Difference between revisions of "2010 IMO Problems/Problem 4"
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Let the tangent at <math>M</math> to <math>\Gamma</math> intersect <math>SC</math> at <math>X</math>. We now have that since <math>\triangle{XMC}</math> and <math>\triangle{SPC}</math> are both isosceles, <math>\angle{SPC}=\angle{SCP}=\angle{XMC}</math>. This yields that <math>MX \| PS</math>. | Let the tangent at <math>M</math> to <math>\Gamma</math> intersect <math>SC</math> at <math>X</math>. We now have that since <math>\triangle{XMC}</math> and <math>\triangle{SPC}</math> are both isosceles, <math>\angle{SPC}=\angle{SCP}=\angle{XMC}</math>. This yields that <math>MX \| PS</math>. |
Revision as of 23:08, 15 July 2010
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 |