2019 AIME I Problems/Problem 15
Contents
Problem
Let be a chord of a circle , and let be a point on the chord . Circle passes through and and is internally tangent to . Circle passes through and and is internally tangent to . Circles and intersect at points and . Line intersects at and . Assume that , , , and , where and are relatively prime positive integers. Find .
Solution 1
Let and be the centers of and , respectively. There is a homothety at sending to that sends to and to , so . Similarly, , so is a parallelogram. Moreover, whence is cyclic. However, so is an isosceles trapezoid. Since , , so is the midpoint of .
By Power of a Point, . Since and , and the requested sum is .
(Solution by TheUltimate123)
Note
One may solve for first using PoAP, . Then, notice that is rational but is not, also . The most likely explanation for this is that is the midpoint of , so that and . Then our answer is . One can rigorously prove this using the methods above
Solution 2
Let the tangents to at and intersect at . Then, since , lies on the radical axis of and , which is . It follows that Let denote the midpoint of . By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), whence . Like above, . Since , we establish that , from which , and the requested sum is .
(Solution by TheUltimate123)
Solution 3
Firstly we need to notice that is the middle point of . Assume the center of circle are , respectively. Then are collinear and are collinear. Link . Notice that, . As a result, and . So we have parallelogram . So Notice that, and divides into two equal length pieces, So we have . As a result, lie on one circle. So . Notice that since , we have . As a result, . So is the middle point of .
Back to our problem. Assume , and . Then we have , that is, . Also, . Solve these above, we have . As a result, we have . So, we have . As a result, our answer is .
Solution By BladeRunnerAUG (Fanyuchen20020715). Edited by bgn4493.
Solution 4
Note that the tangents to the circles at and intersect at a point on by radical axis theorem. Since and , we have so is cyclic.
But if is the center of , clearly is cyclic with diameter , so implies that is the midpoint of . Then, by power of point , whereas it is given that . Thus so , i.e. and the answer is .
Solution 5
Connect , since , so then, so are concyclic
We let , it is clear that , which leads to the conclusion which tells is the midpoint of
Then it is clear, , the answer is
~bluesoul
Solution 6(lazy)
by power of a point. Also, , so and are solutions to the quadratic so and is in some order. Now, because we want and it is known to be rational, we can guess that is irrational or the problem would simply ask for . , and chances are low that is some number with a square root plus or minus to cancel out the in , so one can see that is most likely to be , and our answer is
Note : If our answer is correct, then , which made the midpoint of , a feature that occurs often in AIME problems, so that again made our answer probable. Midpoints have many properties and there is a lot of ways to show if a point is the midpoint of a segment. Even if the answer is wrong, it's still the same as leaving it blank and 065 is a good guess. ~ddk001
Note 2 : I looked at the history of this page, and this was the same as the once deleted solution 5.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
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