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

Revision as of 21:13, 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.

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

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

@@ Line 31: / Line 31: @@
 then the rest is similar to the above solution by power of points.
-===Solution 3===
-Let <math>x</math> represent <math>CX</math>, and let <math>y</math> represent <math>BX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86.
-Then by Stewart's Theorem,
-<math>xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.</math>
-<math>x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x</math>
-<math>x^2 + xy + 86^2 = 97^2</math>
-(Since <math>y</math> cannot be equal to 0, dividing both sides of the equation by <math>y</math> is allowed.)
-<math>x(x+y) = (97+86)(97-86)</math>
-<math>x(x+y) = 2013</math>
-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>
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