Difference between revisions of "2013 AMC 12A Problems/Problem 19"
Fowlmaster (talk | contribs) |
Fowlmaster (talk | contribs) (→Solution 2) |
||
Line 12: | Line 12: | ||
===Solution 2=== | ===Solution 2=== | ||
+ | Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86. | ||
+ | Then by Stewart's Theorem, | ||
+ | |||
+ | <math>xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.</math> | ||
+ | |||
+ | <math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math> | ||
+ | |||
+ | <math>x^2 + xy + 86^2 = 97^2</math> | ||
+ | |||
+ | (Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.) | ||
+ | |||
+ | <math>x(x+y) = (97+86)(97-86)</math> | ||
+ | |||
+ | <math>x(x+y) = 2013</math> | ||
+ | |||
+ | The prime factors of 2013 are 3, 11, and 61. Obviously <math>x < x+y</math>. In addition, by the Triangle Inequality, <math>BC < AB + AC</math>, so <math>x+y < 183</math>. Therefore, <math>x</math> must equal 33, and <math>x+y</math> must equal 61. <math>\boxed{D}</math> |
Revision as of 18:06, 8 February 2013
Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E. Use power of a point on point C to the circle centered at A.
So CX*CB=CD*CE x(x+y)=(97-86)(97+86) x(x+y)=3*11*61.
Obviously x+y>x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61). By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.
Therefore, the answer is D) 61.
Solution 2
Let represent , and let represent . Since the circle goes through and , = = 86. Then by Stewart's Theorem,
(Since cannot be equal to 0, dividing both sides of the equation by is allowed.)
The prime factors of 2013 are 3, 11, and 61. Obviously . In addition, by the Triangle Inequality, , so . Therefore, must equal 33, and must equal 61.