# 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> | ||

+ | The answer is $\textbf{(B)}. | ||

-brainiacmaniac31 | -brainiacmaniac31 |

## Revision as of 22:39, 6 January 2020

Let . Then, by Stewart's Theorem, we have The answer is $\textbf{(B)}. -brainiacmaniac31