2013 AMC 10A Problems/Problem 23

Problem

In $\triangle ABC$ , $AB = 86$ , and $AC=97$ . A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$ . Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$ ?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$

Solution

Let $BX = q$ , $CX = p$ , and $AC$ meet the circle at $Y$ and $Z$ , with $Y$ on $BC$ . Then $AZ = AY = 86$ . Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$ . We know that $p+q>p$ , and that $p>13$ by the triangle inequality on $\triangle ACX$ . Thus, we get that $BC = p+q = \boxed{61 (D)}$

2013 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2013 AMC 10A Problems/Problem 23

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