Difference between revisions of "2002 AIME II Problems/Problem 14"
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== Problem == | == Problem == | ||
− | 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 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>. | + | 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>. |
− | == Solution == | + | == Solution 1 == |
− | 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. So we | + | 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 |
+ | <cmath>\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}</cmath> | ||
+ | Solving, <math>AM = 38</math>. So the ratio of the side lengths of the triangles is 2. Therefore, | ||
+ | <cmath>\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2</cmath> | ||
+ | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <math>[PAM] = r \cdot s</math> | ||
+ | |||
+ | <math>[PAM] = 19 \cdot (114 - x)</math> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <math>\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}</math> | ||
+ | |||
+ | Solving for <math>AM</math>, it yields <math>19 \cdot \sqrt{\frac{x + 38}{x}}</math>. | ||
+ | |||
+ | <math>19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}</math> | ||
+ | |||
+ | The <math>x^3</math> cancels, yielding a quadratic. Solving yields <math>x = \frac{38}{3}</math>. | ||
+ | Add <math>19</math> to find <math>OP</math>, yielding <math>\frac{95}{3}</math> or <math>\boxed{098}</math>. | ||
− | + | == Solution 3 == | |
− | + | 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> | |
− | <math>\frac{ | + | We have that <math>\tan(\angle AMO)=\frac{19}{x},</math> so |
+ | <cmath>\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.</cmath> | ||
− | <math> | + | However <math>\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}</math>, so |
+ | <cmath>\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}</cmath> | ||
+ | <cmath>\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1</cmath> | ||
+ | <cmath>\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.</cmath> | ||
+ | <cmath>x\cdot \frac{x^{2}+361}{x^{2}-361}=y</cmath> | ||
− | <math> | + | We now use the fact that the perimeter of <math>\triangle PAM</math> is <math>152</math>: |
+ | <cmath>PO+OA+AM+MP=152</cmath> | ||
+ | <cmath>\frac{y}{x}\cdot 19+19+x+y=152</cmath> | ||
+ | <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> | ||
+ | <cmath>(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152</cmath> | ||
+ | <cmath>\frac{2x^{2}}{x-19}=152</cmath> | ||
+ | <cmath>x^{2}-76x+19\cdot 76=0.</cmath> | ||
+ | This quadratic factors as <math>(x-38)^{2}=0,</math> so <math>x=38</math>, and | ||
+ | <cmath>\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}</cmath> | ||
+ | <cmath>OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}</cmath> | ||
− | |||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=13|num-a=15}} | {{AIME box|year=2002|n=II|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:15, 19 December 2021
Problem
The perimeter of triangle is , and the angle is a right angle. A circle of radius with center on is drawn so that it is tangent to and . Given that where and are relatively prime positive integers, find .
Solution 1
Let the circle intersect at . Then note and are similar. Also note that 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 Solving, . So the ratio of the side lengths of the triangles is 2. Therefore, so and Substituting for , we see that , so and the answer is .
Solution 2
Reflect triangle across line , creating an isoceles triangle. Let be the distance from the top of the circle to point , with as . Given the perimeter is 152, subtracting the altitude yields the semiperimeter of the isoceles triangle, as . The area of the isoceles triangle is:
Now use similarity, draw perpendicular from to , name the new point . Triangle is similar to triangle , by AA Similarity. Equating the legs, we get:
Solving for , it yields .
The cancels, yielding a quadratic. Solving yields . Add to find , yielding or .
Solution 3
Let the foot of the perpendicular from to be now Also let and This means that , since is on the angle bisector of
We have that so
However , so
We now use the fact that the perimeter of is : This quadratic factors as so , and
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 | ||
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