American geo is back? Let $T=\overline{EF}\cap \overline{BC}$, and let $L$ and $L'$ denote the arc midpoints of minor and major arc $\widehat{BC}$, respectively. It is well-known that $(AEIFQ)$ is cyclic.

Claim 1: $QPDT$ is cyclic with diameter $DT$. Proof: Clearly $\angle TPD=90^\circ$, so $P\in (DT)$. Notice that $P$ and $Q$ swap under incircle inversion. Hence \[ \angle PQD = \angle IQD = \angle ID^*Q^* = \angle IDP \]since $D^*=D$, where $\bullet^*$ denotes the inverse of $\bullet$. Now $\angle IDP$ can be easily angle chased to equal $\tfrac{B-C}{2}$. And $\angle PTD = \angle(\overline{EF},\overline{BC})$ is also $\tfrac{B-C}{2}$, finishing. $\blacksquare$ Claim 2: $Q,D,L$ are collinear. Proof: Note that $T$ and $D$ are harmonic conjugates by Ceva-Menelaus. Hence $(TD;BC)=-1$. From the previous claim, $\angle TQD=90^\circ$. By the Right Angles and Angle Bisectors Lemma, combining these two implies $\overline{QD}$ bisects $\angle BQC$, i.e. $Q,D,L$ collinear. $\blacksquare$ Claim 3: $T,L,G$ are collinear and $G,D,L'$ are collinear. Proof: We claim $GQBC$ is a harmonic quadrilateral. Indeed, \[ (G,Q;B,C) \stackrel{A}{=} (P,\overline{AQ}\cap \overline{EF}; F,E) \stackrel{Q}{=} (I,A;F,E)=-1 \]where we projected finally onto $(AI)$; the final is $-1$ since $AFIE$ is a kite. Note that $T,Q,L'$ collinear since $\angle TQL=90^\circ$ and $\angle L'QL=90^\circ$. Therefore, \begin{align*} -1&=(G,Q;B,C)\stackrel{D}{=}(\overline{GD}\cap (ABC),L;B,C) \\ -1 &= (G,Q;B,C) \stackrel{T}{=} (\overline{GT}\cap (ABC), L';BC). \end{align*}Since $L'LBC$ is a kite, it is harmonic, so $L'\in \overline{GD}$ and $L\in \overline{GT}$ from the above. $\blacksquare$ Claim 4: $QMLT$ and $DMLG$ are cyclic. Proof: We have $QMLT$ is cyclic since $DL\cdot DQ = DB\cdot DC = DM\cdot DT$, where the final equality follows from $D$ and $T$ being harmonic conjugates. And $DMLG$ is cyclic since $\angle DGL=\angle L'GL=90^\circ$ and $\angle DML=90^\circ$. $\blacksquare$ Now, we finish the problem. We have \begin{align*} \angle MQL &= \angle MTL=90^\circ - \angle MLT = 90^\circ-\angle L'LG=\angle GL'L = \angle GQL, \end{align*}proving $\overline{QD}$ bisects $\angle MQG$. Also, \[ \angle QGD = \angle QGL'=\angle QLL'=\angle DLM = \angle DGM,\]proving $\overline{GD}$ bisects $\angle QGM$. These combined prove $D$ is the incenter of $\triangle GQM$.

[asy][asy] size(7cm); defaultpen(fontsize(9pt)); pen tfil=invisible; pair A,B,C,D,E,F,M,G,Q,P,I,X,Y,Z; A=dir(125); B=dir(220); C=dir(320); X=dir(270); Y=dir(172.5); Z=dir(42.5); I=intersectionpoints(A--X,C--Y)[0]; draw(A--B--C--cycle); dot("$I$",I); dot("$A$",A,dir(125)); dot("$B$",B,dir(220)); dot("$C$",C,dir(320)); filldraw(circumcircle(A,B,C),tfil); D=foot(I,B,C); E=foot(I,A,C); F=foot(I,A,B); draw(D--E--F--cycle); dot("$D$",D,dir(X)); dot("$E$",E,dir(Z)); dot("$F$",F,dir(Y)); filldraw(circumcircle(D,E,F),tfil); P=foot(D,E,F); dot("$P$",P,dir(70)); draw(D--P); G=intersectionpoints(A--(3P-2A),circumcircle(A,B,C))[1]; dot("$G$",G,dir(260)); draw(A--G); Q=intersectionpoints(I--(4P-3I),circumcircle(A,B,C))[0]; dot("$Q$",Q,dir(180)); draw(I--Q); draw(B--G--C--Q--cycle,dotted); dot("$M_A$",X,dir(270)); draw(Q--X,dotted); M=(B+C)/2; dot("$M$",M,dir(270)); [/asy][/asy] Claim 1: $DQ$ passes through $M_A$, the midpoint of arc $BC$ Proof: We invert about the incircle. Note that $A^*$ is the midpoint of $EF$, and similar for $B^*, C^*$. So, $(ABC)$ inverts to the $9$-point circle of $DEF$. As $P$ lies on this circle as well as line $IQ$, we must have that $Q,P$ are inverses wrt $(I)$. On the other hand, $I_A$, the A-excenter, maps to the foot from $I$ to $B^*C^*$, and as $M_A$ is the midpoint of $II_A$, it maps to the reflection of $I$ over $B^*C^*$. Hence, $B^*C^*$ is the perpendicular bisector of both segments $DP=D^*Q^*$ and $IM_A^*$. So, $D^*Q^*IM_A^*$ is an isosceles trapezoid $\implies (D^*Q^*IM_A^*)\implies\overline{DQM_A}$, as desired. Claim 2: $BQCG$ is harmonic Proof: Note that $Q^*C^*=B^*A^*$, so $Q^*C^*B^*A^*$ is an isosceles trapezoid. Hence, $$\frac{BQ}{CQ}=\frac{B^*Q^*\cdot\frac{BI\cdot IQ}{r^2}}{C^*Q^*\cdot\frac{CI\cdot IQ}{r^2}}=\frac{B^*Q^*}{C^*Q^*}\cdot\frac{BI}{CI}=\frac{A^*C^*}{A^*B^*}\cdot\frac{BI}{CI}=\frac{DF}{DE}\cdot\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}$$On the other hand, Ratio Lemma gives that $$\frac{BG}{GC}=\frac{\sin\angle BAG}{\sin\angle GAC}=\frac{FP}{PE}\cdot\frac{AE}{AF}=\frac{FP}{PE}=\frac{DF\cdot\sin\angle FDP}{DE\cdot\sin\angle PDE}=\frac{DF}{DE}\cdot\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}$$Hence, $\frac{BQ}{CQ}=\frac{BG}{GC}$, giving $BQCG$ is harmonic, as desired. Now, $QG$ is a symmedian in $\triangle BQC$, so $QG,QM$ are isogonal, and we have $$\angle BQG=\angle MQC\implies \angle BQM_A-\angle BQG=\angle M_AQC-\angle MQC\implies GQM_A=\angle M_AQM$$By Claim 1, $M_A$ lies on $DQ$, so this implies that $DQ$ is an angle bisector of $\angle BQM$. In addition, let $T$ be the intersection of the tangents to the circumcircle at $B,C$. We have $\angle BMC=90^o$, and $(Q,G;QG\cap BC,T)=-1$. Hence, by external angle bisector harmonic lemma, $MD$ bisects $\angle QMG$. As $D$ lies on two angle bisectors of $\triangle BQC$, it is the incenter, as desired.

[asy][asy] unitsize(2inch); pair A, B, C, D, E, F, Q, P, I, G, N, X, M, L, a, q; A = dir(120); B = dir(210); C = dir(330); N = dir(270); L = dir(45); I = extension(A, N, B, L); D = foot(I, B, C); E = foot(I, C, A); F = foot(I, A, B); X = -A; Q = foot(A, I, X); P = extension(I, Q, E, F); a = foot(I, E, F); M = .5B + .5C; G = 2*foot(0, A, P) - A; q = 2*foot(0, Q, M) - Q; draw(A--B--C--cycle); draw(circumcircle(A, B, C), dashed+mediumblue); draw(circumcircle(A, E, F), dotted+lightblue); draw(E--F); draw(Q--X, lightred); draw(Q--q, magenta); draw(G--M, magenta); draw(A--G); draw(Q--N, dashed+lightred); draw(Q--G, magenta); draw(A--N, lightred); draw(I--D); draw(A--X, lightred); draw(Q--a, magenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(20)); dot("$D$", D, dir(D)); dot("$E$", E, dir(40)); dot("$F$", F, dir(200)); dot("$X$", X, dir(X)); dot("$Q$", Q, dir(Q)); dot("$N$", N, dir(N)); dot("$P$", P, dir(P)); dot("$A'$", a, dir(60)); dot("$G$", G, dir(G)); dot("$Q'$", q, dir(q)); dot("$M$", M, dir(M)); dot("$O$", 0, dir(60)); [/asy][/asy] Inversion about the incircle sends the 9 point circle of $DEF$ to $(ABC)$, so $IP\cdot IQ = ID^2$. Then $\triangle IPD\sim\triangle IDQ$ and \[\angle IQD = \angle IDP = |\angle DFE - \angle DEF| = \left|\left(\tfrac{\angle A}{2} + \tfrac{\angle B}{2}\right) - (\tfrac{\angle A}{2} + \tfrac{\angle C}{2})\right| = |\tfrac{\angle B}{2} - \tfrac{\angle C}{2}|.\]Let the midpoint of $EF$ be $A'$; then $\angle IQA = \angle IA'P = 90^{\circ}$. Let $N$ be the midpoint of $\overarc{BC}$ not containing $A$, and let $X$ be the intersection of $QI$ with $(ABC)$ (also the antipode of $A$). We can show that $\angle NAX = |\tfrac{\angle B}{2} - \tfrac{\angle C}{2}|$ as well, so $Q, D, N$ are collinear. From $\angle IQA = 90^{\circ}$, we also have that $Q$ lies on $(AI)$ and is the center of spiral similarity from $EF$ to $BC$. This spiral similarity also sends $(QFE)$ to $(QBC)$, so $A'\mapsto M$ and $I\mapsto N$. Then, \[\angle NQM = \angle IQA' = \angle IAP = \angle NAG = \angle NQG,\]so $QD$ bisects $\angle GDM$. Now if $QM$ intersects $(ABC)$ again at $Q'$, then $NG = NQ'$. By symmetry, we have \[\angle GMB = \angle Q'MC = \angle QMB,\]so $MD$ bisects $\angle GMQ$, and we are done.

This literally took me 3 seconds to solve. Lol. The configuration is super well known, just recall those concurrencies and angle chase blindly. nukelauncher wrote: Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$, respectively. Let $P$ be the foot of the altitude from $D$ to $\overline{EF}$, and let $M$ be the midpoint of $\overline{BC}$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$, respectively. Show that the incenter of triangle $GQM$ coincides with $D$. Zack Chroman and Daniel Liu Let $T,K$ be the midpoints of Major Arc and Minor Arc $BC$ respectively. From EMMO Jr 2016 we get $G,D,T$ are collinear. Let $\overline{EF}\cap\overline{BC}=\{S\}$. It's well known that $K,D,Q$ are collinear. By Shooting Lemma $M,D,Q,T$ is concyclic also $M,D,G,K$ are concyclic. Hence, By Radical Axis Theorem on $\odot(ABC),\odot(MDQT),\odot(MDKG)$ we get that $S\in\overline{GK}$. Hence, $\measuredangle QMD=\measuredangle QTG=\measuredangle QKG=\measuredangle DMG$ also $\measuredangle GQD=\measuredangle DTM=\measuredangle DQM$. Hence, $D$ is the Incenter of $\Delta GQM$. $\blacksquare$

wait how do people type up their solutions and post it in less than a minute. also first post in HSO

Standard :/ [asy][asy] size(8cm); defaultpen(fontsize(9pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(130), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), P = foot(D,E,F), Q = foot(A,I,P), M = (B+C)/2, L = dir(90), N = dir(270), G = D+dir(L--D)*abs(B-D)*abs(C-D)/abs(L-D), G1 = M+dir(G--M)*abs(B-M)*abs(B-M)/abs(G-M), Q1 = M+dir(Q--M)*abs(B-M)*abs(B-M)/abs(Q-M); draw(A--B--C--A); draw(unitcircle); draw(Q--G1^^G--Q1, linewidth(1)); draw(G--L^^Q--N, dashed); draw(G--G1^^Q--Q1); draw(incircle(A,B,C), gray); draw(A--G^^E--F^^I--Q, gray); draw(A--N--L, dotted); draw(circumcircle(A,E,F)); dot("$A$", A, dir(130)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(215)); dot("$L$", L, dir(90)); dot("$N$", N, dir(270)); dot("$Q$", Q, dir(150)); dot("$G_1$", G1, dir(30)); dot("$G$", G, dir(240)); dot("$Q_1$", Q1, dir(300)); dot("$M$", M, dir(270)); dot("$E$", E, dir(30)); dot("$F$", F, dir(200)); dot("$P$", P, dir(50)); dot("$I$", I, dir(60)); [/asy][/asy] Let $N = \overline{AI} \cap (ABC)$ and $L = \overline{NM} \cap (ABC)$. Claim: $Q$, $D$, $N$ are collinear. Proof. Incircle inversion gives $Q \in (AEIF)$, so $Q$ is the center of the spiral similarity sending $\overline{BF}$ to $\overline{CE}$. This means $$\frac{QB}{QC} = \frac{FB}{EC} = \frac{DB}{DC}.$$$\square$ Claim: $G$, $D$, $L$ are collinear. Proof. By Ratio lemma, $$\frac{GB}{GC} = \frac{\sin \angle BAG}{\sin \angle CAG} = \frac{PF}{PE} \cdot \frac{AE}{AF} = \frac{QF}{QE} = \frac{QB}{QC} = \frac{DB}{DC}.$$$\square$ We also get $QBGC$ is harmonic, so if $G_1 = \overline{GM} \cap (ABC)$ and $Q_1 = \overline{QM} \cap (ABC)$, we must have $\overline{QG_1} \parallel \overline{GQ_1} \parallel \overline{BC}$ by symmedian properties. This is enough to imply $\overline{QD}$ bisects $\angle GQQ_1$ and $\overline{GD}$ bisects $\angle QGG_1$. $\blacksquare$

amar_04 wrote: This literally took me 3 seconds to solve. Lol. The configuration is super well known, just recall those concurrencies and angle chase blindly. Agree. It is somewhat too well-known, standard incenter structures. Very weird that it became a USA TST problem. To #16: I recalled incorrectly, sorry.

We have the following characterizations. Characterization of $Q$. $Q$ is the center of the spiral similarity taking $EF$ to $BC$. Define $Q'$ as the intersection of $(AEFI)$ and $(ABC)$, and note that $R'$ is the Miquel Point of $EFBC$ and thus is the center of the spiral similarity taking $EF$ to $BC$. Now, we invert about the incircle. Note that $(AEIF)\leftrightarrow EF$ and $(ABC)$ goes to the nine-point circle of $\triangle DEF$. Thus, we can see that $Q'\to S$, and thus $Q',S,I$ are collinear, as desired. We also have the following useful lemmas. Let $R$ be the other intersection of $DP$ with the incircle: Claim 1. $Q,D,M_A$ are collinear. Proof. Consider the above spiral similarity, and note $I\to M_A$ and $A\to M_{BC}$. Now, we can easily angle chase to get \[\angle RFE=\angle RDE=\angle PDE=90^\circ-\angle PED=90^\circ-\angle FED=\frac12\angle B=\angle IBC\]and similarly with $\angle REF$ gives that $\triangle RFE\sim\triangle IBC$. In particular, if the spiral similarity $f$ centered at $Q$ maps $F\mapsto B$ and $E\mapsto C$, then \[f(R)=I\qquad f(P)=D\qquad f(I)=M_A\]However, we know that $PD\parallel IM_A$ (both are perpendicular to $EF$). Since $R$ is the spiral centre taking $PI$ to $DM_A$ , hence it is also the centre of a spiral similarity $g$ which takes $PD$ to $IM_A$. So $g$ is a homothety for the two mentioned lines. Now, as we know $R,P,I$ are collinear, their images under the homothety are collinear, so thus $R,D,M_A$ are collinear. Claim 2. Define $X$ as the intersection of $EF$ and $BC$. Then, $(QPDGX)$ is cyclic and $GDM_{BC}$ are collinear. Proof. Notice that $(PXD)$ is the $P$-appolonius incircle of $\triangle PBC$, so taking a spiral similarity of $Q$ gives \[\frac{QB}{QC}=\frac{FB}{EC}=\frac{BP}{PC}\]which implies that $Q$ is also on the $S$-Appolonius incircle. Thus, $X,P,Q,D$ are concyclic. In addition, $DX$ is the diameter, so we have $DQ\perp QX$, As $DQ=DM_A\perp QM_{BC}$, we see that $X,Q,M_{BC}$ are collinear. Now, we can note that $GQBC$ is harmonic, as by projecting it through $A$ and $Q$, we get \[(GQ;BC)=(IA;FE)=-1\]This means that $Q$ also lies on this circle. In addition, by this logic, we can get that $BCM_AM_{BC}$ is harmonic, but also $(GD\cap(ABC)M_A;BC)=-1$, so thus $G,D,M_{BC}$ are collinear. Claim 3. $DGMM_A$ is cyclic. Proof. This is immediate as we remarked $DX$ was a diameter, and $X,G,L$ collinear. To finish, we can note that if we let $Q'$ be the intersection of $QM$ with $ABC$, we get $Q'G\parallel BC$, which implies \[\measuredangle GMD=\measuredangle Q'MC=\measuredangle QMD\]and \[\measuredangle QGD=\measuredangle QM_AM_{BC}=\measuredangle QM_{BC}M=\measuredangle DGM\]completing the proof.

Why can't the proposers think of any other configuration except this? I am sick and tired of seeing the same Incircle , same points, same concurrencies in almost each and every Oly Geo problem. Rickyminer wrote: standard incenter structures and a little mixtilinear related I can't find any Mixtillinear stuffs in this problem?

show some respect

jj_ca888 wrote: show some respect Sorry if my sentence came off that way. But it is really not legit to put such standard-ish configurations which literally every contestants are aware of and I am quite sure this problem is not original.

bestzack66 wrote: wait how do people type up their solutions and post it in less than a minute. also first post in HSO Practically everyone that took TST probably pre-typed up their latex sols

Abhaysingh2003 wrote: But it is really not legit to put such standard-ish configurations which literally every contestants are aware of i see the point of this, but i want to add that think another issue might come from the fact that some contestants might not be aware of the config and get put at a disadvantage? [this part is removed because i also now agree with the quoted post because this is just not-super-interesting when its standard ]

mira74 wrote: but i want to add that think another issue might come from the fact that some contestants might not be aware of the config and get put at a disadvantage? Well I think quite well trained contestants appear for this contest, so IMO almost all contestants are aware of this Config as this has appeared infinitely many times in past contests.

Solution. We start showing the following Claim 1. $QBGC$ is a harmonic quadrilateral. First, since $(F,D;B,C)\overset{P}{=}-1$ and $\angle DPE=90^\circ$, we infer that $EF$ is the external bisector of $\angle BPC$. It straightforward follows that $\bigtriangleup BFP\sim \bigtriangleup PEC$, thus $\frac{FP}{PE}=\frac{BD}{DC}$. Therefore, by defining $Q'$ as the center of spiral similarity taking $\overline{BC}$ to $\overline{FE}$ (which coincides with the second intersection of the circumcircles of $AEIF$ and $ABC$), we conclude that $$\dfrac{Q'F}{Q'E}=\dfrac{FB}{EC}=\dfrac{FP}{PE}$$then $Q'D$ is the internal bisector of $\angle BQ'C$ (1) and $Q',\ P$ and $I$ are collinear, thus $Q'\equiv Q$. Let $R=\overline{AG}\cap (AEIFQ), R\neq A$. Taking into account that $AQ$ is the external bisector of $\angle FQE$, notice that $$-1=(Q,R;F,E)\overset{A}{=}(Q,G;B,C)$$thus the claim is proved. $\square$ By Claim 1, we recognize that $QG$ is a symmedian of $\bigtriangleup BQC$. By (1), $QD$ is the internal bisector of $\angle GQM$. Moreover, $M$ is the center of spiral similarity mapping $\overline{QC}$ to $\overline{CG}$, thus $\angle QMD=\angle DMG$. The solution is complete. $\blacksquare$

Note $Q$ is the well-known miquel point of $BCEF$. Letting $AG$ intersect $(AEF)$ again at $G'$ and noting $Q$ is the spiral center between $GD$ and $G'P$, we see that $GD$ passes through the midpoint of $\widehat{BAC}$. Hence the $Q$-apollonius circle of $\triangle QBC$ contains $G$, so $QBGC$ is harmonic. Now note In $\triangle QBC$ since $QD, QG,QM$ are angle bisector, symmedian, median respectively, $QD$ bisects $\angle GQM$ Similarly, in $\triangle GBC$ since $GD,GQ,GM$ are angle bisector, symmedian, median respectively, $GQ$ bisects $\angle QGM$ as desired.

Here is the proof (projective): Note that $QD$ is the angle bisector of $\angle BQC$ (by Sharkydevil). Claim: $QBGC$ is harmonic. Proof: Notice that: $$(QG; BC) \stackrel{A}{=} (AQ \cap EF P; FE) \stackrel{Q}{=} (AI; FE) = -1$$ Hence, $QG$ is the $Q$-symmedian of $\triangle ABC$. Therefore, $QD$ is the angle bisector of $\angle GDM$. Since $QG$ is the $Q$-symmedian of $\triangle ABC$, $GQ$ is the $G$-symmedian of $\triangle GBC$ (by construction) and hence; $$ \angle MGC = \angle QGB = \angle QCB $$$$ \angle MCG = \angle BCG = \angle BQG = \angle MQC $$Therefore, $\angle DMG = \angle DMQ$. Hence, $D$ is the incenter of $\triangle GQM$.

Nice, $1$ hour solve. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -9.116487190844866, xmax = 9.574824765444056, ymin = -4.650102266591628, ymax = 9.384952007453306; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqzzcc = rgb(0,0.6,0.8); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqttcc = rgb(0,0.2,0.8); pen qqwwzz = rgb(0,0.4,0.6); /* draw figures */ draw((-4.29454,7.51148)--(-6.1394249100381275,-1.0693989277188678), linewidth(0.7) + ccqqqq); draw((-6.1394249100381275,-1.0693989277188678)--(4.972813358774656,-0.9621379406078273), linewidth(0.7) + ccqqqq); draw((4.972813358774656,-0.9621379406078273)--(-4.29454,7.51148), linewidth(0.7) + ccqqqq); draw(circle((-2.501697057638439,1.8984733499773532), 2.9326224600834285), linewidth(0.7) + qqzzcc); draw((-5.368802412571072,2.5148996449246095)--(-0.5227762609764418,4.062762372710783), linewidth(0.7) + yqqqyq); draw(circle((-0.6146504645159545,2.231541335442473), 6.435785833513612), linewidth(0.7) + qqttcc); draw(circle((-3.398118528819219,4.704976674988674), 2.9461894656137653), linewidth(0.7) + qqwwzz); draw((-6.290535093986112,5.265296923455273)--(-2.501697057638439,1.8984733499773532), linewidth(0.7) + yqqqyq); draw((-4.29454,7.51148)--(-2.9772932242371786,-3.7548815100297177), linewidth(0.7) + yqqqyq); draw((-3.622947896635701,1.7673783153314582)--(-6.290535093986112,5.265296923455273), linewidth(0.7) + ccqqqq); draw((-3.622947896635701,1.7673783153314582)--(-2.945789336773757,3.2888310088176964), linewidth(0.7) + ccqqqq); draw((-2.945789336773757,3.2888310088176964)--(-6.290535093986112,5.265296923455273), linewidth(0.7) + ccqqqq); draw(circle((-3.1358745437655826,2.462013857536252), 0.8483861076009478), linewidth(0.7) + dotted + yqqqyq); draw(circle((-4.032296014946362,5.268517182547576), 2.258241375090564), linewidth(0.7) + dotted + yqqqyq); /* dots and labels */ dot((-4.29454,7.51148),dotstyle); label("$A$", (-4.476799825808611,7.727920806267363), NE * labelscalefactor); dot((-6.1394249100381275,-1.0693989277188678),dotstyle); label("$B$", (-6.64866695594902,-1.61889142427904), NE * labelscalefactor); dot((4.972813358774656,-0.9621379406078273),dotstyle); label("$C$", (5.033981144623637,-1.420047680136727), NE * labelscalefactor); dot((-2.501697057638439,1.8984733499773532),linewidth(4pt) + dotstyle); label("$I$", (-2.5049326956682014,1.4477725537726407), NE * labelscalefactor); dot((-2.473391209656339,-1.0340125020525919),linewidth(4pt) + dotstyle); label("$D$", (-2.6546436317221612,-1.602032048302759), NE * labelscalefactor); dot((-0.5227762609764418,4.062762372710783),linewidth(4pt) + dotstyle); label("$E$", (-0.31765150929396574,4.115592787682009), NE * labelscalefactor); dot((-5.368802412571072,2.5148996449246095),linewidth(4pt) + dotstyle); label("$F$", (-5.952713851193581,2.2591397144249093), NE * labelscalefactor); dot((-3.770052029892727,3.0255543650951537),linewidth(4pt) + dotstyle); label("$P$", (-4.179401401322971,2.4242647066146603), NE * labelscalefactor); dot((-6.290535093986112,5.265296923455273),linewidth(4pt) + dotstyle); label("$Q$", (-6.913791948236806,5.192663068452871), NE * labelscalefactor); dot((-2.9772932242371786,-3.7548815100297177),linewidth(4pt) + dotstyle); label("$G$", (-3.1843154884056535,-4.369563218247705), NE * labelscalefactor); dot((-0.5833057756317359,-1.0157684341633475),linewidth(4pt) + dotstyle); label("$M$", (-0.7821983736896717,-1.51889142427904), NE * labelscalefactor); dot((-3.622947896635701,1.7673783153314582),linewidth(4pt) + dotstyle); label("$T$", (-4.0968389051790786,1.2317803696896247), NE * labelscalefactor); dot((-2.945789336773757,3.2888310088176964),linewidth(4pt) + dotstyle); label("$S$", (-3.019190496117867,2.6728193867925516), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Clearly $Q$ is the $A$-sharkydevil point. Then it is well known that $Q$ is the Miquel point of $BFEC$ and $\overline{QD}$ is the angle bisector of $\angle BQC$. Now define $T = \overline{AP} \cap (AEF)$. Let $S$ be the midpoint of $\overline{EF}$. Now consider the spiral similarity $\phi$ mapping $\overline{BC} \mapsto \overline{FE}$. Note that under $\phi$ we find, $\phi(M) = S$ as $S$ is the midpoint of $\overline{EF}$ $\phi(D) = P$ as $\angle DQB = \angle DQC$ and $\angle IQF = \angle IQE$. $\phi(G) = T$ as $\angle BQG = \angle BAG = \angle FAT = \angle FQT$. Then it suffices to show that $P$ is the incenter of $\triangle QST$. Claim: $AQPS$ and $SPTI$ are cyclic. Proof. Note that $\angle ISP = \angle IQA = 90$ and hence $\triangle IPS \sim \triangle IAQ$. Then from power of a point we find $AQPS$ cyclic. Similarly we may find that $SPTI$ is cyclic as $\angle PSI = \angle ATI = 90$. $\square$ Now to conclude we do a little angle chasing. Claim: $\angle TQS$ is bisected by $\overline{IQ}$. Proof. Note $\angle TQI = \angle TAI = \angle PAS = \angle PQS$. $\square$ Claim: $\angle QST$ is bisected by $\overline{SP}$. Proof. Note $\angle QSP = \angle QAT = \angle QIT = \angle PST$. $\square$ Then $P$ is the incenter of $\triangle QST$ then our claim follows. $\blacksquare$ Remark: I feel like my solution presents a new main idea, besides the usage of the Sharkydevil point, in that the spiral similarity may be abused to turn the problem into simpler triangle geo, rather than the complicated mess of the original problem. For me personally, this was the biggest leap in the problem, and allowed it to become much simpler.

We note the well-known properties of $Q$, the $A$-Sharkydevil point. As $D$ is the foot of the angle bisector of $\angle BQC$, we can use symmedian properties to conclude if we prove the following: Claim: $QBGC$ is a harmonic quadrilateral. Let $X = QE \cap (ABC)$, $Y = QF \cap (ABC)$, and the antipode of $A$ be $A'$. Then \[(QG;BC) \overset{A}{=} (AQ \cap EF, P; FE) \overset{Q}{=} (AA';YX),\] so it suffices to show that arcs $AX$ and $AY$ have equal measure. Assuming WLOG that $AB<AC$, spiral similarity at $Q$ tells us \begin{align*} \angle EQC = \angle FQB = \angle (EF,BC) &\implies \overarc{CX} = \overarc{BY} = \angle B - \angle C \\ &\implies \overarc{AX} = \overarc{AY} = \angle B + \angle C. \quad \blacksquare \end{align*}

PoP at $I$, we get that $Q$ lies on $(AEF)$. Thus $Q$ is the Miquel point of $BFEC$, so $\frac{BQ}{CQ} = \frac{BF}{CE} = \frac{BD}{CD}$, so $QD$ is bisector of $\angle BQC$. Thus in order to prove $\angle GQD = \angle MQD$, we only need to prove $\angle BQG = \angle CQM$, or equivalently $(B, C; Q, G) = -1$. Note that $(B, C; Q, G) \overset{A}{=} (F, E; AQ \cap EF, P) \overset{Q}{=} (F, E; A, I) = -1$ (Note that $Q, P, I$ are collinear) $QG$ is symmedian of $BQC$, which means $\angle GQD = \angle MQD$. Since $GQ$ is symmedian of $BGC$, we have $\angle DMG = \angle BMG = \angle GCQ = \angle BMQ = \angle DMQ$. Therefore $D$ is the incenter of $GQM$, as required. $\blacksquare$

Point $Q$ is the $A-$Sharkydevil point, which is the second intersection of $(AEF)$ and $(ABC)$. It is well-known that $QD$ is the bisector of $\angle BQC$, and hence $\frac{BQ}{QC}=\frac{BD}{DC}$. We claim that this ratio equals $\frac{BD}{DC}$. Note that $\frac{BG}{GC}=\frac{\sin(\angle BAG)}{\sin(\angle GAC)}$.

gives $\frac{FP}{PE}=\frac{BD}{DC}$, which means that $\frac{BG}{GC}=\frac{BD}{DC}$, hence $GD$ is the angle bisector of $\angle BGC$. Now, we have $\frac{BG}{GC}=\frac{BD}{DC}=\frac{BQ}{QC}$, which means that $BG\times QC=BQ\times GC$. Hence, $(BQCG)$ is a harmonic bundle. So, $QG$ is the $Q-$symmedian of $\triangle QBC$, and $QM$ is the median, while $QD$ is the angle bisector. So, $\angle GQD=\angle DQM$ and $QD$ is the angle bisector of $\angle GQM$. Similarly, $GM$ is the $G-$median of $\triangle GBC$, $GQ$ is the symmedian and $GD$ is the angle bisector. So, $GD$ is the angle bisector of $\angle MGQ$. Hence, $D$ is the incenter of $\triangle GMQ$, as required. $\blacksquare$

I think It could be more shorter

Hategeometry wrote: I think It could be shorter by use the appollonian cricle

Interesting. Note that $Q$ is the $A$-Sharkydevil point and thus three things: (1) $P$ and $Q$ are inverses with respect to the incircle $\omega$, (2) $A,E,F,Q$ are concyclic, and (3) $QD$ bisects $\angle BQC.$ Now we state our main claim: Main Claim: $(B,C;Q,G) = -1.$ If we prove this claim, then this implies that $QG$ is a symmedian in $\triangle BQC,$ so that $PD$ bisects $\angle GQM.$ By the Angle-Bisector Theorem and harmonic quadrilateral condition, we see that $GD$ bisects $\angle BGC$ and $GQ$ is a symmedian in $\triangle BGC,$ so $GD$ bisects $\angle QGM$, which would solve the problem. All that is left to do is prove our claim. Let $Z$ denote the $D$-queue point of $\triangle DEF.$ Lemma: $A,Q,Z,D$ are concyclic. Proof: Invert with respect to $\omega.$ $A$ gets sent to the midpoint $N$ of $EF$ while $Q$ gets sent to $P,$ so it suffices to show that $N,P,Z,D$ are concyclic, or equivalently that $\angle DZN = 90^\circ,$ which is a well-known property of the $D$-queue point. Now, applying the radical center theorem on $(AQFIE), \omega,$ and $(AQZD),$ we get that $DZ$ and $AQ$ meet on $EF,$ say at point $X.$ However, since $X$ lies on $DZ,$ it is known that $(F,E;X,P) = -1.$ Therefore, $$(F,E;X,P) \stackrel{A}{=} (B,C;Q,G) = -1,$$and we conclude our proof.

Claim I: $Q=(ABC)\cap (AEF)$ is the Miquel Point and Sharky-Devil Point of $FBCE$. Proof: Inverting around the incircle will send ABC to $N_9$ circle, it will also send $A,B,C$ to midpoints of $EF,FD,DE$ and $P\rightarrow Q$ w.r.t to the incircle. Claim II: $QDM_A$ are collinear. Also $QD$ bisects $\angle BQC$. Proof: Properties of Sharky-Devil Point. Claim III:$CQGB$ is harmonic. Proof: $$\frac{QC}{QB}=\frac{FP}{FE}=\frac{\sin(PAF)}{\sin(PAE)}=\frac{GC}{BC}$$Claim IV: $GQ$ is symmedian of $\triangle GBC$ and $BM$ bisects $\angle QMG$. Proof:$GQ$ is symmedian so $\displaystyle \angle MQD=\angle DQC -\angle MQC=\angle DQB-\angle GQD=\angle GQD$. $BM$ bisects $\angle QMG$ because of EGMO lemma $4.26.$ $\blacksquare$

All points labelled as in the diagram. Notice that two projections from $A$ and $Q$ send $(G,Q;BC)$ to $(IA;EF)$ which is harmonic. Since $(TD;BC)=-1$, Projecting from $L$ gives that $G$ lies on $LD$. It is also well known that $D$ lies on $QM_a$ so $(PQTD)$ is cyclic with diamter $TD$. Notice that $\sqrt{bc}$ inversion in $\Delta LBC$ sends $D$ to $G$ and $Q$ to $T$ implying $(DGQT)$ is cyclic due to Power of point. Hence $(PQTDG)$ is cyclic and due to right angles, $(DMM_aG)$ and $(QMM_aT)$ are cyclic and the required follows by angle chasing.

Let $N,R$ be the midpoints of arcs $BC$ not containing $A$ and containing $A$ respectively and $A'$ be the antipode of $A$ on $(ABC)$. $AQ\cap BC=T$. $Q$ is $A-$Sharky-Devil point. Claim: $R,D,G$ are collinear. Proof: Note that \[(QR,QD;QB,QC)=(EF\cap BC,D;B,C)=-1\]yields $Q,D,N$ are collinear. \[\measuredangle ANR=\frac{\measuredangle B-\measuredangle C}{2}=\frac{\measuredangle A}{2}-(90-\measuredangle B)=\measuredangle A'AN\]Hence $A'R\parallel AN$. Since $\measuredangle AQI=90=\measuredangle AQA'$ we conclude that $Q,P,I,A'$ are collinear. By Pascal at $AGRA'QN, \ P,GR\cap QN,AN\cap A'R$ are collinear. Since $AN\cap A'R=AN_{\infty},$ the line passes through $P,GR\cap QN$ must be parallel to $AN$ which is equavilent that line to be perpendicular to $EF$. Since $DP\perp EF,\ D$ is on this line. $D\in QN$ results in the collinearity of $R,D,G$.$\square$ Claim: $D$ is the incenter of $\triangle GMQ$. Proof: Let $K$ be the midpoint of $RD$. $GN\cap BC=S$. $K$ is the circumcenter of $(AQDM)$. Since $SD\perp RN$ and $RD\perp SN, \ D$ is the orthocenter of $RSN$. Thus, $S,Q,R$ are collinear. Nine point circle of $RSN$ passes through $K,Q,G,M$ hence $K\in (GQM)$. $K$ is the midpoint of the arc $QM$ on $(GQM)$ and $KQ=KM=KD$ so $D$ is the incenter of $GMQ$ as desired.$\blacksquare$

First, we will show that $\angle DQG=\angle DQM$. This is equivalent to $\angle BQG=\angle CQM$ since $QD$ bisects $\angle BQC$. But this is equivalent to $QG$ being the $Q$-symmedian of triangle $QBC$, so we want $(QG;BC)=-1$. Projecting through $A$ onto $EF$, we want to prove $(AQ\cap EF,P;F,E)=-1$. Let $N$ be the midpoint of arc $BAC$, and let $L$ be the midpoint of the other arc $BC$. Now note that since $AF=AE$ and $NB=NC$, the spiral similarity centered at $Q$ taking $FE$ to $BC$ also takes $A$ to $N$, so it takes $QA$ to $QN$. Also, \[\angle FQP=\angle FQI=\angle FAI=\frac12\angle A=\angle BQD,\]so the spiral similarity also takes $P$ to $D$. Taking the desired cross ratio through the spiral similarity, we want $(NQ\cap BC,D;B,C)=-1$. Projecting through $Q$ onto the circumcircle, we just want $(N,L;B,C)=-1$, which is true. Now let's prove $\angle QMD=\angle GMD$. Let $Q'$ be the reflection of $Q$ over the perpendicular bisector of $BC$; then we want to show that $Q',M,G$ are collinear. But note that \[-1=(B,C;M,\infty_{BC})\stackrel{Q'}{=}(B,C;Q'M\cap (ABC),Q)\Longrightarrow Q'M\cap (ABC)=G,\]as desired. $\blacksquare$

Let $R$ denotes the intersection of ray $EF$ and $BC$ and $M_A,M_B,M_C$ denote the midpoints of sides $EF,DF,DE$, respectively. Also, define $L$ and $N$ as the midpoints of the major and minor arc $BC$ in $(ABC)$, respectively. Claim 1. $Q$ is the miquel point of complete quadrilateral $BCEF$. Proof. It's equivalent to show that $Q$ lies on $(AEIF)$ or $\angle{ADI}=90^\circ$. Let $Q'$ be the second intersection of $(AEIF)$ and $(ABC)$. Perform inversion w.r.t the incircle of $\triangle{ABC}$, denote it as $\varphi$. Note that $BC,CA,AB$ are tangents to the incircle at $P,Q,R$, respectively. This implies that $A,B,C$ are the poles of $EF,DE,DF$, respectively i.e. the images of $M_A,M_B,M_C$, respectively. This means that $\varphi:(ABC)\leftrightarrow(M_AM_BM_C)$ a.k.a. $(ABC)$ is inverted to the nine-point circle of $\triangle{DEF}$. Note that $\varphi:(AEIF)\leftrightarrow$ $EF$ since that $(AEIF)$ passes through the center of the inversion circle, which is the incircle of $\triangle{ABC}$ centered at $I$. Now simply see that $Q'=(AEIF)\cap(ABC)$ and $P=EF\cap (M_AM_BM_C)$, so $Q'$ is the image of $P$ under this inversion, meaning that $I,P,Q'$ are collinear. In other words, $Q'=IP\cap (ABC)$, but $Q$ is also defined in a similar way, so $Q'=Q$. From here, it follows that $Q$ is the miquel point of complete quadrilateral $BCEF$. $\square$ Claim 2. $Q,D,N$ are collinear. Proof. Define $\vartheta$ as the inversion w.r.t. $(BIC)$. Clearly $N$ is the center of this circle. It's easy to prove that $LB$ and $LC$ are tangents to this circle at points $B,C$, respectively. This implies that $L$ is the image of $M$ under this inversion. Let $AN$ cut $BC$ at $K$. Easy to see that $ADML$ is cyclic, so we have $NK\cdot NA=NM\cdot NL=(r_{(BIC)})^2$, so $A$ is mapped to $K$ by $\vartheta$. Point $I$ stays under this inversion. So $\vartheta:(AEIFQ)\leftrightarrow(IK)$ where $(IK)$ represents the circle with diameter $K$ that's clearly pass through $D$. Note that $\vartheta:(ABC)\leftrightarrow BC$, so we have $\vartheta:(AEIFQ)\cap(ABC)\leftrightarrow(IK)\cap BC$ or equivalently $\vartheta:Q\leftrightarrow D$, meaning that $N,D,Q$ are collinear, as wanted. $\square$ Claim 3. $L,Q,R$ are collinear. Proof. Let $R'$ be the intersection point of $LQ$ and $BC$. Since that $AD$ is the angle bisector of $\angle{BAC}$ and $AD,BE,CF$ are concurrent at the Gergonne point of $\triangle{ABC}$, thus we have $(R,D;B,C)=-1$. But $BNCL$ is a harmonic quadrilateral (since it's a cyclic kite), so $(R',D;B,C)\overset{Q}{=}(L,N;B,C)=-1$. This forces $R'=R$, so the conclusion follows from it. $\square$ Claim 4. $GQBC$ is a harmonic quadrilateral. Proof. Observe that \[(Q,G;B,C)\overset{A}{=}(AQ\cap EF,P;F,E)\overset{Q}{=}(A,I;F,E)=-1\]The last cross-ratio comes from the fact that $AEIF$ is a cyclic kite and hence it's also a harmonic quadrilateral. Thus, $GQBC$ is a harmonic quadrilateral. $\square$ Claim 5. $N,G,R$ are collinear. Proof. Once again, let $R'$ be the intersection of $NG$ with $BC$. Notice that $(R',D;B,C)\overset{N}{=}(G,Q;B,C)=-1$. Recall that we also have $(R,D;B,C)=-1$, so $R'=R$. This proves the claim. $\square$ Claim 6. $G,D,L$ are collinear. There are two proofs that we may present here. Proof 1 (Quick Projective). Let $L'$ be the intersection of $GD$ and $(ABC)$ with $L'\ne G$. Next, by noticing that $(N,L';B,C)\overset{G}{=}(R,D;B,C)=-1$, we can conclude that $L'=L$ since that $BNLC$ is a cyclic kite (which is once again, a harmonic quadrilateral). So $G,D,L$ are collinear. $\square$ Proof 2 (Synthetic). Before that, we'll prove this subclaim first. Subclaim. $R,Q,P,D,G$ are concylic. Note that since $Q$ is the miquel point of complete quadrilateral $(BCEF)$ and $R=BC\cap EF$, then $RQFB$ is cyclic. It's very easy to see that $\angle{RPD}=\angle{RQD}=90^\circ$, so $RQPD$ is cyclic. So now it suffices to prove that $RQPG$ is cyclic. To do this, observe that $\angle{PGQ}=\angle{AGQ}=\angle{ABQ}=\angle{FBQ}=\angle{FRQ}=\angle{PRQ}$, so the result follows from it. $\circ$ By Subclaim., we may have $\angle{DGN}=180^\circ-\angle{DGR}=\angle{DPR}=90^\circ$. But it's obvious that $\angle{LGN}=90^\circ$, so $G,D,L$ are collinear. $\square$ After all, we may conclude that $QG$ is a $Q$-symmedian of $\triangle{BQC}$. Since that $QM$ is the $Q$-median of $\triangle{BQC}$, hence both lines are isogonal w.r.t. $\angle{BQC}$, so $\angle{GQD}=\angle{MQD}\Longleftrightarrow$ $\angle{QD}$ is the angle bisector of $\angle{GQM}$. Similarly, $GQ$ and $GM$ are isogonal w.r.t. $\angle{BQC}$ since they are $Q$-symmedian and $Q$-median of $\triangle{BGC}$, respectively. And so we can have $\angle{QGD}=\angle{MGD}\Longleftrightarrow$ $GD$ is the angle bisector of $\angle{QGM}$. Therefore, we have $D$ to be the incenter of $\triangle{GQM}$, as desired. $\blacksquare$ Q.E.D. Remark. The main idea to prove the first two claims at the beginning of the solution is to notice that $I$ and $N$ are the centers of the incircle of $\triangle{ABC}$ and $(BIC)$, respectively. This motivates us to invert w.r.t. those circles, and after some computation we obtained that the desired collinearities have polar relationships (idk how to say it haha).

great problem!

Just did this as first problem in 2025. Killed it in approximately 3 minutes. My solution: Notice that $Q$ is the Sharky-Devil point. Let $N_1$ and $N_2$ be the midpoints of long and short arc $BC$ respectively and $A'$ be the $A$-antipode. Well-known Sharky-Devil things and Pascal on $AGN_1A'QN_2$ gives that $Q, D, N_1$ are collinear. Obviously $\angle DGN_2=90^\circ=\angle DMN_2$ which implies $DGN_2M$ cyclic. Now angle chase $$\angle MGD=\angle MN_2D=\angle N_1N_2Q=\angle QGD$$so $GD$ is angle bisector of $\angle MGQ$. Also $$\angle QDM=180^\circ-\angle N_2DM=90^\circ+\angle DN_2M=90^\circ+\angle DGM$$Combining these 2 facts implies $D$ is incenter of $\triangle GMQ$ and we are done.