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

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.

@@ Line 12: / Line 12: @@
 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+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^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 ==

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
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