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

 
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Latest revision as of 16:32, 19 January 2021

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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