2025 SSMO Speed Round Problems/Problem 6

Problem

The centroid $G$ of $\triangle{ABC}$ has distances $21,$ $60,$ and $28$ from sides $AB,$ $BC,$ and $CA,$ respectively. Find the perimeter of $\triangle{ABC}$ .

Solution

$[asy] unitsize(0.15cm); pair A,B,C,G,M; B=(0,0); A=(60,0); C=(24.17,17.78); G=(28.06,5.93); M=(30,0); fill(A--B--G--cycle,palegreen); dot(A); dot(B); dot(C); dot(G); dot(M); draw(A--B--C--cycle); label("$A$",A,dir(0)); label("$B$",B,dir(180)); label("$C$",C,dir(90)); label("$G$",G,dir(45)); label("$M$",(31,0),dir(270)); draw(A--G); draw(B--G); draw(C--G); draw(G--M,dashed); [/asy]$

Let $a=BC$ , $b=CA$ , and $c=AB$ . If $M$ denotes the midpoint of $AB$ , note that by centroid properties, $GM:CM = 1:3$ , and hence, $[ABG] = [BGM] + [AGM] = \tfrac{1}{3}[BCM]+\tfrac{1}{3}[ACM] = \tfrac{1}{3}[ABC]$ by same-altitude triangles. Using analogous reasoning, we can show that $[ABG]=[BCG]=[CAG] = \tfrac{1}{3}[ABC]$ .

We know that the length of the altitude from $G$ to $AB$ is $21$ , and thus $[ABG] = \tfrac{21}{2}c$ . Similarly, we can find that $[BCG] = 30a$ and $[CAG] = 14b$ . Equating the areas of these triangles, we have $30a = 14b = \tfrac{21}{2}c$ , which is equivalent to $a:b:c = 7:15:20$ .

Let $a = 7x$ , $b=15x$ , and $c=20x$ for some positive real $x$ ; we now solve for $x$ by computing the area of $ABC$ in two different ways. On one hand, we have $[ABC] = [ABG]+[BCG]+[CAG] = 630x$ . On the other hand, by Heron's formula, $[ABC] = 42x^2$ . Hence, $630x = 42x^2$ , which gives $x = 15$ . The perimeter of $ABC$ is $42x = \boxed{630}$ .

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2025 SSMO Speed Round Problems/Problem 6

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