Difference between revisions of "2000 AIME II Problems/Problem 10"
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Captainsnake (talk | contribs) (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+ | + | <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
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 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.