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Difference between revisions of "2002 AIME II Problems/Problem 14"

 
 
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== Problem ==
 
== Problem ==
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The [[perimeter]] of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a [[right angle]]. A [[circle]] of [[radius]] <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Given that <math>OP=m/n</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers, find <math>m+n</math>.
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== Solution 1 ==
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Let the circle intersect <math>\overline{PM}</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> 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
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<cmath>\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}</cmath>
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Solving, <math>AM = 38</math>. So the ratio of the side lengths of the triangles is 2. Therefore,
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<cmath>\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2</cmath>
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so <math>2OP = PB+38</math> and <math>2PB = OP+19.</math> Substituting for <math>PB</math>, we see that <math>4OP-76 = OP+19</math>, so <math>OP = \frac{95}3</math> and the answer is <math>\boxed{098}</math>.
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== Solution 2 ==
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Reflect triangle <math>PAM</math> across line <math>AP</math>, creating an isoceles triangle. Let <math>x</math> be the distance from the top of the circle to point <math>P</math>, with <math>x + 38</math> as <math>AP</math>. Given the perimeter is 152, subtracting the altitude yields the semiperimeter <math>s</math> of the isoceles triangle, as <math>114 - x</math>. The area of the isoceles triangle is:
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<math>[PAM] = r \cdot s</math>
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<math>[PAM] = 19 \cdot (114 - x)</math>
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Now use similarity, draw perpendicular from <math>O</math> to <math>PM</math>, name the new point <math>D</math>. Triangle <math>PDO</math> is similar to triangle <math>PAM</math>, by AA Similarity. Equating the legs, we get:
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<math>\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}</math>
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Solving for <math>AM</math>,  it yields <math>19 \cdot \sqrt{\frac{x + 38}{x}}</math>.
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<math>19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}</math>
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The <math>x^3</math> cancels, yielding a quadratic. Solving yields <math>x = \frac{38}{3}</math>.
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Add <math>19</math> to find <math>OP</math>, yielding <math>\frac{95}{3}</math> or <math>\boxed{098}</math>.
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== Solution 3 ==
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Let the foot of the perpendicular from <math>O</math> to <math>PM</math> be <math>D;</math> now <math>OD=19.</math> Also let <math>AM=x</math> and <math>PM=y.</math> This means that <math>OP=\frac{y}{x}\cdot 19</math>, since <math>O</math> is on the angle bisector of <math>\angle M.</math>
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We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so
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<cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath>
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However <math>\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}</math>, so
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<cmath>\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}</cmath>
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<cmath>\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1</cmath>
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<cmath>\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.</cmath>
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<cmath>x\cdot \frac{x^{2}+361}{x^{2}-361}=y</cmath>
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We now use the fact that the perimeter of <math>\triangle PAM</math> is <math>152</math>:
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<cmath>PO+OA+AM+MP=152</cmath>
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<cmath>\frac{y}{x}\cdot 19+19+x+y=152</cmath>
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<cmath>19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152</cmath>
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<cmath>(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152</cmath>
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<cmath>\frac{2x^{2}}{x-19}=152</cmath>
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<cmath>x^{2}-76x+19\cdot 76=0.</cmath>
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This quadratic factors as <math>(x-38)^{2}=0,</math> so <math>x=38</math>, and
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<cmath>\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}</cmath>
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<cmath>OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}</cmath>
  
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== Solution ==
 
  
 
== See also ==
 
== See also ==
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* [[2002 AIME II Problems]]
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{{AIME box|year=2002|n=II|num-b=13|num-a=15}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 18:15, 19 December 2021

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 \[\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}\] \[OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}\]


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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