Difference between revisions of "2012 IMO Problems/Problem 5"
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Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here. | Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here. | ||
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+ | ~ Latex edit by Kscv | ||
Latest revision as of 07:30, 20 November 2023
Problem
Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let . Prove that .
Solution
Let , , be the circumcircle of triangle , the circle with its center as and radius as , and the circle with its center as and radius as , Respectively. Since the center of lies on , the three circles above are coaxial to line .
Let Line and Line collide with on () and (), Respectively. Also let .
Then, since , by ceva's theorem, lies on .
Since triangles and are similar, , Thus . In the same way, .
Therefore, making concyclic. In the same way, is concyclic.
So , and in the same way . Therefore, the lines and are tangent to and , respectively.
Since is on , and is the concentric line of and , , Thus . Since is in the middle and ,
we can say triangles and are congruent. Therefore, .
Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.
~ Latex edit by Kscv
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2012 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |