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

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Therefore, the answer is '''D) 61.'''
 
Therefore, the answer is '''D) 61.'''
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===Solution 2===
 
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Let <math>h</math> be the perpendicular from <math>B</math> to <math>AC</math>, <math>AX=x</math>, <math>XC=y</math>, then by Pythagorean Theorem,
 
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<math>h^2 + (x/2)^2 = 86^2</math>
 
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<math>h^2 + (x/2 + y)^2 = 97^2</math>
 
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Subtracting the two equations, we get <math>(x+y)y = (97-86)(97+86)</math>,
 
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then the rest is similar to the above solution by power of points.
 
  
 
== See also ==
 
== See also ==

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.

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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