# 2003 JBMO Problems/Problem 3

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## 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 $\triangle DMN$:

Sincel $\angle QNM = \angle B/2, \angle DNM = 90^{\circ} - \angle B/2$. 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 $\angle RNM = \angle RPM = \angle A/2$.

Similarly, $\angle RMN = \angle RPN = \angle A/2$. Thus $RN$ = $RM$.

Now, $\angle RND = 90^{\circ} - (\angle A/2 + \angle B/2) = \angle C/2$

and $\angle RDN = \angle ADF = \angle C/2$. Thus $RN$ = $RD$.

Thus we have $RN$ = $RM$ = $RD$. So $R$ is the circumcenter of $\triangle DMN$.