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Difference between revisions of "2008 AIME II Problems/Problem 11"

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To find <math>XY</math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>.
 
To find <math>XY</math>, consider the right triangle <math>PCX</math>. Since <math>\odot P</math> is tangent to <math>\overline{AC},\overline{BC}</math>, then <math>PC</math> [[bisect]]s <math>\angle ACB</math>. Let <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we recognize the <math>7 - 24 - 25</math> [[right triangle]], except scaled by <math>4</math>.
  
−
So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {4r}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {64}{3}</math>.
+
So we get that <math>\tan(2\theta) = 24/7</math>. From the [[trigonometric identity|half-angle identity]], we find that <math>\tan(\theta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3}</math>.
  
 
We conclude that <math>XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}</math>.
 
We conclude that <math>XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}</math>.

Revision as of 19:52, 22 December 2012

Problem

In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$, where $m$, $n$, and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$.

Solution

[asy] size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f)); [/asy]

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle.

To find $XY$, consider the right triangle $PCX$. Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$, then $PC$ bisects $\angle ACB$. Let $\angle ACB = 2\theta$; then $\angle PCX = \angle QBX = \theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$.

So we get that $\tan(2\theta) = 24/7$. From the half-angle identity, we find that $\tan(\theta) = \frac {3}{4}$. Therefore, $XC = \frac {64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY = \frac {4r}{3}$.

We conclude that $XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}$.

So our right triangle $QPM$ has sides $r + 16$, $r - 16$, and $\frac {104 - 4r}{3}$.

By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\sqrt {35}$, for a final answer of $\fbox{254}$.

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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