2014 AIME I Problems/Problem 15
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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
Solution 4
See inside the , we can find that since if , we can see that Ptolemy Theorem inside cyclic quadrilateral doesn't work. Now let's see when , since , we can assume that , since we know so is isosceles right triangle. We can denote .Applying Ptolemy Theorem inside the cyclic quadrilateral we can get the length of can be represented as . After observing, we can see , whereas so we can see is isosceles triangle. Since is a triangle so we can directly know that the length of AF can be written in the form of . Denoting a point on side with that is perpendicular to side . Now with the same reason, we can see that is a isosceles right triangle, so we can get while the segment is since its 3-4-5 again. Now adding all those segments together we can find that and and the desired which our answer is ~bluesoul
Solution 5
The main element of the solution is the proof that is bisector of
Let be the midpoint of
is the center of the circle BF is bisector of Let vladimir.shelomovskii@gmail.com, vvsss
Solution 6
The main element of the solution is the proof that is midpoint of
As in Solution 5 we get
is isosceles triangle with
Similarly
Let vladimir.shelomovskii@gmail.com, vvsss
Solution 7
Let . By Incenter-Excenter(Fact ), is the angle bisector of . Then by Ratio Lemma we have Thus, is the midpoint of .
We can calculate and to be and respectively. And then by Power of a Point, we have And then similarly, we have .
Then and and by Pythagorean we have , so our answer is
~dolphinday
Video Solution by mop 2024
~r00tsOfUnity
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
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