2012 IMO Problems/Problem 1
Problem
Given triangle the point is the centre of the excircle opposite the vertex This excircle is tangent to the side at , and to the lines and at and , respectively. The lines and meet at , and the lines and meet at Let be the point of intersection of the lines and , and let be the point of intersection of the lines and Prove that is the midpoint of .
Solution
First, because and are both tangents from to the excircle . Then . Call the the intersection between and . Similarly, let the intersection between the perpendicular line segments and be . We have and . We then have, . So . We also have . Then . Notice that . Then, . . Similarly, . Draw the line segments and . and are congruent and and are congruent. Quadrilateral is cyclic because . Quadrilateral is also cyclic because . The circumcircle of also contains the points and because there is a circle around the quadrilaterals and . Therefore, pentagon is also cyclic. Finally, quadrilateral is cyclic because . Again, is common in both the cyclic pentagon and cyclic quadrilateral , so the circumcircle of also contains the points , , and . Therefore, hexagon is cyclic. Since and are both right angles, is the diameter of the circle around cyclic hexagon . Therefore, and are both right angles. and are congruent by ASA congruency, and so are and . We have , , , and . Since and are tangents from to the circle , . Then, we have , which becomes , which is , or . This means that is the midpoint of .
QED
--Aopsqwerty 21:19, 19 July 2012 (EDT)
Solution 2
For simplicity, let written alone denote the angles of triangle , and , , denote its sides.
Let be the radius of the A-excircle. Because , we have isosceles and so by the Exterior Angle Theorem. Then because , we have , again by the Exterior Angle Theorem.
Notice that and , and so after converting tangents to sine and cosine. Thus, It follows that . By the Law of Sines on triangle and and the double-angle formula for sine, we have Therefore, triangle is congruent to a right triangle with hypotenuse length and one angle of measure by SAS Congruence, and so . It then follows that triangles and are congruent by , and so . Thus, lies on the perpendicular bisector of . Similarly, lies on the perpendicular bisector of , and so is the circumcenter of . In particular, lies on the perpendicular bisector of , and so, because is perpendicular to , must be the midpoint of , as desired.
--Suli 17:53, 8 February 2015 (EST)
Solution 3
Same as Solution 2, except noticing that (letting be the semi-perimeter):
--Suli 18:21, 8 February 2015 (EST)
Solution 4
As before in Solution 2, we find that But it is clear that bisects , so and hence is cyclic. In particular, , and continue as in Solution 2.
See Also
2012 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |