1987 IMO Problems/Problem 2
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 .
Solution 2
Clearly, is a kite, so its diagonals are perpendicular. Furthermore, we have triangles and similar because two corresponding angles are equal.
Hence, we have Notice that we used the fact that a quadrilateral's area is equal to half the product of its perpendicular diagonals (if they are, in fact, perpendicular).
But in (right) triangle , we have . Furthermore, if is the intersection of diagonals and we have the midpoint of and an altitude of , so so . Hence as desired.
Solution 3
Proceed as in Solution 2. To prove that , consider that
via usage of definition of sine, equal angles in a right triangle if its altitude is drawn, and the Law of Sines.
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 |