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Difference between revisions of "2013 AMC 12A Problems/Problem 19"
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===Solution 3===
 
===Solution 3===
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Let <math>x</math> represent <math>BX</math>, and let <math>y</math> represent <math>CX</math>. Since the circle goes through <math>B</math> and <math>X</math>, <math>AB</math> = <math>AX</math> = 86.
+
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,
 
Then by Stewart's Theorem,
  
Line 51: Line 51:
 
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>
 
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>
  
−
 
−
-Solution 3 by '''fowlmaster'''
 
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}

Revision as of 16:19, 23 February 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. $\boxed{D}$


See also

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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