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PrintSAUDI ARABIAN MATHEMATICAL COMPETITIONS
Saudi Arabia geometry
Problem
The triangle () is inscribed in the circle . On the sides and , the points and are chosen, respectively, so that . The lines and intersect at point . Let be the center of the inscribed circle of triangle , and the center of the excircle of the triangle tangent to side . Prove that the midpoint of the of the circle is equidistant from and .
Solution
Let be the second intersection of two circles and . We have and , so . Then , which means that is the midpoint of the arc of circle .
On the other hand, we also have then is cyclic. Similarly, is also cyclic.
Since , it is easy to see that and are equal. So, if we call the midpoints of and of circles and , respectively, then two isosceles triangles and are congruent. But we know that are also the circumcenters of and , so .
At last, from and being equal, we get , then .
On the other hand, we also have then is cyclic. Similarly, is also cyclic.
Since , it is easy to see that and are equal. So, if we call the midpoints of and of circles and , respectively, then two isosceles triangles and are congruent. But we know that are also the circumcenters of and , so .
At last, from and being equal, we get , then .
Techniques
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleCyclic quadrilateralsSpiral similarityAngle chasing