## A Geometry Problem from the 2000 Bay Area Mathematical Olympiad

January 30, 2009

Problem: Let $ABC$ be a triangle with $D$ as the midpoint of side $\overline{AB}$, $E$ as the midpoint of side $\overline{BC}$, and $F$ as the midpoint of side $\overline{AC}$. Let ${k}_{1}$ be the circle passing through points $A$, $D$ and $F$; let ${k}_{2}$ be the circle passing through points $B$, $E$ and $D$; Let ${k}_{3}$ be the circle passing through points $C$, $F$ and $E$. Prove that ${k}_{1}$, ${k}_{2}$ and ${k}_{3}$ intersect in a point.

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Solution: Let the midpoints of $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$ be $D$, $E$ and $F$ respectively. Then $△ADF$, $△FEC$, $△DBE$ and $△DEF$ are all congruent, and each is similar (with similarity ratio $1∕2$) to the original $△ABC$. The $△DEF$ is the medial triangle. The circle ${k}_{1}$ that passes through points $A$, $D$, and $F$ is the circumscribing circle for $△ADF$ and similarly for the circles ${k}_{2}$ and ${k}_{3}$ 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 $\overline{DB}$, $\overline{BE}$, $\overline{ED}$. Let $I$ be the circumcenter of $△DBE$, and let $H$ be the opposite end of the diameter of ${k}_{2}$ through $B$.

Now consider the diameter of ${k}_{2}$ through $E$. Let the opposite end of the diameter on ${k}_{2}$ be ${H}_{1}$. 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.