Difference between revisions of "2000 AIME II Problems/Problem 10"
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== Problem == | == Problem == | ||
− | A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[square]] of the [[radius]] of the circle. | + | A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math>\overline{AB}</math> at <math>P</math> and to <math>\overline{CD}</math> at <math>Q</math>. Given that <math>AP=19</math>, <math>PB=26</math>, <math>CQ=37</math>, and <math>QD=23</math>, find the [[Perfect square|square]] of the [[radius]] of the circle. |
== Solution == | == Solution == | ||
− | Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, | + | Call the [[center]] of the circle <math>O</math>. By drawing the lines from <math>O</math> tangent to the sides and from <math>O</math> to the vertices of the quadrilateral, four pairs of congruent [[right triangle]]s are formed. |
Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180</math>. | Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180</math>. | ||
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Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get <math>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0</math>. | Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get <math>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0</math>. | ||
− | Use the identity for <math>\tan(A+B)</math> again to get <math>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2 | + | Use the identity for <math>\tan(A+B)</math> again to get <math>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0</math>. |
Solving gives <math>r^2=\boxed{647}</math>. | Solving gives <math>r^2=\boxed{647}</math>. | ||
+ | == Solution 2== | ||
+ | Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on (<math>a, b, c,</math> and <math>d</math> are the tangent lengths, not the side lengths). | ||
+ | <cmath>A = \sqrt{(a+b+c+d)(abc+bcd+cda+dac)} = 105\sqrt{647}</cmath> | ||
+ | <math>r^2=\frac{A}{a+b+c+d} = \boxed{647}</math>. | ||
+ | == See also == | ||
{{AIME box|year=2000|n=II|num-b=9|num-a=11}} | {{AIME box|year=2000|n=II|num-b=9|num-a=11}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 21:15, 8 April 2019
Contents
Problem
A circle is inscribed in quadrilateral , tangent to at and to at . Given that , , , and , find the square of the radius of the circle.
Solution
Call the center of the circle . By drawing the lines from tangent to the sides and from to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.
Thus, , or .
Take the of both sides and use the identity for to get .
Use the identity for again to get .
Solving gives .
Solution 2
Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ( and are the tangent lengths, not the side lengths). .
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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