# Difference between revisions of "1973 AHSME Problems/Problem 9"

## Problem

In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$, then the area of $\triangle ABC$ is

$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$

## Solution

$[asy] pair A=(-6,0),B=(6,0),C=(-3,5.196),M=(0,0),H=(-3,0); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-3,5.196)--(0,0)); draw(C--H); dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,N); dot(M); label("M",M,S); dot(H); label("H",H,S); markscalefactor=0.1; draw(anglemark((-6,0),C,(6,0))); draw((-3,0.5)--(-2.5,0.5)--(-2.5,0)); [/asy]$

Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).

Note that because $CM$ is a median, $AM = BM$. Also, by ASA Congruency, $\triangle CHA = \triangle CHM$, so $AH = HM$. That means $HM = \tfrac{1}{4} \cdot AB$, and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{\textbf{(E)}\ 4K}$.