lt;math&gt;G_2G_3&lt;/math&gt; and &lt;math&gt;AB = 3G_2G_3&lt;/math&gt;. Because &lt;math&gt;3G_1G_2 = 3G_2G_3&lt;/math&gt; (due to the triangle being equilateral), &lt;math&gt;AB = AD&lt;/math&gt;, and the pair of parallel lines preserve the &lt;math&gt;60^{\circ}&lt;/math&gt; angle, meaning &lt;math&gt;\angle BAD = 60^\circ&lt;/math&gt;. Therefore &lt;math&gt;\triangle BAD&lt;/math&gt; is equilateral.<br /> <br /> At this point, we can finish as in Solution 1, or, to avoid using trigonometry, we can continue as follows:<br /> <br /> Let &lt;math&gt;BD = 2x&lt;/math&gt;, where &lt;math&gt;2 &lt; x &lt; 4&lt;/math&gt; due to the Triangle Inequality in &lt;math&gt;\triangle BCD&lt;/math&gt;. By breaking the quadrilateral into &lt;math&gt;\triangle ABD&lt;/math&gt; and &lt;math&gt;\triangle BCD&lt;/math&gt;, we can create an expression for the area of &lt;math&gt;ABCD&lt;/math&gt;. We use the formula for the area of an equilateral triangle given its side length to find the area of &lt;math&gt;\triangle ABD&lt;/math&gt; and Heron's formula to find the area of &lt;math&gt;\triangle BCD&lt;/math&gt;.<br /> <br /> After simplifying,<br /> <br /> &lt;cmath&gt;[ABCD] = x^2\sqrt 3 + \sqrt{36 - (x^2-10)^2}&lt;/cmath&gt;<br /> <br /> Substituting &lt;math&gt;k = x^2 - 10&lt;/math&gt;, the expression becomes<br /> <br /> &lt;cmath&gt;[ABCD] = k\sqrt{3} + \sqrt{36 - k^2} + 10\sqrt{3}&lt;/cmath&gt;<br /> <br /> We can ignore the &lt;math&gt;10\sqrt{3}&lt;/math&gt; for now and focus on &lt;math&gt;k\sqrt{3} + \sqrt{36 - k^2}&lt;/math&gt;.<br /> <br /> By the Cauchy-Schwarz inequality,<br /> <br /> &lt;cmath&gt;\left(k\sqrt 3 + \sqrt{36-k^2}\right)^2 \leq \left(\left(\sqrt{3}\right)^2+1^2\right)\left(\left(k\right)^2 + \left(\sqrt{36-k^2}\right)^2\right)&lt;/cmath&gt;<br /> <br /> The RHS simplifies to &lt;math&gt;12^2&lt;/math&gt;, meaning the maximum value of &lt;math&gt;k\sqrt{3} + \sqrt{36 - k^2}&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt;.<br /> <br /> Thus the maximum possible area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C) }12 + 10\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2019|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Hi13