2010 IMO Problems/Problem 4
Problem
Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows .
Solution
Solution 1
Without loss of generality, suppose that . By Power of a Point, , so is tangent to the circumcircle of . Thus, . It follows that after some angle-chasing,
so as desired.
Solution 2
Let the tangent at to intersect at . We now have that since and are both isosceles, . This yields that .
Now consider the power of point with respect to .
Hence by AA similarity, we have that . Combining this with the arc angle theorem yields that . Hence .
This implies that the tangent at is parallel to and therefore that is the midpoint of arc . Hence .
Solution 3
Since , we can construct circle tangent to and at and respectively. Then, we see that the circumcircle of must be tangent to at by the converse of the radical axis theorem, since is internally tangent to at , and meets at and . Thus, , so lines and are parallel.
It then follows that . Since is tangent to , we see that , and thus . Thus since bisects in cyclic quad , we have .
See Also
2010 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |