Difference between revisions of "2014 AIME I Problems/Problem 15"
(→Solution) |
(→Solution) |
||
Line 22: | Line 22: | ||
dot("$F$",F,dir(90)); | dot("$F$",F,dir(90)); | ||
dot("$G$",G,dir(0)); | dot("$G$",G,dir(0)); | ||
− | draw(Circle((1. | + | draw(Circle((1.109, 0.609), 1.28)); |
draw(D--E); | draw(D--E); | ||
draw(E--F); | draw(E--F); |
Revision as of 00:45, 19 March 2014
Problem 15
In , , , and . Circle intersects at and , at and , and at and . Given that and , length , where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find .
Solution
First we note that is an isosceles right triangle with hypotenuse the same as the diameter of . We also note that since is a right angle and the ratios of the sides are .
From congruent arc intersections, we know that , and that from similar triangles is also congruent to . Thus, is an isosceles triangle with , so is the midpoint of and . Similarly, we can find from angle chasing that . Therefore, is the angle bisector of . From the angle bisector theorem, we have , so and .
Lastly, we apply power of a point from points and with respect to and have and , so we can compute that and . From the Pythagorean Theorem, we result in , so
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.