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 <math>\textbf{(B)}.</math> | ||
-brainiacmaniac31 | -brainiacmaniac31 | ||
Revision as of 23:39, 6 January 2020
Let 
. Then, by Stewart's Theorem, we have
![\[2x^3+18x=16x+64x\]](http://latex.artofproblemsolving.com/a/f/d/afd3cf1a1a368ef79905d4e8d3fa41029f467bc9.png)
![\[\implies x^2+9=40\]](http://latex.artofproblemsolving.com/e/3/9/e391428c480e31e2bb9e9ff00e008e3893142584.png)
![\[\implies x=\sqrt{31}\implies BC=\boxed{2\sqrt{31}}.\]](http://latex.artofproblemsolving.com/1/8/1/181584bcc40552dafcf9b933095297d4ac21c89c.png)
The answer is 
-brainiacmaniac31
