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Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 2"

(Created page with "==Problem== Let <math>ABC</math> be a triangle with circumcircle <math>\omega</math>. The midpoint of <math>AB</math> is <math>M,</math> and the line <math>CM</math> intersec...")
 
(→‎Solution)
 
(One intermediate revision by the same user not shown)
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==Solution==
 
==Solution==
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<asy>
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import geometry;
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unitsize(2.5cm);
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point A=dir(00);
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point B=dir(180);
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point C=dir(60);
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point M=0.5*A+0.5*B;
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point P=-C;
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draw(A--B--C--cycle);
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draw(unitcircle,blue);
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draw(C--P,red+dashed);
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dot(A); dot(B); dot(C); dot(M); dot(P);
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label("$A$", A, dir(0));
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label("$B$", B, dir(180));
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label("$C$", C, dir(60));
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label("$M$", M+(0.05,0), dir(-90));
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label("$P$", P, dir(240));
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label("$\omega$", (0,-0.9));
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</asy>
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To begin, note that <math>\angle AMC = 60^\circ</math> and <math>AM = BM = MC</math>. This means that <math>\triangle AMC</math> is equilateral. Now, observe that <math>\angle BCM = \tfrac{1}{2}(180^\circ - 120^\circ) = 30^\circ</math>. Thus, <math>\angle ACB = 90^\circ</math>. Because <math>\triangle AMC</math> is equilateral, <math>\angle BAC = 60^\circ</math>, implying that <math>\triangle ABC</math> is a 30-60-90 triangle.
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The circumcenter of a right triangle is the midpoint of its hypotenuse, so <math>M</math> is the center of <math>\omega</math> and <math>PM</math> is a radius of <math>\omega</math>. Therefore, it suffices to find the length of <math>AM</math> since <math>AM</math> is also a radius of <math>\omega</math>. By 30-60-90 triangle properties, <math>AB = \tfrac{2}{\sqrt{3}} \cdot BC = \tfrac{40\sqrt{3}}{3}</math>, so <math>AM = \tfrac{1}{2}\cdot AB = \tfrac{20\sqrt{3}}{3}</math>. We extract <math>20+3+3 = \boxed{26}</math>.
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~Sedro

Latest revision as of 11:57, 11 September 2025

Problem

Let $ABC$ be a triangle with circumcircle $\omega$. The midpoint of $AB$ is $M,$ and the line $CM$ intersects $\omega$ again at $P$. Given $\angle BMC = 120^\circ,$ $\triangle BMC$ is isosceles, and $BC = 20,$ the length of $PM$ can be written as $\tfrac{a\sqrt{b}}{c},$ where $a,$ $b,$ and $c$ are positive integers such that $a$ and $c$ are relatively prime and $b$ is square-free. Find $a+b+c$.

Solution

[asy] import geometry; unitsize(2.5cm); point A=dir(00); point B=dir(180); point C=dir(60); point M=0.5*A+0.5*B; point P=-C; draw(A--B--C--cycle); draw(unitcircle,blue); draw(C--P,red+dashed); dot(A); dot(B); dot(C); dot(M); dot(P); label("$A$", A, dir(0)); label("$B$", B, dir(180)); label("$C$", C, dir(60)); label("$M$", M+(0.05,0), dir(-90)); label("$P$", P, dir(240)); label("$\omega$", (0,-0.9)); [/asy]

To begin, note that $\angle AMC = 60^\circ$ and $AM = BM = MC$. This means that $\triangle AMC$ is equilateral. Now, observe that $\angle BCM = \tfrac{1}{2}(180^\circ - 120^\circ) = 30^\circ$. Thus, $\angle ACB = 90^\circ$. Because $\triangle AMC$ is equilateral, $\angle BAC = 60^\circ$, implying that $\triangle ABC$ is a 30-60-90 triangle.

The circumcenter of a right triangle is the midpoint of its hypotenuse, so $M$ is the center of $\omega$ and $PM$ is a radius of $\omega$. Therefore, it suffices to find the length of $AM$ since $AM$ is also a radius of $\omega$. By 30-60-90 triangle properties, $AB = \tfrac{2}{\sqrt{3}} \cdot BC = \tfrac{40\sqrt{3}}{3}$, so $AM = \tfrac{1}{2}\cdot AB = \tfrac{20\sqrt{3}}{3}$. We extract $20+3+3 = \boxed{26}$.

~Sedro

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