2015 IMO Problems/Problem 4
Problem
Triangle has circumcircle and circumcenter . A circle with center intersects the segment at points and , such that , , , and are all different and lie on line in this order. Let and be the points of intersection of and , such that , , , , and lie on in this order. Let be the second point of intersection of the circumcircle of triangle and the segment . Let be the second point of intersection of the circumcircle of triangle and the segment .
Suppose that the lines and are different and intersect at the point . Prove that lies on the line .
Proposed by Silouanos Brazitikos and Evangelos Psychas, Greece
Solution
Lemma (On three chords). If two lines pass through different endpoints of two circles' common chord, then the other two chords cut by these lines on the circles are parallel.
Proof The second and the third chords are anti-parallel to the first (common) chord with respect to the given lines, so they are parallel to each other.
To solve this problem, it is sufficient to apply the lemma 5 times. Indeed, let the lines meet second time at respectively. One of the circles that figure in lemma is always , while the other is one of three other circles from the problem statement. Applying the lemma to the lines and , and , and , and , and , we get , , , , , respectively. From this, , , . Therefore, . This means that and are symmetric wrt , a diameter of through . So are and , as . Therefore, the lines and are symmetric wrt and meet on it.
See Also
2015 IMO (Problems) • Resources) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |