Difference between revisions of "2014 AIME I Problems/Problem 15"
m (→Solution 3) |
Tempaccount (talk | contribs) (Adding problem section) |
||
Line 1: | Line 1: | ||
+ | |||
+ | ==Problem== | ||
== Problem 15 == | == Problem 15 == | ||
Revision as of 15:41, 9 August 2018
Problem
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 1
Since , is the diameter of . Then . But , so is a 45-45-90 triangle. Letting , we have that , , and .
Note that by SAS similarity, so and . Since is a cyclic quadrilateral, and , implying that and are isosceles. As a result, , so and .
Finally, using the Pythagorean Theorem on , Solving for , we get that , so . Thus, the answer is .
Solution 2
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
Also: . We can also use Ptolemy's Theorem on quadrilateral to figure what is in terms of :
Thus .
Solution 3
Call and as a result . Since is cyclic we just need to get and using LoS(for more detail see the nd paragraph of Solution ) we get and using a similar argument(use LoS again) and subtracting you get so you can use Ptolemy to get . ~First
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.