1975 AHSME Problems/Problem 20

Revision as of 22:32, 6 January 2020 by Brainiacmaniac31 (talk | contribs) (Created page with "Let <math>BM=CM=x</math>. Then, by Stewart's Theorem, we have <cmath>2x^3+18x=16x+64x</cmath> <cmath>\implies x^2+9=40</cmath> <cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\...")
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Let $BM=CM=x$. Then, by Stewart's Theorem, we have \[2x^3+18x=16x+64x\] \[\implies x^2+9=40\] \[\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.\] -brainiacmaniac31

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