Difference between revisions of "2013 AMC 12A Problems/Problem 19"

Revision as of 20:11, 7 September 2013

Problem

In $\bigtriangleup ABC$ , $AB = 86$ , 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

Solution 1

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 $h$ be the perpendicular from $B$ to $AC$ , $AX=x$ , $XC=y$ , then by Pythagorean Theorem,

$h^2 + (x/2)^2 = 86^2$

$h^2 + (x/2 + y)^2 = 97^2$

Subtracting the two equations, we get $(x+y)y = (97-86)(97+86)$ ,

then the rest is similar to the above solution by power of points.

Solution 3

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 61. $\boxed{D}$

@@ Line 9: / Line 9: @@
 ===Solution 1===
-Let CX=x, BX=y. Let the circle intersect AC at D and the diameter including AD intersect the circle again at E.
+Let <math>CX=x, BX=y</math>. Let the circle intersect <math>AC</math> at <math>D</math> and the diameter including <math>AD</math> intersect the circle again at <math>E</math>.
 Use power of a point on point C to the circle centered at A.
-So CX*CB=CD*CE
+So <math>CX*CB=CD*CE</math>
-x(x+y)=(97-86)(97+86)
+<math>x(x+y)=(97-86)(97+86)</math>
-x(x+y)=3*11*61.
+<math>x(x+y)=3*11*61</math>.
-Obviously x+y>x so we have three solution pairs for (x,x+y)=(1,2013),(3,671),(11,183),(33,61).
+Obviously <math>x+y>x</math> so we have three solution pairs for <math>(x,x+y)=(1,2013),(3,671),(11,183),(33,61)</math>.
-By the Triangle Inequality, only x+y=61 yields a possible length of BX+CX=BC.
+By the Triangle Inequality, only<math> x+y=61</math> yields a possible length of <math>BX+CX=BC</math>.
 Therefore, the answer is '''D) 61.'''
 ===Solution 2===

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