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 .
Solution 3
WLOG, let the diameter of be
We see that and from right triangles and
We now look at By the Extended Law of Sines on we get that Similarly,
We now look at By Ptolemy's Theorem, we have which gives us This means that We now seek to relate the lengths computed with the areas.
To do this, we consider the altitude from to This is to find the area of Finding the area of is similar.
We claim that In order to prove this, we will prove that In other words, we wish to prove that This is equivalent to proving that
Note that and Therefore, we get that Thus, In this way, we get that the altidude from to has length Therefore, we see that and so the two areas are equal.
Solution by Ilikeapos
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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