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

(See also)
(Solution 1)
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===Solution 1===
 
===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.
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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.
 
Use power of a point on point C to the circle centered at A.
  
So CX*CB=CD*CE
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So <math>CX*CB=CD*CE</math>
x(x+y)=(97-86)(97+86)
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<math>x(x+y)=(97-86)(97+86)</math>
x(x+y)=3*11*61.
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<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).
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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.
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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.'''
 
Therefore, the answer is '''D) 61.'''
 
  
 
===Solution 2===
 
===Solution 2===

Revision as of 21:11, 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. $\boxed{D}$


See also

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

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