Difference between revisions of "2000 AIME II Problems/Problem 10"

m (Solution 1)
(Added fancier version of solution 2 that involves less bashing, also fixed some typos.)
 
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== Solution 2==
 
== 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).
 
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>
+
<cmath>A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}</cmath>
<math>r^2=\frac{A}{a+b+c+d} = \boxed{647}</math>.
+
<math>r^2=\frac{A}{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}</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 14:16, 26 February 2021

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). \[A = \sqrt{(a+b+c+d)(abc+bcd+cda+dab)} = 105\sqrt{647}\] $r^2=\frac{A}{a+b+c+d}^2 = \boxed{647}$.

Solution 3 (Smart algebra to make 2 less annoying)

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

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 (ProblemsAnswer KeyResources)
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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