Difference between revisions of "1989 IMO Problems/Problem 2"
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+ | [[Category: Olympiad Geometry Problems]] |
Latest revision as of 11:50, 30 January 2021
is a triangle, the bisector of angle meets the circumcircle of triangle in , points and are defined similarly. Let meet the lines that bisect the two external angles at and in . Define and similarly. Prove that the area of triangle area of hexagon area of triangle .
Solution
Notice that since and substend the same angle in the circle, and so are equal. Thus is iscoceles, and similarly for triangles and . Also, since is a cyclic quadrilatera, , and similarly for the other triangles. Thus, the area of triangle and similarly for the other triangles. Thus, the area of the hexagon is equal to .
Now we shall find the area of triangle . It is obvious that the points are collinear since the angle the make at is , and similarly for the other points. Thus, it suffices to find the area of each of the triangles . It is well known that is the center of the excircle with radius . Thus the area of triangle is and so the area of triangel is .
Now we can prove that area of hexagon by simplifying. We have
area of hexagon
as desired.
As for the inequality, notice that it is equivalent to
Letting for positive reals, the inequality becomes
which is true by AM-HM.
See Also
1989 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |