# 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 $BC$?

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

$\textbf{(A)}\ 2\sqrt{26} \qquad \textbf{(B)}\ 2\sqrt{31} \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 4 + 2\sqrt{13}\\ \textbf{(E)}\ \text{not enough information given to solve the problem}$

## 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

## See Also

 1975 AHSME (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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