1977 AHSME Problems/Problem 19
Problem 19
Let be the point of intersection of the diagonals of convex quadrilateral , and let , and be the centers of the circles circumscribing triangles , and , respectively. Then
Solution
Let , , , and be the intersections of and , and , and , and and respectively. Since the circumcenter of a triangle is determined by the intersection of its perpendicular bisectors, each of the pairs mentioned earlier are perpendicular to each other. Let . Then since . This means , which similarly implies . Through a similar method, we find that .
Since the opposite angles of are congruent, is indeed a parallelogram - and we showed this without any assumption about the type of quadrilateral was. Therefore, is always a parallelogram, and so our answer is .
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
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