1973 AHSME Problems/Problem 9

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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, $[ABC] = \boxed{\textbf{(E)}\ 4K}$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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