Difference between revisions of "2013 AMC 10A Problems/Problem 23"

m (Solution 2)
(Solution 1 (Power of a Point))
Line 10: Line 10:
 
==Solution 1 (Power of a Point)==
 
==Solution 1 (Power of a Point)==
  
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>AC</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>.
+
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>AC</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>11</math> by the triangle inequality on <math>\triangle ACX</math>.  Thus, we get that <math>BC = p+q = \boxed{\textbf{(D) }61}</math>.
  
 
==Solution 2 (Stewart's Theorem)==
 
==Solution 2 (Stewart's Theorem)==

Revision as of 23:44, 31 July 2015

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 1 (Power of a Point)

Let $BX = q$, $CX = p$, and $AC$ meet the circle at $Y$ and $Z$, with $Y$ on $AC$. 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 $p>11$ by the triangle inequality on $\triangle ACX$. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$.

Solution 2 (Stewart's Theorem)

Let $x$ represent $CX$, and let $y$ represent $BX$. 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 $\boxed{\textbf{(D) }61}$.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png