1985 IMO Problems/Problem 5
A circle with center passes through the vertices and of the triangle and intersects the segments and again at distinct points and respectively. Let be the point of intersection of the circumcircles of triangles and (apart from ). Prove that .
is the Miquel Point of quadrilateral , so there is a spiral similarity centered at that takes to . Let be the midpoint of and be the midpoint of . Thus the spiral similarity must also send to and so is cyclic. is also cyclic with diameter and thus must lie on the same circumcircle as , , and so .
Let and be the circumcircles and circumcenters of respectively.
Let is cyclic
The radius of is
Let and be midpoints of and respectively.
is the Miquel Point of quadrilateral so is cyclic. is trapezium as desired.
Solution 3 (No Miquel's point)
Consider and , they are similar because = , and also .
Now draw , and intersecting at ; , at . Naturally bisects , and bisects . We claim , because
Thus , this implies . Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have . (by gougutheorem)
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