http://rmm.lbi.ro/index.php?id=solutions_math

Here's an outline of my solution from the contest. It's enough to show that $\measuredangle{GPD}=\measuredangle{GQF}$. But $DE\parallel BA$ and $ABQE$ is cyclic, so $\triangle{GDE}\sim\triangle{GEQ}$ with opposite orientation; similarly, $\triangle{GEF}\sim\triangle{GFP}$ with opposite orientation. Using complex numbers, set $g=0$ so we have $d+e+f=0$. It's easy to see that $q=e\overline{e}/\overline{d}$ and $p=f\overline{f}/\overline{e}$, so using $d=-e-f$, it's easy to show that \[\frac{d-p}{g-p}\bigg{/}\frac{f-q}{g-q}=|e|^2/|f|^2\in\mathbb{R},\]as desired.

Note that the medians $ AD ~,~ BE ~,~ CF $ meet at $ G$ . Suppose that the circle $ BCE $ and the line $ CF $ intersect at $ S $ . We now show that $ \angle DPG = \angle DSG $ . Let $ X ~,~ Y $ be points on the rays of $ BG $ and $ CG $ respectively such that $ BP = PX $ and $ CS = SY $ . Therefore , $ DP \parallel CX $ and $ DS \parallel BY $ . $ BG \cdot GX = BG \cdot GP + BG \cdot PX = BG \cdot GP + BG \cdot BP $ $ = 2 BG \cdot GP + BG^2 = BG^2 + CG^2 $ . Similarly , $ CG \cdot GY = BG^2 + CG^2 $ . Therefore , $BCXY$ are concyclic and $ \angle DPG = \angle CXB = \angle BYC = \angle DSG $ as desired . As $ AG \cdot GQ = BG \cdot GE = CG \cdot GS $ , so $ ACQS $ are concyclic and because $ \angle FDG = \angle CAG = \angle QSG $ , $ FDQS $ are concyclic as well . $ \angle FQG = \angle DSG = \angle DPG $ , we get the result .

we quickly translate to an equality of angles in Q and in the point analogous to Q (intersection of A-Median with AFC circle) we invert in A We use trigonometry to finish Sine Law, noting that the A-Median converts to the A-symmedian

Let $AG=2a$, $BG=2b$, $CG=2c$, so we get $DG=a$, $EG=b$, $FG=c$. And some calculation, we know $AB^2=8a^2+8b^2-4c^2$. So we take $H$ on $AD$ such that $FH$ is perpendicular to $AD$, we get $AH=(3a^2+b^2-c^2)/2a$. And $GQ*AG=BG*GF$ so $GQ=b^2/a$, $tan\angle FQG=\sqrt{-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2}/a^2+b^2+c^2$. Similarly, $tan\angle DPG=\sqrt{-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2}/a^2+b^2+c^2$. so we get $\angle FQG = \angle DPG$ as desired.

It suffices to prove $\angle GPD = \angle GQF$, or rather, $\angle BPD = \angle AQF$. From http://www.artofproblemsolving.com/community/c6h621987p3718575 From my solution to KMO 2014 P5 wrote: There are many midpoints in this configuration. This calls for a lemma. Lemma. Let there be a triangle $\triangle ABC$. Let the midpoint of $BC$ be $M$. $\cot \angle BAM = 2 \cot A + \cot B$. Proof. We bash using LoC and $S=\frac{1}{2}ab\sin C$ and find $\cot \angle BAM = \frac{3c^2+b^2-a^2}{4S}, \cot A = \frac{b^2+c^2-a^2}{4S}, \cot B = \frac{a^2+c^2-b^2}{4S}$. The proof is complete. $\blacksquare$. Now also, we note that $$\cot \angle AMC = \cot (\angle MAB + \angle ABM) = \frac{(2\cot A + \cot B)\cot B -1}{2\cot A +2\cot B} = \frac{\cot B}{2} + \frac{\cot A \cot B - 1}{2(\cot A +\cot B)} = \frac{\cot B - \cot C}{2}$$ So we use this lemma, to find that $$\cot \angle BPD = 2\cot \angle BPC + \cot \angle PBC = 2\cot \angle BFC + \cot \angle EBC$$$$= 2\left(\frac{\cot A - \cot B}{2}\right)+ (2\cot B+\cot C)= \cot A + \cot B + \cot C$$and $$\cot \angle AQF = 2\cot \angle AQB + \cot \angle DAB = 2 \cot \angle AEB + \cot \angle DAB$$$$=2\left(\frac{\cot C - \cot A}{2}\right) + (2\cot A + \cot B) = \cot A + \cot B + \cot C$$which is enough to say that $\angle BPD =\angle AQF$ as desired.

It's sufficient to show that $\measuredangle GPD = \measuredangle GQF.$ Let $K$ be the symmedian point and denote $X \equiv AK \cap BG$ and $Y \equiv BK \cap CG$ and $Z \equiv CK \cap BG.$ The inversion with pole $B$ and power $\tfrac{1}{2} \cdot BA \cdot BC$ composed with a reflection in the bisector of $\angle B$ sends $\odot(BCF) \mapsto CF$ and $BG \mapsto BK.$ Hence, this inversion sends $P \mapsto Y.$ Moreover, $D \mapsto A$, so it follows that $\measuredangle GPD = \measuredangle BPD = \measuredangle YAB.$ Similarly, $\measuredangle GQF = \measuredangle XCA.$ Thus, we need only show that $\measuredangle YAB = \measuredangle XCA.$ Since, $Y$ and $Z$ are isogonal conjugates (because $BY, BZ$ and $CY, CZ$ are isogonal), this is equivalent to $\measuredangle CAZ = \measuredangle XCA.$ In other words, we must show that $R \equiv AZ \cap CX$ lies on the perpendicular bisector of $\overline{AC}.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 60.40375928404985, xmax = 178.98634130954153, ymin = 62.142101886012526, ymax = 124.53069914549678; /* image dimensions */ draw((129.42059191217,73.87130566156932)--(120.14047243325963,113.13334961068217)--(77.13068792523266,73.69284182543699)--cycle, red); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((129.42059191217,73.87130566156932)--(120.14047243325963,113.13334961068217), red); draw((120.14047243325963,113.13334961068217)--(77.13068792523266,73.69284182543699), red); draw((77.13068792523266,73.69284182543699)--(129.42059191217,73.87130566156932), red); draw(circle((95.41489883298036,96.92524143956406), 29.564450956334248), linetype("2 2") + blue); draw((120.14047243325963,113.13334961068217)--(100.7301126402281,67.84251009375102)); draw((77.13068792523266,73.69284182543699)--(124.78053217271483,93.50232763612576)); draw((77.13068792523266,73.69284182543699)--(115.19647182406923,87.65505747461411)); draw((129.42059191217,73.87130566156932)--(110.97457254257012,91.74624986580724)); draw((120.14047243325963,113.13334961068217)--(115.19647182406923,87.65505747461411)); /* dots and labels */ dot((129.42059191217,73.87130566156932),linewidth(3.pt) + dotstyle); label("$A$", (129.87191367958042,72.5402014292599), SE * labelscalefactor); dot((120.14047243325963,113.13334961068217),linewidth(3.pt) + dotstyle); label("$B$", (120.43250416511343,113.54263650759708), NE * labelscalefactor); dot((77.13068792523266,73.69284182543699),linewidth(3.pt) + dotstyle); label("$C$", (75.96403590555404,72.68769220292299), SW * labelscalefactor); dot((115.19647182406923,87.65505747461411),linewidth(3.pt) + dotstyle); label("$K$", (114.90160015273042,86.03560721943202), S * labelscalefactor); dot((98.63558017924615,93.41309571805958),linewidth(3.pt) + dotstyle); label("$D$", (97.7189250209272,94.1475997709016), NW * labelscalefactor); dot((103.27563991870133,73.78207374350316),linewidth(3.pt) + dotstyle); label("$E$", (103.10233825964667,72.46645604242836), SE * labelscalefactor); dot((124.78053217271483,93.50232763612576),linewidth(3.pt) + dotstyle); label("$F$", (125.52093585650579,92.89392819476541), NE * labelscalefactor); dot((110.97457254257012,91.74624986580724),linewidth(3.pt) + dotstyle); label("$X$", (110.10815000866515,92.15647432645), NW * labelscalefactor); dot((108.1067205255129,85.05459515936145),linewidth(3.pt) + dotstyle); label("$Z$", (108.11702456420726,83.60200945399116), SE * labelscalefactor); dot((115.5896809852469,89.681411973932),linewidth(3.pt) + dotstyle); label("$Y$", (116.15527172887056,90.53407581615608), NE * labelscalefactor); dot((100.7301126402281,67.84251009375102),linewidth(3.pt) + dotstyle); label("$P$", (100.74248588102992,66.49307970907348), S * labelscalefactor); dot((108.89725075688743,86.89916569922951),linewidth(3.pt) + dotstyle); label("$G$", (108.19076995103903,87.80549650338902), NW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Assign barycentric coordinates. Since $K = \left(a^2 : b^2 : c^2\right)$ and $X \in AK$, we deduce that $X = \left(c^2 : b^2 : c^2\right).$ Similarly, $Z = \left(a^2 : b^2 : a^2\right).$ Hence, as $R \in AZ$ and $R \in CX$, we obtain $R = \left(c^2 : b^2 : a^2\right).$ Thus, it's clear that $R$ verifies the equation of the perpendicular bisector of $\overline{AC}: b^2(z - x) + y\left(c^2 - a^2\right) = 0.$ This completes the proof. $\square$

Also discussed at http://www.artofproblemsolving.com/community/c6h1087522.

An elegant solution with practically zero computation RMM 2012/5 wrote: Given a non-isosceles triangle $ABC$, let $D,E$, and $F$ denote the midpoints of the sides $BC,CA$, and $AB$ respectively. The circle $BCF$ and the line $BE$ meet again at $P$, and the circle $ABE$ and the line $AD$ meet again at $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$. (United Kingdom) David Monk Notice that it suffices to show $\measuredangle AQF=\measuredangle BPD$. Let $L$ be the symmedian point of $ABC$; set $M=\overline{AL} \cap \overline{BE}$ and $N=\overline{BL} \cap \overline{AD}$. Claim: $\measuredangle AQF=\measuredangle ACM$. (Proof) Apply inversion at $A$ with radius $\sqrt{\frac{1}{2}AB \cdot AC}$ followed by reflection in the bisector of angle $BAC$. Notice that this is an involution on the plane; and $B \mapsto E$, so $\odot(ABE) \mapsto \overline{EB}$. Also, $\overline{AD} \mapsto \overline{AL}$ and $F \mapsto C$. Evidently, we have $Q \mapsto M$ so $\measuredangle AQF=\measuredangle ACM$. $\blacksquare$ Analogously, we may show that $\measuredangle BPD=\measuredangle BCN$. However, since $L, G$ are isogonal conjugates in triangle $ABC$; apply the isogonality lemma on $\{A, B, L, G\}$; so, lines $\overline{CM}, \overline{CN}$ are isogonal in angle $ACB$, hence we conclude that $P, Q, R, G$ are indeed concyclic. $\blacksquare$

The above solution seems to gloss over the fact that $\measuredangle BPD=\measuredangle BCN$ is not analogous (the figure is cyclic, not symmetric). Thanks to niyu for pointing it out So years later, let’s finish the business. Clearly, from above reasoning, we need to prove the following lemma: wrote: In triangle $ABC$ with centroid $G$, Lemoine point $K$, lines $BG$ and $AK$ meet at $U$ and lines $CG$ and $AK$ meet at $V$. Then $\measuredangle VBA=\measuredangle ACU$. By taking isogonal conjugates, this is equivalent to the following: let lines $BK$ and $CK$ meet line $AG$ at $X$ and $Y$ respectively. Let lines $CX$ and $BY$ meet at $L$. Then $LB=LC$. Note that in complete quadrilateral $KXLYBC$, line $XY$ bisects $BC$. Thus, $KL \parallel BC$ and $XY$ bisects $KL$. Let $D$ be the midpoint of $BC$ and $\ell$ be the line through $D$ perpendicular to $BC$. Let $T$ be the intersection of tangents to $\odot(ABC)$ at $B$ and $C$, and note that $T$ lies on $AK$. Let $J=AK \cap BC$. It is clear that $(AJ; KT)=-1$. Redefine $L$ as point on $\ell$ with $KL \parallel BC$. It suffices to show that $AG$ bisects $KL$; projecting $KL$ through $A$ onto $\ell$, it suffices to show that $(TL; DR)=-1$. Projecting on $AK$ through $BC_{\infty}$, we conclude that $(TL;DR)=(TK; JA)=-1$, completing the proof. $\blacksquare$

Solved with anser. We use barycentrics on $DEF$. The point $(x, y, z)$ with reference triangle $ABC$ is $(y + z - x, z + x - y, x + y - z)$ with reference triangle $DEF$ (since $A = E + F - D$ and so on), so we can calculate $P$ and $Q$ in $ABC$ first. If the equation of $(BCF)$ is $[u, v, w]$, then $v = w = 0$ by plugging in $B$ and $C$, and $u = \frac{1}{2}c^2$ by plugging in $F$. Then if $P = (1 : t :1)$, $-a^2t - b^2 - c^2t + \frac{1}{2}c^2(t + 2) = 0$, so $t = \frac{c^2 - b^2}{a^2 + \frac{1}{2}c^2}$ and $P = (a^2 + \frac{1}{2}c^2 : c^2 - b^2 : a^2 + \frac{1}{2}c^2)$. Converting to $DEF$ (since $d : e : f = a : b : c$, we can keep the side lengths) we get $$P = (c^2 - b^2 : 2a^2 + b^2 : c^2 - b^2),$$and cyclically shifting gets $Q = (2c^2 + a^2 : b^2 - a^2 : b^2 - a^2)$. Then if $R = (x : y : z)$, from $R$ on $DP$ we have $\frac{y}{z} = \frac{2a^2 + b^2}{c^2 - b^2}$, and from $R$ on $FQ$ we have $\frac{x}{y} = \frac{2c^2 + a^2}{b^2 - a^2}$. Combining these gets $$R = ((a^2+2c^2)(2a^2+b^2) : (2a^2+b^2)(-a^2+b^2) : (-a^2+b^2)(-b^2+c^2)).$$ Let $(GPQ) = [u, v, w]$. Then $u$ is the power of $D$, which is $DQ \cdot DG$. We have $\frac{DQ}{DM} = 1 - x_q = \frac{2(b^2 - a^2)}{2c^2 + 2b^2 - a^2}$, and since $DM^2 = \frac{1}{2}(b^2 + c^2) - \frac{1}{4}a^2$ this gets $u = \frac{b^2 - a^2}{3}$. Similarly $v = \frac{c^2 - b^2}{3}$. Then to get $w$, we plug in $G = (1 : 1 : 1)$, and solving gets $w = \frac{2a^2 + b^2}{3}$, so the equation is $$-a^2yz-b^2zx-c^2xy + \left(\frac{-a^2+b^2}{3}x + \frac{-b^2+c^2}{3}y + \frac{2a^2+b^2}{3}z\right)(x+y+z).$$ Note that when plugging $R$ into the equation for $(GPQ)$, we get that $(2a^2+b^2)(-a^2+b^2)$ divides all terms. Factoring out $(2a^2+b^2)(-a^2+b^2)$ and simplifying shows that the expression evaluates to 0, and $R$ lies on $(GPQ)$.

The key is to introduce symmedians into our diagram. Suppose $K$ is the intersection of the $A$-median and $B$-symmedian and $L$ is the intersection of the $A$-symmedian and $B$-median. Then $K$ and $L$ are isogonal conjugates, so \[\measuredangle ACL = \measuredangle KCB.\] The overlay inversion centered at $A$ with radius $\sqrt{bc/2}$ tells us $\measuredangle ACL = \measuredangle AQF$, and a similar argument gives $\measuredangle KCB = \measuredangle BPD$. Equating our new angles, we get the desired. $\blacksquare$

Incredible. I have no words. Let $M$ be the intersection of the $B$-median with the $A$-symmedian and $N$ the intersection of the $A$-median with the $B$-symmedian. As $M$ and $N$ are isogonal conjugates, $\overline{CM}$ and $\overline{CN}$ are isogonal. But $\sqrt{bc/2}$ inversion swaps $\{G, M\}$ and $\{C, F\}$, hence $\measuredangle AQF = -\measuredangle MCA$. It follows symmetrically by isogonality that $\measuredangle AQF = \measuredangle BPD$, implying the result.