1978 IMO Problems/Problem 4
Problem
In a triangle we have A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides in the points respectively Prove that the midpoint of is the center of the inscribed circle of the triangle
Solution
Denote the triangle sides, the triangle inradius and circumradius and , the triangle semiperimeter and area. Let be the tangency points of the incircle (I) with the sides . Let be the midpoints of of and the midpoint of the arc of the circumcircle (O) opposite to the vertex . The isosceles triangles are all centrally similar with the homothety center . The homothety coefficient for the triangles is . The homothety coefficient for the triangles is . The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles are also centrally similar with the homothety center A and the same homothety coefficient as the triangles : , where h = AK is the A-altitude of the triangle . The homothety coefficient of the triangles is then Denote the -altitude of the triangle . Then
Substituting , and , we get
(substituting 2s = 2b + a and for an isosceles triangle)
which means that the point is identical with the incenter of the triangle .
The above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]
See Also
1978 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |