## 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/$\sim $sdunbar1

sdunbar1 at unl dot edu

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.

Solution: Let the midpoints of $\overline{AB}$,
$\overline{BC}$, and
$\overline{AC}$ be
$D$,
$E$ and
$F$ respectively.
Then $\u25b3ADF$,
$\u25b3FEC$,
$\u25b3DBE$ and
$\u25b3DEF$
are all congruent, and each is similar (with similarity ratio
$1\u22152$) to the
original $\u25b3ABC$. The
$\u25b3DEF$ is the medial triangle.
The circle ${k}_{1}$ that passes
through points $A$,
$D$, and
$F$ is the circumscribing
circle for $\u25b3ADF$ 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 $\u25b3DBE$
is at the intersection of the perpendicular bisectors of
$\overline{DB}$,
$\overline{BE}$,
$\overline{ED}$. Let
$I$ be the circumcenter
of $\u25b3DBE$, 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,
$\u25b3DBE$, its circumcise and the
diameter through $E$ can be
rigidly translated to $\u25b3FEC$, and the
diameter will pass through $H$.
Similarly for $\u25b3ADE$ 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.