2018 AIME II Problems/Problem 14
Contents
Problem
The incircle of triangle is tangent to at . Let be the other intersection of with . Points and lie on and , respectively, so that is tangent to at . Assume that , , , and , where and are relatively prime positive integers. Find .
Diagram
Solution 1
Let the sides and be tangent to at and , respectively. Let and . Because and are both tangent to and and subtend the same arc of , it follows that . By equal tangents, . Applying the Law of Sines to yields Similarly, applying the Law of Sines to gives It follows that implying . Applying the same argument to yields from which . The requested sum is .
Solution 2 (Projective)
Let the incircle of be tangent to and at and . By Brianchon's theorem on tangential hexagons and , we know that and are concurrent at a point . Let . Then by La Hire's lies on the polar of so lies on the polar of . Therefore, also passes through . Then projecting through , we have Therefore, . Since we know that and . Therefore, and . Since , we also have . Solving for , we obtain . 😃 -Vfire
Solution 3 (Combination of Law of Sine and Law of Cosine)
Let the center of the incircle of be . Link and . Then we have
Let the incircle of be tangent to and at and , let and .
Use Law of Sine in and , we have
therefore we have
Solve this equation, we have
As a result, , , , ,
So,
Use Law of Cosine in and , we have
And we have
So
Solve this equation, we have
As a result,
So, the final answer of this question is
~Solution by (Frank FYC)
Solution 4 (Projective geometry)
Claim
Let the sides and be tangent to at and , respectively. Then lines and are concurrent and lines and are concurrent.
Proof
Let be point of crossing and We make projective transformation such that circle maps into the circle and point maps into the center of new circle point We denote images using notification
maps into , so lines and be the diameters. This implies be a square.
Therefore be the diameter be diagonals of the square. and be midlines which crossing in the center Therefore ines and are concurrent.
Lines and are concurrent.
Solution The cross-ratio associated with a list of four collinear points is defined as The cross-ratio be projective invariant of a quadruple of collinear points, so
For visuals only, I will show how one can find the perceptor and the image’s plane. is image of inversion with respect is the diameter of are collinear. is diameter of .
Plane of images is perpendicular to
Last diagram shows the result of transformation. Transformation is possible. The end.
vladimir.shelomovskii@gmail.com, vvsss
See Also
2018 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 |
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