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

(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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<cmath>\implies x^2+9=40</cmath>
 
<cmath>\implies x^2+9=40</cmath>
 
<cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.</cmath>
 
<cmath>\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.</cmath>
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The answer is $\textbf{(B)}.
 
-brainiacmaniac31
 
-brainiacmaniac31

Revision as of 22:39, 6 January 2020

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 $\textbf{(B)}. -brainiacmaniac31

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