Difference between revisions of "2025 SSMO Speed Round Problems/Problem 6"

Revision as of 16:51, 9 September 2025

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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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 6"

Revision as of 16:51, 9 September 2025

Problem

Solution

@@ Line 4: / Line 4: @@
 ==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 <math>a=BC</math>, <math>b=CA</math>, and <math>c=AB</math>. If <math>M</math> denotes the midpoint of <math>AB</math>, note that by centroid properties, <math>GM:CM = 1:3</math>, and hence, <math>[ABG] = [BGM] + [AGM] = \tfrac{1}{3}[BCM]+\tfrac{1}{3}[ACM] = \tfrac{1}{3}[ABC]</math> by same-altitude triangles. Using analogous reasoning, we can show that <math>[ABG]=[BCG]=[CAG] = \tfrac{1}{3}[ABC]</math>.
+We know that the length of the altitude from <math>G</math> to <math>AB</math> is <math>21</math>, and thus <math>[ABG] = \tfrac{21}{2}c</math>. Similarly, we can find that <math>[BCG] = 30a</math> and <math>[CAG] = 14b</math>. Equating the areas of these triangles, we have <math>30a = 14b = \tfrac{21}{2}c</math>, which is equivalent to <math>a:b:c = 7:15:20</math>.
+Let <math>a = 7x</math>, <math>b=15x</math>, and <math>c=20x</math> for some positive real <math>x</math>; we now solve for <math>x</math> by computing the area of <math>ABC</math> in two different ways. On one hand, we have <math>[ABC] = [ABG]+[BCG]+[CAG] = 630x</math>. On the other hand, by Heron's formula, <math>[ABC] = 42x^2</math>. Hence, <math>630x = 42x^2</math>, which gives <math>x = 15</math>. The perimeter of <math>ABC</math> is <math>42x =  \boxed{630}</math>.