2002 AIME II Problems/Problem 14

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. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have $\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}$ 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 $\boxed{098}$ .

Solution 3

Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\frac{y}{x}\cdot 19$ , since $O$ is on the angle bisector of $\angle M.$

We have that $\tan(\angle AMO)=\frac{19}{x},$ so $\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.$

However $\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}$ , so $\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}$ $\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1$ $\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.$ $x\cdot \frac{x^{2}+361}{x^{2}-361}=y$

We now use the fact that the perimeter of $\triangle PAM$ is $152$ : $PO+OA+AM+MP=152$ $\frac{y}{x}\cdot 19+19+x+y=152$ $19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152$ $(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152$ $\frac{2x^{2}}{x-19}=152$ $x^{2}-76x+19\cdot 76=0.$ This quadratic factors as $(x-38)^{2}=0,$ so $x=38$ , and $\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}$ $OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}$

2002 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2002 AIME II Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3

See also