2021 AMC 12A Problems/Problem 9

Problem

Triangle $ABC$ lies in a plane with $AB=13$ , $AC=14$ , and $BC=15$ . For any point $X$ in the plane of $\triangle ABC$ , let $f(X)$ denote the sum of the three distances from $X$ to the three vertices of $\triangle ABC$ . Let $P$ be the unique point in the plane of $\triangle ABC$ where $f(X)$ is minimized. Then $AP^2+BP^2+CP^2=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m+n$ ?

Solution

Point $P$ is the Brocard point of $\triangle ABC$ , where $\angle APB = \angle BPC = \angle APC = 120^\circ$ and $\triangle APB, \triangle BPC, \triangle APC$ are all $30-30-120$ triangles, and the squares of the side lengths are in the ratio $\frac{\frac{1}{1}}{3}$ which can easily be seen by dividing this triangle into two smaller $30-60-90$ triangles. It follows $AP^2+BP^2=\frac{2}{3}AB^2$ , $AP^2+CP^2=\frac{2}{3}AC^2$ , and $BP^2+CP^2=\frac{2}{3}BC^2$ . Now $2AP^2+2BP^2+2CP^2=\frac{2}{3}(AB^2+AC^2+BC^2)$ because each got counted twice, so $AP^2+BP^2+CP^2=\frac{1}{3}(13^2+14^2+15^2)=\frac{590}{3}$ , and $m+n=\boxed{593}$ . ~icematrix2

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 AMC 12A Problems/Problem 9

Problem

Solution

See also