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Difference between revisions of "2013 AMC 10A Problems/Problem 23"
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==Solution==
 
==Solution==
  
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Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>BC</math>.  Then <math>AZ = AY = 86</math>.  Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>.  We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, we get that <math>BC = p+q = \boxed{61 (D)}</math>
+
Let <math>BX = q</math>, <math>CX = p</math>, and <math>AC</math> meet the circle at <math>Y</math> and <math>Z</math>, with <math>Y</math> on <math>BC</math>.  Then <math>AZ = AY = 86</math>.  Using the Power of a Point, we get that <math>p(p+q) = 11(183) = 11 * 3 * 61</math>.  We know that <math>p+q>p</math>, and that <math>p>13</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>
  
  

Revision as of 15:31, 8 February 2013

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 $BC = p+q = \boxed{\textbf{(D) }61}$


See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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