2003 JBMO Problems/Problem 3

Problem

Let $D$ , $E$ , $F$ be the midpoints of the arcs $BC$ , $CA$ , $AB$ on the circumcircle of a triangle $ABC$ not containing the points $A$ , $B$ , $C$ , respectively. Let the line $DE$ meets $BC$ and $CA$ at $G$ and $H$ , and let $M$ be the midpoint of the segment $GH$ . Let the line $FD$ meet $BC$ and $AB$ at $K$ and $J$ , and let $N$ be the midpoint of the segment $KJ$ .

a) Find the angles of triangle $DMN$ ;

b) Prove that if $P$ is the point of intersection of the lines $AD$ and $EF$ , then the circumcenter of triangle $DMN$ lies on the circumcircle of triangle $PMN$ .

Solution

Let $FC$ , $EB$ intersect $DE$ , $FD$ at $M'$ , $N'$ respectively. We will prove first that $M' = M, N = N'$ and that lines $AD$ , $BE$ , $FC$ are altitudes of the $\triangle FDC$ .

It's easy to see that lines $AD$ , $BE$ and $CF$ form the internal angle bisectors of $\triangle ABC$ .

Consequently, we can determine the $\angle FDC$ of $\triangle FDC$ as being equal to $\angle FDA + \angle ADE = \angle C/2 + \angle B/2 = 90^{\circ} - \angle A/2$

Also we have $\angle DFC = \angle A/2$ , thus $\angle FM'D = 90^{\circ}$ . Similarly $\angle EN'D = 90^{\circ}$ .

Thus $AD$ , $BE$ , $FC$ are altitudes of the $\triangle FDC$ with $P$ , $N'$ , $M'$ respectively being the feet of the altitudes.

Now since $M'C$ is internal bisector of $\angle HCG$ and $CM'$ is perpendicular to $GH$ , we have that $CM'$ is the perpendicular bisector of $GH$ . Hence $M' = M$ .

Similarly it can be shown that $BN'$ is the perpendicular bisector of $KJ$ , and hence $N' = N$ .

Now lines $AD$ , $BE$ and $CF$ intersect at point $Q$ . So $Q$ is the incenter of $\triangle ABC$ and orthocenter of $\triangle DEF$ .

Clearly, $QNDM$ is a cyclic quadrilateral as the $N$ , $M$ are the feet of perpendiculars from $E$ and $F$ .

So, we have $\angle QNM = \angle QDM = \angle ADE = \angle B/2$ .

Similarly, since $PQNF$ is also a cyclic-quadrilateral, reasoning as above, $\angle QNP = \angle PFQ = \angle CFE = \angle B/2$ .

Thus we have that $\angle QNM = \angle QNP$ and so $NQ$ is an internal bisector of $\angle PNM$ . Reasoning in a similar fashion it can be proven that $PQ$ and $MQ$ are internal bisectors of other 2 angles of $\triangle PNM$ .

Thus $Q$ also happens to be the incenter of $\triangle PNM$ in addition to being that of $\triangle ABC$ .

$Part 1$ : Angles of $\triangle DMN$ :

Since $\angle QNM = \angle B/2, \angle DNM = 90^{\circ} - \angle B/2$ . Similarly $\angle NMD = 90^{\circ} - \angle C/2$ . Finally $\angle NDM = 90^{\circ} - \angle A/2$ .

$Part 2$ :

Let circumcircle of $\triangle PMN$ cut line $DQ$ at point $R$ . Since $PMRN$ is a cyclic quadrilateral, we have $\angle RNM = \angle RPM = \angle A/2$ .