Difference between revisions of "2000 AIME II Problems/Problem 10"
Grolarbear (talk | contribs) m (→Solution 3 (Smart algebra to make 2 less annoying)) |
|||
(10 intermediate revisions by 7 users not shown) | |||
Line 2: | Line 2: | ||
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. | 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 1== |
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. | 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>. | ||
− | Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get < | + | Take the <math>\tan</math> of both sides and use the identity for <math>\tan(A+B)</math> to get <cmath>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.</cmath> |
− | Use the identity for <math>\tan(A+B)</math> again to get < | + | Use the identity for <math>\tan(A+B)</math> again to get <cmath>\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.</cmath> |
Solving gives <math>r^2=\boxed{647}</math>. | Solving gives <math>r^2=\boxed{647}</math>. | ||
+ | |||
+ | Note: the equation may seem nasty at first, but once you cancel the <math>r</math>s and other factors, you are just left with <math>r^2</math>. That gives us <math>647</math> quite easily. | ||
+ | |||
+ | == 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+dab)} = 105\sqrt{647}</cmath> | ||
+ | <math>r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}</math>. | ||
+ | |||
+ | == Solution 3 (Smart algebra to make 2 less annoying) == | ||
+ | |||
+ | Using the formulas established in solution 2, one notices: | ||
+ | <cmath>r^2=\frac{A^2}{(a+b+c+d)^2}</cmath> | ||
+ | <cmath>r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}</cmath> | ||
+ | <cmath>r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}</cmath> | ||
+ | <cmath>r^2=\boxed{647}</cmath> | ||
+ | |||
+ | which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand. | ||
== See also == | == See also == |
Latest revision as of 16:37, 2 January 2024
Contents
[hide]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 1
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 .
Note: the equation may seem nasty at first, but once you cancel the s and other factors, you are just left with . That gives us quite easily.
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). .
Solution 3 (Smart algebra to make 2 less annoying)
Using the formulas established in solution 2, one notices:
which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just with less opportunity to cancel coefficients beforehand.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.