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2002 AIME II Problems/Problem 14

Revision as of 15:50, 16 June 2018 by Mudhaniu (talk | contribs)

Problem

The perimeter of triangle $APM$ is $152$, and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$. Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution 1

Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. So we have \[\frac{19}{AM} = \frac{152-2AM}{152}\] Solving, $AM = 38$. So the ratio of the side lengths of the triangles is 2. Therefore, \[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\] so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$, we see that $4OP-76 = OP+19$, so $OP = \frac{95}3$ and the answer is $\boxed{098}$.

Solution 2

Reflect triangle $PAM$ across line $AP$, creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point P, with $x + 38$ as $AP$. Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$. The area of the isoceles triangle is:

$[PAM] = r \cdot s$

$[PAM] = 19 \cdot (114 - x)$

Now use similarity, draw perpendicular from $O$ to $PM$, name the new point $D$. Triangle $PDO$ is similar to triangle $PAM$, by AA Similarity. Equating the legs, we get:

$\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}$

Solving for $AM$, it yields $19 \cdot \sqrt{\frac{x + 38}{x}}$.

$19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}$

The $x^3$ cancels, yielding a quadratic. Solving yields $x = \frac{38}{3}$. Add $19$ to find $OP$, yielding $\frac{95}{3}$ or $\boxed{098}$.

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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