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Difference between revisions of "2000 AIME II Problems/Problem 10"
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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.
  
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== 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>.
  
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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 <cmath>\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.</cmath>
  
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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>.
+
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

Problem

A circle is inscribed in quadrilateral $ABCD$, tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q$. Given that $AP=19$, $PB=26$, $CQ=37$, and $QD=23$, find the square of the radius of the circle.

Solution 1

Call the center of the circle $O$. By drawing the lines from $O$ tangent to the sides and from $O$ to the vertices of the quadrilateral, four pairs of congruent right triangles are formed.

Thus, $\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180$, or $(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=180$.

Take the $\tan$ of both sides and use the identity for $\tan(A+B)$ to get \[\tan(\arctan(\tfrac{19}{r})+\arctan(\tfrac{26}{r}))+\tan(\arctan(\tfrac{37}{r})+\arctan(\tfrac{23}{r}))=n\cdot0=0.\]

Use the identity for $\tan(A+B)$ again to get \[\frac{\tfrac{45}{r}}{1-19\cdot\tfrac{26}{r^2}}+\frac{\tfrac{60}{r}}{1-37\cdot\tfrac{23}{r^2}}=0.\]

Solving gives $r^2=\boxed{647}$.

Note: the equation may seem nasty at first, but once you cancel the $r$s and other factors, you are just left with $r^2$. That gives us $647$ quite easily.

Solution 2

Just use the area formula for tangential quadrilaterals. The numbers are really big. A terrible problem to work on ($a, b, c,$ and $d$ are the tangent lengths, not the side lengths). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A^2}{(a+b+c+d)^2} = \boxed{647}$.

Solution 3 (Smart algebra to make 2 less annoying)

Using the formulas established in solution 2, one notices: \[r^2=\frac{A^2}{(a+b+c+d)^2}\] \[r^2=\frac{(a+b+c+d)(abc+bcd+cda+abd) }{(a+b+c+d)^2}\] \[r^2=\frac{abc+bcd+acd+abd}{a+b+c+d}\] \[r^2=\boxed{647}\]

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

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