2008 IMO Problems/Problem 1
Problem
An acute-angled triangle has orthocentre . The circle passing through with centre the midpoint of intersects the line at and . Similarly, the circle passing through with centre the midpoint of intersects the line at and , and the circle passing through with centre the midpoint of intersects the line at and . Show that , , , , , lie on a circle.
Solution
Let , , and be the midpoints of sides , , and , respectively. It's not hard to see that . We also have that , so . Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that is on the radical axis of the circles centered at and , so is too. We then have . This implies that , so . Therefore . This shows that quadrilateral is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments and . However, these are just the perpendicular bisectors of and , which meet at the circumcenter of , so the circumcenter of is the circumcenter of triangle . Similarly, the circumcenters of and are coincident with the circumcenter of . The desired result follows.
See Also
2008 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |