# Difference between revisions of "1975 AHSME Problems/Problem 20"

## Problem

In the adjoining figure triangle $ABC$ is such that $AB = 4$ and $AC = 8$. IF $M$ is the midpoint of $BC$ and $AM = 3$, what is the length of $BC$?

$[asy] draw((-4,0)--(4,0)--(-1,4)--cycle); draw((-1, 4)--(0, 0.00001)); label("B", (-4,0), S); label("C", (4,0), S); label("A", (-1, 4), N); label("M", (0, 0.0001), S); [/asy]$

## Solution

Let $BM=CM=x$. Then, by Stewart's Theorem, we have $$2x^3+18x=16x+64x$$ $$\implies x^2+9=40$$ $$\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.$$ The answer is $\boxed{B}.$ -brainiacmaniac31