2010 IMO Problems/Problem 4
Contents
[hide]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 .
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 |