1986 AHSME Problems/Problem 11

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In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$. Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$. The length of $HM$ is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$


In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is $13$, the median must be $6.5$.

Solution 2 (Self Torture)

Warning: this solution is very intensive in calculation. Please do NOT try this on the test!

Let's start by finding $AH$. By Heron's Formula, $s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$. Using the area formula $A=0.5bh$, $AH=12$. Now using the Pythagorean Theorem, $BH=5, HC=9$.

Now $AM=MB=6.5$. Using Stewart's Theorem on $\triangle{ABH}$, letting $HM=x$:

$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$).

Thus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$. Select $\boxed{B}$.


P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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