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

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

Latest revision as of 11:57, 11 September 2025

Problem

Solution

@@ Line 26: / Line 26: @@
 label("$M$", M+(0.05,0), dir(-90));
 label("$P$", P, dir(240));
+label("$\omega$", (0,-0.9));
 </asy>