Lemma: CDEI and BDFI are cyclic (I is the incenter of ABC) Proof: draw the perpendiculars from I Then <RNI=<BDI=90 and <RMI=<CDI=90 Then since RN and RM are chords, and I is the center, M and N are the midpoints of PR, QR and MN parallel to QR Also B,D,C,G are harmonic (wlog AB>AC) and since DR perpendicular to RG because DG is diameter, then DR is bisector of <BDC. Then arc PD=arc QD and ID is perpendicular to PQ, but ID is perpendicular to BC Therefore MN, PQ, BC are parallel and RX, PM, QN are concurrent (Desargues Theorem)

Quote: The circle $ \Gamma $ is inscribed to the scalene triangle $ABC$. $ \Gamma $ is tangent to the sides $BC, CA$ and $AB$ at $D, E$ and $F$ respectively. The line $EF$ intersects the line $BC$ at $G$. The circle of diameter $GD$ intersects $ \Gamma $ in $R$ ($ R\neq D $). Let $P$, $Q$ ($ P\neq R , Q\neq R $) be the intersections of $ \Gamma $ with $BR$ and $CR$, respectively. The lines $BQ$ and $CP$ intersects at $X$. The circuncircle of $CDE$ meets $QR$ at $M$, and the circuncircle of $BDF$ meet $PR$ at $N$. Prove that $PM$, $QN$ and $RX$ are concurrents. Author: Arnoldo Aguilar, El Salvador Let $K$ be the midpoint of $GD$. Since $(G,D,B,C)$ is a harmonic division; thus $KD^2=$ $KR^2=$ $\overline {KB}.$ $\overline {KC}$, which imples $KR$ is a tangent to the circumcircle of $\triangle RBC$. Hence, $\angle KRB=$ $\angle RCB$. On the other hand, since $KD$ has already been a tangent from $K$ to $\Gamma$. As a result, $KR$ is also another tangent from $K$ to $\Gamma$ $\Longrightarrow$ $\angle KRB=$ $\angle RQP$. Therefore, $\angle RQP=$ $\angle RCB$, which implies that $RQ$ $\|$ $BC$. As a consequence, $RX$ will pass through the midpoint of $PQ$. For that argument, $RX,$ $PM$ and $QN$ will concur at the centroid of $\triangle RQP$. Our proof is completed then. $\square$

If I am not mistaken, second year in a row that Arnoldo gets one of his geometry problems selected as problem 3 in Ibero, so double congrats!!!

Solution by Johan Gurandi: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2093045#p2093045

it's trivial that $C,D,B,G$ form harmonic sequence.hence $R$ is on the Apollonius circle $(B,C,\frac{BD}{DC})$ by $\angle GRD=90$ we get $RD$ bisects $\angle BRC$ hence $PD=DQ$,$PQ$ parallel to $BC$ then it's trivial that $PM,QN,RX$ are medians of three sides of triangle $QPR$.

Let $I$ be the center of $\Gamma$ and $Y$ the second intersection of $\overleftrightarrow{GR}$ and $\Gamma$. Since $\overline{GD}$ is a diameter of $(GRD)$, we know that $\angle GRD = 90^{\circ}$. Thus, $\angle YRD = 90^{\circ}$, so $\overline{YD}$ is a diameter of $\Gamma$. Now, note that by Ceva+Menelaus, $(GD;BC)=-1$. Thus, \[ -1=(GD;BC)\stackrel{R}= (YD;PQ). \]Since $\overline{YD}$ is a diameter of $\Gamma$, it must be the case that $Y$ and $D$ are midpoints of the major and minor arcs $\widehat{PQ}$. Hence, $\overline{PQ}\parallel\overline{BC}$ since both are perpendicular to $\overline{ID}$. Now, since $R$ and $X$ are the centers of homothety taking $\overline{PQ}$ to $\overline{BC}$ and $\overline{QP}$ to $\overline{BC}$, respectively, it is well known that $\overline{RX}$ passes through the midpoint of $\overline{PQ}$. It is easy to see $I\in(CDE), (BDF)$, since $\overline{IC}$ and $\overline{IB}$ are the diameters of these circles, respectively. Thus, $\overline{IM}\perp\overline{RQ}$. Since $I$ is the center of $\Gamma$, $M$ is the midpoint of $\overline{RQ}$. Analogously, $N$ is the midpoint of $\overline{RP}$. This implies $\overline{PM}, \overline{QN}, \overline{RX}$ intersect at the centroid of $\triangle PQR$. $\blacksquare$

Its well known that if $I$ is the incenter of $\triangle ABC$ then: $DIEC,DIFB$ are cyclic and $-1=(G, B; D, C)$ A few claims will be shown here to be used on the resolution. Claim 1: $BC \parallel PQ$ Proof: Note that since $\angle GRD=90$ and $-1=(G, B; D, C)$ we have that $RD$ bisects $\angle PRQ$ thus $ID \perp PQ$ but since $ID \perp BC$ we have the desired result. Claim 2: $MN \parallel BC \parallel PQ$ Proof: By Claim 1 we have that $BC \parallel PQ$, now note that $PI=RI=IQ$ and $\angle BNI=\angle IMC=90$ thus $M,N$ are midpoints of $RP, RQ$ respectvily meaning that by mid-base $MN \parallel PQ$ and we are done!. Claim 3: Let $K$ the midpoint of $BC$ then if $Y, Z$ are points on $BR, CR$ respectivily such that $YZ \parallel BC$ then $AK,BZ,CY $are concurrent. Proof: Trivial by Ceva's theorem. Now use Claim 3 to show that $A,X,K$ are colinear and then note that $A,X$ and the midpoint of $PQ$ are colinear, thus now the concurrency point is the baricenter of $\triangle PRQ$ thus we are done