https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Correcthorsebatterystaple&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-28T11:34:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Intermediate_Value_Theorem&diff=144457 Intermediate Value Theorem 2021-02-01T22:08:21Z <p>Correcthorsebatterystaple: Added grammar and punctuation</p> <hr /> <div>The '''Intermediate Value Theorem''' is one of the very interesting properties of continous functions.<br /> <br /> ==Statement==<br /> Take a function &lt;math&gt;f&lt;/math&gt; and interval &lt;math&gt;[a,b]&lt;/math&gt; such that the following hold:<br /> <br /> &lt;math&gt;f:[a,b]\rightarrow\mathbb{R},&lt;/math&gt;<br /> <br /> &lt;math&gt;f&lt;/math&gt; is continuous on &lt;math&gt;[a,b],&lt;/math&gt;<br /> <br /> &lt;math&gt;f(a)&lt;k&lt;f(b).&lt;/math&gt;<br /> <br /> Then, &lt;math&gt;\exists c\in (a,b)&lt;/math&gt; such that &lt;math&gt;f(c)=k.&lt;/math&gt;<br /> <br /> ==Proof==<br /> Consider &lt;math&gt;g:[a,b]\rightarrow\mathbb{R}&lt;/math&gt; such that &lt;math&gt;g(x)=f(x)-k.&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;g(a)&lt;0&lt;/math&gt; and &lt;math&gt;g(b)&gt;0&lt;/math&gt;<br /> <br /> By the [[Location of roots theorem]], &lt;math&gt;\exists c\in (a,b)&lt;/math&gt; such that &lt;math&gt;g(c)=0&lt;/math&gt; or &lt;math&gt;f(c)=k.&lt;/math&gt;<br /> &lt;p align=right&gt;QED&lt;/p&gt;<br /> <br /> ==See Also==<br /> *[[Continuity]]<br /> *[[Location of roots theorem]]<br /> <br /> [[Category:Analysis]]<br /> [[Category:Theorems]]</div> Correcthorsebatterystaple https://artofproblemsolving.com/wiki/index.php?title=Bolzano-Weierstrass_Theorem&diff=144454 Bolzano-Weierstrass Theorem 2021-02-01T21:48:34Z <p>Correcthorsebatterystaple: Removed an inequality which was irrelevant to the article</p> <hr /> <div>Every bounded [[sequence]] of reals contains a convergent subsequence.<br /> <br /> == Proof ==<br /> Lemma 1: A bounded increasing sequence of reals converges to its least upper bound.<br /> <br /> Proof: Suppose &lt;math&gt;(p_n)&lt;/math&gt; is a bounded increasing sequence of reals, so &lt;math&gt;p_1\leq p_2\leq p_3\leq\cdots&lt;/math&gt;. Let &lt;math&gt;p=\sup(\{p_1,p_2,p_3,\ldots\})&lt;/math&gt;. We want to show that &lt;math&gt;(p_n)\rightarrow p&lt;/math&gt;. Assume &lt;math&gt;\epsilon&gt;0&lt;/math&gt;. By the definition of convergence, we need to produce some &lt;math&gt;n\in\mathbb{N}&lt;/math&gt; such that for all &lt;math&gt;m\geq n&lt;/math&gt;, &lt;math&gt;|p_m-p|&lt;\epsilon&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; is the least upper bound of the sequence, there exists an &lt;math&gt;n\in\mathbb{N}&lt;/math&gt; such that &lt;math&gt;p_n&gt;p-\epsilon&lt;/math&gt;, and since &lt;math&gt;(p_n)&lt;/math&gt; is increasing, we know that for all &lt;math&gt;m\geq n&lt;/math&gt;, &lt;math&gt;p_m\geq p_n&gt;p-\epsilon&lt;/math&gt;. Therefore for all &lt;math&gt;m\geq n&lt;/math&gt;, &lt;math&gt;|p_m-p|&lt;\epsilon&lt;/math&gt;. This shows that &lt;math&gt;(p_n)&lt;/math&gt; converges to &lt;math&gt;p&lt;/math&gt;. This completes the proof of Lemma 1.<br /> <br /> We can prove analogously that a bounded decreasing sequence of reals converges to its greatest lower bound.<br /> <br /> <br /> <br /> Lemma 2: Every sequence of reals has a monotone subsequence.<br /> <br /> Proof: Given a sequence &lt;math&gt;(p_n)&lt;/math&gt; of reals, we consider two cases:<br /> <br /> * Case 1: Every infinite subsequence of &lt;math&gt;(p_n)&lt;/math&gt; contains a term that is strictly smaller than infinitely many later terms in the subsequence.<br /> <br /> In this case, we build a strictly increasing subsequence &lt;math&gt;(i_n)&lt;/math&gt;.<br /> <br /> Note that we can apply the assumption of Case 1 to &lt;math&gt;(p_n)&lt;/math&gt; itself. Thus there exists some &lt;math&gt;p_a&lt;/math&gt; such that there exist infinitely many &lt;math&gt;b&gt;a&lt;/math&gt; such that &lt;math&gt;p_b&gt;p_a&lt;/math&gt;. Let &lt;math&gt;i_1=p_a&lt;/math&gt;, and consider the subsequence &lt;math&gt;(q_n)&lt;/math&gt; where &lt;math&gt;q_1=p_a&lt;/math&gt;, and &lt;math&gt;q_k&lt;/math&gt; is the &lt;math&gt;k-1&lt;/math&gt;th term in &lt;math&gt;(p_n)&lt;/math&gt; after &lt;math&gt;p_a&lt;/math&gt; that is greater than &lt;math&gt;p_a&lt;/math&gt;. This exists by the assumption.<br /> <br /> Now we can apply the assumption of Case 1 to &lt;math&gt;(q_n)&lt;/math&gt;. Thus there exists some &lt;math&gt;q_c&lt;/math&gt; such that there exist infinitely many &lt;math&gt;d&gt;c&lt;/math&gt; such that &lt;math&gt;q_d&gt;q_c&lt;/math&gt;. Since &lt;math&gt;(q_n)&lt;/math&gt; is a subsequence of &lt;math&gt;(p_n)&lt;/math&gt;, there exists an &lt;math&gt;e\in\mathbb{N}&lt;/math&gt; such that &lt;math&gt;p_e=q_c&lt;/math&gt;. We then let &lt;math&gt;i_2=p_e&lt;/math&gt;, and let &lt;math&gt;(r_n)&lt;/math&gt; be a subsequence of &lt;math&gt;(q_n)&lt;/math&gt; such that &lt;math&gt;r_1=q_c&lt;/math&gt;, and &lt;math&gt;r_k&lt;/math&gt; is the &lt;math&gt;k-1&lt;/math&gt;th term in &lt;math&gt;(q_n)&lt;/math&gt; after &lt;math&gt;q_c&lt;/math&gt; that is gerater tha &lt;math&gt;q_c&lt;/math&gt;. Note that &lt;math&gt;(r_n)&lt;/math&gt; is a subsequence of &lt;math&gt;(p_n)&lt;/math&gt;.<br /> <br /> We can continue in this fashion to construct a strictly increasing subsequence of &lt;math&gt;(p_n)&lt;/math&gt;.<br /> <br /> * Case 2: Case 1 fails.<br /> <br /> In other words, there exists a subsequence &lt;math&gt;(q_n)&lt;/math&gt; of &lt;math&gt;(p_n)&lt;/math&gt; such that every term in &lt;math&gt;(q_n)&lt;/math&gt; is at least as large as some later term in &lt;math&gt;(q_n)&lt;/math&gt;. In this case, we can build a decreasing subsequence &lt;math&gt;(d_n)&lt;/math&gt;.<br /> <br /> Consider such a sequence &lt;math&gt;(q_n)&lt;/math&gt;. Let &lt;math&gt;d_1=q_1&lt;/math&gt;. By the assumption of Case 2, there exists some &lt;math&gt;q_a\leq q_1&lt;/math&gt;. Let &lt;math&gt;d_2=q_a&lt;/math&gt;. By the assumption again, there exists some &lt;math&gt;q_a\leq q_b&lt;/math&gt;. Let &lt;math&gt;d_3=q_b&lt;/math&gt;. We can continue in this fashion to build a decreasing subsequence of &lt;math&gt;(q_n)&lt;/math&gt;. Since &lt;math&gt;(q_n)&lt;/math&gt; is a subsequence of &lt;math&gt;(p_n)&lt;/math&gt;, we have that &lt;math&gt;(d_n)&lt;/math&gt; is a decreasing subsequence of &lt;math&gt;(p_n)&lt;/math&gt;. This completes the proof of Lemma 2.<br /> <br /> <br /> <br /> The Bolzano-Weierstrass Theorem follows immediately: every bounded sequence of reals contains some monotone subsequence by Lemma 2, which is in turn bounded. This subsequence is convergent by Lemma 1, which completes the proof.<br /> <br /> == See also == <br /> <br /> [[Category:Analysis]]<br /> [[Category:Theorems]]<br /> {{stub}}</div> Correcthorsebatterystaple https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_15&diff=140835 2005 AIME II Problems/Problem 15 2020-12-28T17:17:56Z <p>Correcthorsebatterystaple: /*Solution 6*/</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; w_1 &lt;/math&gt; and &lt;math&gt; w_2 &lt;/math&gt; denote the [[circle]]s &lt;math&gt; x^2+y^2+10x-24y-87=0 &lt;/math&gt; and &lt;math&gt; x^2 +y^2-10x-24y+153=0, &lt;/math&gt; respectively. Let &lt;math&gt; m &lt;/math&gt; be the smallest positive value of &lt;math&gt; a &lt;/math&gt; for which the line &lt;math&gt; y=ax &lt;/math&gt; contains the center of a circle that is externally [[tangent (geometry)|tangent]] to &lt;math&gt; w_2 &lt;/math&gt; and internally tangent to &lt;math&gt; w_1. &lt;/math&gt; Given that &lt;math&gt; m^2=\frac pq, &lt;/math&gt; where &lt;math&gt; p &lt;/math&gt; and &lt;math&gt; q &lt;/math&gt; are relatively prime integers, find &lt;math&gt; p+q. &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Rewrite the given equations as &lt;math&gt;(x+5)^2 + (y-12)^2 = 256&lt;/math&gt; and &lt;math&gt;(x-5)^2 + (y-12)^2 = 16&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;w_3&lt;/math&gt; have center &lt;math&gt;(x,y)&lt;/math&gt; and radius &lt;math&gt;r&lt;/math&gt;. Now, if two circles with radii &lt;math&gt;r_1&lt;/math&gt; and &lt;math&gt;r_2&lt;/math&gt; are externally tangent, then the distance between their centers is &lt;math&gt;r_1 + r_2&lt;/math&gt;, and if they are internally tangent, it is &lt;math&gt;|r_1 - r_2|&lt;/math&gt;. So we have<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> r + 4 &amp;= \sqrt{(x-5)^2 + (y-12)^2} \\<br /> 16 - r &amp;= \sqrt{(x+5)^2 + (y-12)^2} \end{align*} &lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;r&lt;/math&gt; in both equations and setting them equal, then simplifying, yields<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> 20 - \sqrt{(x+5)^2 + (y-12)^2} &amp;= \sqrt{(x-5)^2 + (y-12)^2} \\<br /> 20+x &amp;= 2\sqrt{(x+5)^2 + (y-12)^2}<br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Squaring again and canceling yields &lt;math&gt;1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.&lt;/math&gt;<br /> <br /> So the locus of points that can be the center of the circle with the desired properties is an [[ellipse]].<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d; <br /> pair A = (-5, 12), B = (5, 12), C = (0, 0);<br /> D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype(&quot;2 2&quot;) + d + red);<br /> D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP(&quot;y=ax&quot;,(14,14 * (69/100)^.5),E),EndArrow(4));<br /> <br /> void bluecirc (real x) {<br /> pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue);<br /> D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype(&quot;4 4&quot;));<br /> }<br /> <br /> bluecirc(-9.2); bluecirc(-4); bluecirc(3);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since the center lies on the line &lt;math&gt;y = ax&lt;/math&gt;, we substitute for &lt;math&gt;y&lt;/math&gt; and expand: <br /> &lt;cmath&gt;1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.&lt;/cmath&gt;<br /> <br /> We want the value of &lt;math&gt;a&lt;/math&gt; that makes the line &lt;math&gt;y=ax&lt;/math&gt; tangent to the ellipse, which will mean that for that choice of &lt;math&gt;a&lt;/math&gt; there is only one solution to the most recent equation. But a quadratic has one solution [[iff]] its discriminant is &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;(-96a)^2 - 4(3+4a^2)(276) = 0&lt;/math&gt;.<br /> <br /> Solving yields &lt;math&gt;a^2 = \frac{69}{100}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> As above, we rewrite the equations as &lt;math&gt;(x+5)^2 + (y-12)^2 = 256&lt;/math&gt; and &lt;math&gt;(x-5)^2 + (y-12)^2 = 16&lt;/math&gt;. Let &lt;math&gt;F_1=(-5,12)&lt;/math&gt; and &lt;math&gt;F_2=(5,12)&lt;/math&gt;. If a circle with center &lt;math&gt;C=(a,b)&lt;/math&gt; and radius &lt;math&gt;r&lt;/math&gt; is externally tangent to &lt;math&gt;w_2&lt;/math&gt; and internally tangent to &lt;math&gt;w_1&lt;/math&gt;, then &lt;math&gt;CF_1=16-r&lt;/math&gt; and &lt;math&gt;CF_2=4+r&lt;/math&gt;. Therefore, &lt;math&gt;CF_1+CF_2=20&lt;/math&gt;. In particular, the locus of points &lt;math&gt;C&lt;/math&gt; that can be centers of circles must be an ellipse with foci &lt;math&gt;F_1&lt;/math&gt; and &lt;math&gt;F_2&lt;/math&gt; and major axis &lt;math&gt;20&lt;/math&gt;.<br /> <br /> Clearly, the minimum value of the slope &lt;math&gt;a&lt;/math&gt; will occur when the line &lt;math&gt;y=ax&lt;/math&gt; is tangent to this ellipse. Suppose that this point of tangency is denoted by &lt;math&gt;T&lt;/math&gt;, and the line &lt;math&gt;y=ax&lt;/math&gt; is denoted by &lt;math&gt;\ell&lt;/math&gt;. Then we reflect the ellipse over &lt;math&gt;\ell&lt;/math&gt; to a new ellipse with foci &lt;math&gt;F_1'&lt;/math&gt; and &lt;math&gt;F_2'&lt;/math&gt; as shown below.<br /> &lt;center&gt;&lt;asy&gt;<br /> size(220); <br /> pair F1 = (-5, 12), F2 = (5, 12),C=(0,12);<br /> draw(circle(F1,16));<br /> draw(circle(F2,4));<br /> draw(ellipse(C,10,5*sqrt(3)));<br /> xaxis(&quot;$x$&quot;,Arrows);<br /> yaxis(&quot;$y$&quot;,Arrows);<br /> dot(F1^^F2^^C);<br /> <br /> real l(real x) {return sqrt(69)*x/10;}<br /> path g=graph(l,-7,14);<br /> draw(g);<br /> draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3)));<br /> pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10)));<br /> dot(T);<br /> pair F1P=reflect((0,0),(10,l(10)))*F1;<br /> pair F2P=reflect((0,0),(10,l(10)))*F2;<br /> dot(F1P^^F2P);<br /> dot((0,0));<br /> label(&quot;$F_1$&quot;,F1,N,fontsize(9));<br /> label(&quot;$F_2$&quot;,F2,N,fontsize(9));<br /> label(&quot;$F_1'$&quot;,F1P,SE,fontsize(9));<br /> label(&quot;$F_2'$&quot;,F2P,SE,fontsize(9));<br /> label(&quot;$O$&quot;,(0,0),NW,fontsize(9));<br /> label(&quot;$\ell$&quot;,(13,l(13)),SE,fontsize(9));<br /> label(&quot;$T$&quot;,T,NW,fontsize(9));<br /> draw((0,0)--F1--F2--F2P--F1P--cycle);<br /> draw(F1--F2P^^F2--F1P);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that &lt;math&gt;F_1&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt;, and &lt;math&gt;F_2'&lt;/math&gt; are collinear, and similarly, &lt;math&gt;F_2&lt;/math&gt;, &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;F_1'&lt;/math&gt; are collinear. Therefore, &lt;math&gt;OF_1F_2F_2'F_1'&lt;/math&gt; is a pentagon with &lt;math&gt;OF_1=OF_2=OF_1'=OF_2'=13&lt;/math&gt;, &lt;math&gt;F_1F_2=F_1'F_2'=10&lt;/math&gt;, and &lt;math&gt;F_1F_2'=F_1'F_2=20&lt;/math&gt;. Note that &lt;math&gt;\ell&lt;/math&gt; bisects &lt;math&gt;\angle F_1'OF_1&lt;/math&gt;. We can bisect this angle by bisecting &lt;math&gt;\angle F_1'OF_2&lt;/math&gt; and &lt;math&gt;F_2OF_1&lt;/math&gt; separately.<br /> <br /> We proceed using complex numbers. Triangle &lt;math&gt;F_2OF_1'&lt;/math&gt; is isosceles with side lengths &lt;math&gt;13,13,20&lt;/math&gt;. The height of this from the base of &lt;math&gt;20&lt;/math&gt; is &lt;math&gt;\sqrt{69}&lt;/math&gt;. Therefore, the complex number &lt;math&gt;\sqrt{69}+10i&lt;/math&gt; represents the bisection of &lt;math&gt;\angle F_1'OF_2&lt;/math&gt;.<br /> <br /> Similarly, using the 5-12-13 triangles, we easily see that &lt;math&gt;12+5i&lt;/math&gt; represents the bisection of the angle &lt;math&gt;F_2OF_1&lt;/math&gt;. Therefore, we can add these two angles together by multiplying the complex numbers, finding<br /> &lt;cmath&gt;\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.&lt;/cmath&gt;<br /> Now the point &lt;math&gt;F_1&lt;/math&gt; is given by the complex number &lt;math&gt;-5+12i&lt;/math&gt;. Therefore, to find a point on line &lt;math&gt;\ell&lt;/math&gt;, we simply subtract &lt;math&gt;\frac{1}{2}\angle F_1'OF_1&lt;/math&gt;, which is the same as multiplying &lt;math&gt;-5+12i&lt;/math&gt; by the conjugate of &lt;math&gt;(\sqrt{69}+10i)(12+5i)&lt;/math&gt;. We find<br /> &lt;cmath&gt;(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).&lt;/cmath&gt;<br /> In particular, note that the tangent of the argument of this complex number is &lt;math&gt;\sqrt{69}/10&lt;/math&gt;, which must be the slope of the tangent line. Hence &lt;math&gt;a^2=69/100&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> We use the same reflection as in Solution 2. As &lt;math&gt;OF_1'=OF_2=13&lt;/math&gt;, we know that &lt;math&gt;\triangle OF_1'F_2&lt;/math&gt; is isosceles. Hence &lt;math&gt;\angle F_2F_1'O=\angle F_1'F_2O&lt;/math&gt;. But by symmetry, we also know that &lt;math&gt;\angle OF_1T=\angle F_2F_1'O&lt;/math&gt;. Hence &lt;math&gt;\angle OF_1T=\angle F_1'F_2O&lt;/math&gt;. In particular, as &lt;math&gt;\angle OF_1T=\angle OF_2T&lt;/math&gt;, this implies that &lt;math&gt;O, F_1, F_2&lt;/math&gt;, and &lt;math&gt;T&lt;/math&gt; are concyclic.<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;F_2F_1'&lt;/math&gt; with the &lt;math&gt;x&lt;/math&gt;-axis. As &lt;math&gt;F_1F_2&lt;/math&gt; is parallel to the &lt;math&gt;x&lt;/math&gt;-axis, we know that &lt;cmath&gt;\angle TXO=180-\angle F_1F_2T.\tag{1}&lt;/cmath&gt; But &lt;cmath&gt;180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}&lt;/cmath&gt; By the fact that &lt;math&gt;OF_1F_2T&lt;/math&gt; is cyclic, &lt;cmath&gt;\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}&lt;/cmath&gt; Therefore, combining (1), (2), and (3), we find that<br /> &lt;cmath&gt;\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}&lt;/cmath&gt;<br /> <br /> By symmetry, we also know that<br /> &lt;cmath&gt;\angle F_1TO=\angle OTF_1'.\tag{5}&lt;/cmath&gt;<br /> Therefore, (4) and (5) show by AA similarity that &lt;math&gt;\triangle F_1OT\sim \triangle OXT&lt;/math&gt;. Therefore, &lt;math&gt;\angle XOT=\angle OF_1T&lt;/math&gt;.<br /> <br /> Now as &lt;math&gt;OF_1=OF_2'=13&lt;/math&gt;, we know that &lt;math&gt;\triangle OF_1F_2'&lt;/math&gt; is isosceles, and as &lt;math&gt;F_1F_2'=20&lt;/math&gt;, we can drop an altitude to &lt;math&gt;F_1F_2'&lt;/math&gt; to easily find that &lt;math&gt;\tan \angle OF_1T=\sqrt{69}/10&lt;/math&gt;. Therefore, &lt;math&gt;\tan\angle XOT&lt;/math&gt;, which is the desired slope, must also be &lt;math&gt;\sqrt{69}/10&lt;/math&gt;. As before, we conclude that the answer is &lt;math&gt;\boxed{169}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> [[Image:2005_AIME_II_-15.png||center|800px]]<br /> First, rewrite the equations for the circles as &lt;math&gt;(x+5)^2+(y-12)^2=16^2&lt;/math&gt; and &lt;math&gt;(x-5)^2+(y-12)^2=4^2&lt;/math&gt;. <br /> Then, choose a point &lt;math&gt;(a,b)&lt;/math&gt; that is a distance of &lt;math&gt;x&lt;/math&gt; from both circles. Use the distance formula between &lt;math&gt;(a,b)&lt;/math&gt; and each of &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; (in the diagram above). The distances, as can be seen in the diagram above are &lt;math&gt;16-x&lt;/math&gt; and &lt;math&gt;4+x&lt;/math&gt;, respectively.<br /> &lt;cmath&gt;(a-5)^2+(b-12)^2=(4+x)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;(a+5)^2+(b-12)^2=(16-x)^2&lt;/cmath&gt;<br /> Subtracting the first equation from the second gives &lt;cmath&gt;20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2&lt;/cmath&gt;<br /> Substituting this into the first equation gives<br /> &lt;cmath&gt;a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4&lt;/cmath&gt;<br /> &lt;cmath&gt;b^2-24b+69+\frac{3a^2}4=0&lt;/cmath&gt;<br /> Now, instead of converting this to the equation of an eclipse, solve for &lt;math&gt;b&lt;/math&gt; and then divide by &lt;math&gt;a&lt;/math&gt;.<br /> &lt;cmath&gt;b=\frac{24\pm\sqrt{300-3a^2}}{2}&lt;/cmath&gt;<br /> We take the smaller root to minimize &lt;math&gt;\frac b a&lt;/math&gt;.<br /> &lt;cmath&gt;\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}&lt;/cmath&gt;<br /> Now, let &lt;math&gt;10\cos\theta=a&lt;/math&gt;. This way, &lt;math&gt;\sqrt{100-a^2}=10\sin\theta&lt;/math&gt;.<br /> Substitute this in. &lt;math&gt;\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta&lt;/math&gt;<br /> Then, take the derivative of this and set it to 0 to find the minimum value.<br /> &lt;math&gt;\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}&lt;/math&gt;<br /> Then, use this value of &lt;math&gt;\sin\theta&lt;/math&gt; to find the minimum of &lt;math&gt;\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta&lt;/math&gt; to get &lt;math&gt;\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}&lt;/math&gt;<br /> <br /> ==Solution 5 (probably fastest)==<br /> Like before, notice that the distances from the centers of the given circles to the desired center are &lt;math&gt;4+r&lt;/math&gt; and &lt;math&gt;16-r&lt;/math&gt;, which add up to &lt;math&gt;20&lt;/math&gt;. This means that the possible centers of the third circle lie on an ellipse with foci &lt;math&gt;(-5, 12)&lt;/math&gt; and &lt;math&gt;(5, 12)&lt;/math&gt;. Using the fact that the sum of the distances from the foci is &lt;math&gt;20&lt;/math&gt;, we find that the semi-major axis has length &lt;math&gt;10&lt;/math&gt; and the semi-minor axis has length &lt;math&gt;5\sqrt{3}&lt;/math&gt;. Therefore, the equation of the ellipse is &lt;cmath&gt;\dfrac{x^2}{100}+\dfrac{(y-12)^2}{75} = 1,&lt;/cmath&gt; where the numbers &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;75&lt;/math&gt; come from &lt;math&gt;10^2&lt;/math&gt; and &lt;math&gt;(5\sqrt{3})^2&lt;/math&gt; respectively. <br /> <br /> <br /> We proceed to find &lt;math&gt;m&lt;/math&gt; using the same method as Solution 1.<br /> <br /> ==Solution 6==<br /> First, obtain the equation of the ellipse as laid out in previous solutions. We now scale the coordinate plane in the &lt;math&gt;x&lt;/math&gt; direction by a factor of &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt; centered at &lt;math&gt;x=0.&lt;/math&gt; This takes the ellipse to a circle centered at &lt;math&gt;(0,12)&lt;/math&gt; with radius &lt;math&gt;5\sqrt{3}&lt;/math&gt; and takes the line &lt;math&gt;y=ax&lt;/math&gt; to &lt;math&gt;y=\left( \frac{\sqrt{3}}{2} \right)^{-1} ax.&lt;/math&gt; The tangent point of our line to the circle with positive slope forms a right triangle with the origin and the center of the circle. Thus, the distance from this tangent point to the origin is &lt;math&gt;\sqrt{69}.&lt;/math&gt; By similar triangles, the slope of this line is then &lt;math&gt;\frac{\sqrt{69}}{5\sqrt{3}}.&lt;/math&gt; We multiply this by &lt;math&gt;\frac{\sqrt{3}}{2}&lt;/math&gt; to get &lt;math&gt;a=\frac{\sqrt{69}}{10},&lt;/math&gt; so our final answer is &lt;math&gt;\boxed{169.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=14|after=Last Question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Correcthorsebatterystaple https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_6&diff=140739 2003 AIME II Problems/Problem 6 2020-12-27T16:20:29Z <p>Correcthorsebatterystaple: /*Solution 3*/</p> <hr /> <div>== Problem ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; &lt;math&gt;AB = 13,&lt;/math&gt; &lt;math&gt;BC = 14,&lt;/math&gt; &lt;math&gt;AC = 15,&lt;/math&gt; and point &lt;math&gt;G&lt;/math&gt; is the intersection of the medians. Points &lt;math&gt;A',&lt;/math&gt; &lt;math&gt;B',&lt;/math&gt; and &lt;math&gt;C',&lt;/math&gt; are the images of &lt;math&gt;A,&lt;/math&gt; &lt;math&gt;B,&lt;/math&gt; and &lt;math&gt;C,&lt;/math&gt; respectively, after a &lt;math&gt;180^\circ&lt;/math&gt; rotation about &lt;math&gt;G.&lt;/math&gt; What is the area of the union of the two regions enclosed by the triangles &lt;math&gt;ABC&lt;/math&gt; and &lt;math&gt;A'B'C'?&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Since a &lt;math&gt;13-14-15&lt;/math&gt; triangle is a &lt;math&gt;5-12-13&lt;/math&gt; triangle and a &lt;math&gt;9-12-15&lt;/math&gt; triangle &quot;glued&quot; together on the &lt;math&gt;12&lt;/math&gt; side, &lt;math&gt;[ABC]=\frac{1}{2}\cdot12\cdot14=84&lt;/math&gt;. <br /> <br /> There are six points of intersection between &lt;math&gt;\Delta ABC&lt;/math&gt; and &lt;math&gt;\Delta A'B'C'&lt;/math&gt;. Connect each of these points to &lt;math&gt;G&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> size(8cm);<br /> pair A,B,C,G,D,E,F,A_1,A_2,B_1,B_2,C_1,C_2;<br /> B=(0,0);<br /> A=(5,12);<br /> C=(14,0);<br /> E=(12.6667,8);<br /> D=(7.6667,-4);<br /> F=(-1.3333,8);<br /> G=(6.3333,4);<br /> B_1=(4.6667,0);<br /> B_2=(1.6667,4);<br /> A_1=(3.3333,8);<br /> A_2=(8,8);<br /> C_1=(11,4);<br /> C_2=(9.3333,0);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(G);<br /> dot(D);<br /> dot(E);<br /> dot(F);<br /> dot(A_1);<br /> dot(B_1);<br /> dot(C_1);<br /> dot(A_2);<br /> dot(B_2);<br /> dot(C_2);<br /> draw(B--A--C--cycle);<br /> draw(E--D--F--cycle);<br /> draw(B_1--A_2);<br /> draw(A_1--C_2);<br /> draw(C_1--B_2);<br /> label(&quot;$B$&quot;,B,WSW);<br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$G$&quot;,G,S);<br /> label(&quot;$B'$&quot;,E,ENE);<br /> label(&quot;$A'$&quot;,D,S);<br /> label(&quot;$C'$&quot;,F,WNW);<br /> &lt;/asy&gt;<br /> <br /> There are &lt;math&gt;12&lt;/math&gt; smaller congruent triangles which make up the desired area. Also, &lt;math&gt;\Delta ABC&lt;/math&gt; is made up of &lt;math&gt;9&lt;/math&gt; of such triangles. <br /> Therefore, &lt;math&gt;\left[\Delta ABC \bigcup \Delta A'B'C'\right] = \frac{12}{9}[\Delta ABC]= \frac{4}{3}\cdot84=\boxed{112}&lt;/math&gt;.<br /> <br /> ==Solution 2(Doesn’t require good diagram)==<br /> <br /> First, find the area of &lt;math&gt;\Delta ABC&lt;/math&gt; either like the first solution or by using Heron’s Formula. Then, draw the medians from &lt;math&gt;G&lt;/math&gt; to each of &lt;math&gt;A, B, C, A’, B’,&lt;/math&gt; and &lt;math&gt;C’&lt;/math&gt;. Since the medians of a triangle divide the triangle into 6 triangles with equal area, we can find that each of the 6 outer triangles have equal area. (Proof: Since I’m too lazy to draw out a diagram, I’ll just have you borrow the one above. Draw medians &lt;math&gt;GA&lt;/math&gt; and &lt;math&gt;GB’&lt;/math&gt;, and let’s call the points that &lt;math&gt;GA&lt;/math&gt; intersects &lt;math&gt;C’B’&lt;/math&gt; “&lt;math&gt;H&lt;/math&gt;” and the point &lt;math&gt;GB’&lt;/math&gt; intersects &lt;math&gt;AC&lt;/math&gt; “&lt;math&gt;I&lt;/math&gt;”. From the previous property and the fact that both &lt;math&gt;\Delta ABC&lt;/math&gt; and &lt;math&gt;\Delta A’B’C’&lt;/math&gt; are congruent, &lt;math&gt;\Delta GHB’&lt;/math&gt; has the same area as &lt;math&gt;\Delta GIA&lt;/math&gt;. Because of that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles).<br /> Also, since the centroid of a triangle divides each median with the ratio &lt;math&gt;2:1&lt;/math&gt;, along with the previous fact, each outer triangle has &lt;math&gt;1/9&lt;/math&gt; the area of &lt;math&gt;\Delta ABC&lt;/math&gt; and &lt;math&gt;\Delta A’B’C’&lt;/math&gt;. Thus, the area of the region required is &lt;math&gt;\frac{4}{3}&lt;/math&gt; times the area of &lt;math&gt;\Delta ABC&lt;/math&gt; which is &lt;math&gt;\boxed {112}&lt;/math&gt;.<br /> <br /> Solution by Someonenumber011<br /> <br /> ==Solution 3 (Rigorous)==<br /> <br /> &lt;math&gt;[ABC]&lt;/math&gt; can be calculated as 84 using Heron's formula or other methods. Since a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation is equivalent to reflection through a point, we have a homothety with scale factor &lt;math&gt;-1&lt;/math&gt; from &lt;math&gt;ABC&lt;/math&gt; to &lt;math&gt;A'B'C'&lt;/math&gt; through the centroid &lt;math&gt;G&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; be the midpoint of &lt;math&gt;BC&lt;/math&gt; which maps to &lt;math&gt;M'&lt;/math&gt; and note that &lt;math&gt;A'G=AG=2GM,&lt;/math&gt; implying that &lt;math&gt;GM=MA'.&lt;/math&gt; Similarly, we have &lt;math&gt;AM'=M'G=GM=MA'.&lt;/math&gt; Also let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the intersections of &lt;math&gt;BC&lt;/math&gt; with &lt;math&gt;A'B'&lt;/math&gt; and &lt;math&gt;A'C',&lt;/math&gt; respectively. The homothety implies that we must have &lt;math&gt;DE || B'C',&lt;/math&gt; so there is in fact another homothety centered at &lt;math&gt;A'&lt;/math&gt; taking &lt;math&gt;A'DE&lt;/math&gt; to &lt;math&gt;A'B'C'&lt;/math&gt;. Since &lt;math&gt;A'M'=3A'M,&lt;/math&gt; the scale factor of this homothety is 3 and thus &lt;math&gt;[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].&lt;/math&gt; We can apply similar reasoning to the other small triangles in &lt;math&gt;A'B'C'&lt;/math&gt; not contained within &lt;math&gt;ABC&lt;/math&gt;, so our final answer is &lt;math&gt;[ABC]+3\cdot\frac{1}{9}[ABC]=\boxed{112.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2003|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Correcthorsebatterystaple