A Geometry Problem from the
2000 Bay Area Mathematical Olympiad

Steven R. Dunbar
Department of Mathematics
Univesity of Nebraska-Lincoln
Lincoln, Nebraska
www.math.unl.edu/ sdunbar1
sdunbar1 at unl dot edu

January 30, 2009

Problem: Let ABC be a triangle with D as the midpoint of side AB¯, E as the midpoint of side BC¯, and F as the midpoint of side AC¯. Let k1 be the circle passing through points A, D and F; let k2 be the circle passing through points B, E and D; Let k3 be the circle passing through points C, F and E. Prove that k1, k2 and k3 intersect in a point.

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Solution: Let the midpoints of AB¯, BC¯, and AC¯ be D, E and F respectively. Then ADF, FEC, DBE and DEF are all congruent, and each is similar (with similarity ratio 12) to the original ABC. The DEF is the medial triangle. The circle k1 that passes through points A, D, and F is the circumscribing circle for ADF and similarly for the circles k2 and k3 through D, B, E and E, F, C respectively. Since the triangles are all congruent, the circumscribing circles all have the same diameter.

The circumcenter of DBE is at the intersection of the perpendicular bisectors of DB¯, BE¯, ED¯. Let I be the circumcenter of DBE, and let H be the opposite end of the diameter of k2 through B.

Now consider the diameter of k2 through E. Let the opposite end of the diameter on k2 be H1. By congruence, DBE, its circumcise and the diameter through E can be rigidly translated to FEC, and the diameter will pass through H. Similarly for ADE and the diameter through A also passing through H. Hence the circles have the point H in common. Hence the three circumcircles have the point H in common.

Commentary: This is a special case of Miquel’s Theorem and the Pivot Theorem.