Difference between revisions of "2010 IMO Problems/Problem 4"
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 16:49, 3 April 2012
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,
\widehat{ML} &= \widehat{MA} + 2\angle AKL \ &= \left(2\angle CPK - \widehat{KC}\right) + 2\angle ABL \ &= 2\left(\angle CPK + \angle KPS\right) - \widehat{KC} \ &= 2\angle PCS - \widehat{KC} \ &= \widehat{MK},
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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 |