Difference between revisions of "2010 AIME II Problems/Problem 14"
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This gives that the answer is <math>\boxed{007}</math>. | This gives that the answer is <math>\boxed{007}</math>. | ||
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+ | An alternate finish for this problem would be to use Power of a Point on <math>BA</math> and <math>CD</math>. By Power of a Point Theorem, <math>CP\cdot PD=1\cdot 2=BP\cdot PA</math>. Since <math>BP+PA=4</math>, we can solve for <math>BP</math> and <math>PA</math>, giving the same values and answers as above. | ||
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ | <center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ |
Revision as of 23:54, 24 February 2016
Contents
[hide]Problem
Triangle with right angle at , and . Point on is chosen such that and . The ratio can be represented in the form , where , , are positive integers and is not divisible by the square of any prime. Find .
Solution
Let be the circumcenter of and let the intersection of with the circumcircle be . It now follows that . Hence is isosceles and .
Denote the projection of onto . Now . By the pythagorean theorem, . Now note that . By the pythagorean theorem, . Hence it now follows that,
This gives that the answer is .
An alternate finish for this problem would be to use Power of a Point on and . By Power of a Point Theorem, . Since , we can solve for and , giving the same values and answers as above.
Solution 2
Let , by convention. Also, Let and . Finally, let and .
We are then looking for
Now, by arc interceptions and angle chasing we find that , and that therefore Then, since (it intercepts the same arc as ) and is right,
.
Using law of sines on , we additionally find that Simplification by the double angle formula yields
.
We equate these expressions for to find that . Since , we have enough information to solve for and . We obtain
Since we know , we use
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.