Difference between revisions of "1987 IMO Problems/Problem 2"
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Thus we have proven that <math>[AKNM]=[ABC]</math>. | Thus we have proven that <math>[AKNM]=[ABC]</math>. | ||
+ | ==See also== | ||
{{IMO box|num-b=1|num-a=3|year=1987}} | {{IMO box|num-b=1|num-a=3|year=1987}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 12:56, 2 November 2008
Problem
In an acute-angled triangle the interior bisector of the angle intersects at and intersects the circumcircle of again at . From point perpendiculars are drawn to and , the feet of these perpendiculars being and respectively. Prove that the quadrilateral and the triangle have equal areas.
Solution
We are to prove that or equivilently, . Thus, we are to prove that . It is clear that since , the segments and are equal. Thus, we have since cyclic quadrilateral gives . Thus, we are to prove that
From the fact that and that is iscoceles, we find that . So, we have . So we are to prove that
We have ,, , ,, and so we are to prove that
We shall show that this is true: Let the altitude from touch at . Then it is obvious that and and thus .
Thus we have proven that .
See also
1987 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |