2007 IMO Problems/Problem 4
Problem
In the bisector of intersects the circumcircle again at , the perpendicular bisector of at , and the perpendicular bisector of at . The midpoint of is and the midpoint of is . Prove that the triangles and have the same area.
Solution
The area of is given by and the area of is . Let , , and . Now and , thus . , so , or . The ratio of the areas is . The two areas are only equal when the ratio is 1, therefore it suffices to show . Let be the center of the circle. Then , and . Using law of sines on we have: so . by law of sines, and , thus 1) . Similarly, law of sines on results in or . Cross multiplying we have or 2) . Dividing 1) by 2) we have
Solution 2 (Power of a point)
, and similarly , we have . Using triangle area formula , the problem is equivalent to proving , or . Draw line perpendicular to BC and intersects BC at , then , and . Now the problem is equivalent to proving , or . Since , we have . Let the radius of the circumcircle be , then the diameter through is divided by point into lengths of and . By power of point, . Similarly, . Therefore .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2007 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
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