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Vietnamese Mathematical Olympiad

Geometry Difficulty 7.2 National Olympiad, round 2 Prove it Vietnam

Given a triangle $ABC$ with fixed vertices $B$ , $C$ ; point $A$ moves such that triangle $ABC$ is acute. Let $D$ be the midpoint of $BC$ and $E$ , $F$ be the projections of $D$ to $AB$ , $AC$ respectively.

a) Let $O$ be the circumcenter of triangle $ABC$ . $EF$ meets $AO$ and $BC$ at $M$ , $N$ respectively. Prove that the circumcircle of triangle $AMN$ passes through a fixed point.

b) Suppose that the tangents of the circumcircle of triangle $AEF$ at $E$ , $F$ intersect each other at $T$ . Prove that $T$ lies on a fixed line.

Solution

a) Without loss of generality, suppose that $AB < AC$ . Clearly, $N$ lies on the opposite ray of ray $BC$ . Note that $AEDF$ is a cyclic quadrilateral and $\angle OAC = 90^\circ - \angle ABC$ , we have
$\begin{align*} \angle AMN &= \angle MAE + \angle MEA = 90^\circ - \angle ABC + \angle ADF \\ &= \angle BDF + \angle ADF = \angle NDA, \end{align*}$
which implies $AMDN$ is a cyclic quadrilateral. Thus, $(AMN)$ passes through a fixed point $D$ .

b) Let $(K)$ be the circumcircle of triangle $AEF$ , it is obvious that $AD$ is the diameter of $(K)$ . Let $L$ be the point on $(K)$ such that $DL \perp BC$ , we can easily prove $AL \perp BC$ and $AL \parallel BC$ . Note that $D$ is the midpoint of $BC$ , hence
$A(FE, DL) = A(BC, DL) = -1.$
This follows that $LEDF$ is a harmonic quadrilateral. Thus, $DL$ passes through $T$ . Clearly, $DL$ is the perpendicular bisector of $BC$ so $T$ lies on a fixed line. $\square$

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