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Difference between revisions of "2013 AMC 12A Problems/Problem 19"
(→‎Solution 2)
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===Solution 2===
 
===Solution 2===
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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.
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Then by Stewart's Theorem,
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<math>xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.</math>
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<math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math>
 +
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<math>x^2 + xy + 86^2 = 97^2</math>
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(Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.)
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<math>x(x+y) = (97+86)(97-86)</math>
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<math>x(x+y) = 2013</math>
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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 19:06, 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 $x$ represent $BX$, and let $y$ represent $CX$. Since the circle goes through $B$ and $X$, $AB$ = $AX$ = 86. Then by Stewart's Theorem,

$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$

$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$

$x^2 + xy + 86^2 = 97^2$

(Since $y$ cannot be equal to 0, dividing both sides of the equation by $y$ is allowed.)

$x(x+y) = (97+86)(97-86)$

$x(x+y) = 2013$

The prime factors of 2013 are 3, 11, and 61. Obviously $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal 33, and $x+y$ must equal 61. $\boxed{D}$

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