Full solution: First let $A_0,B_0,C_0$ be the reflections of $A, B, C$ to the line $IO$. By symmetry, $F\in A_0B_0$ and $E\in A_0C_0$ and $A_0B_0,A_0C_0$ are tangent to the incircle. Let $A_0X$ intersects the $(ABC)$ second time at $P.$ By DDIT to the degenerate quadrilateral $ABXC$ from the point $A_0$, we have $$(A_0B,A_0C),(A_0A,A_0X) \text{ and } (A_0B_0,A_0C_0) \text{ are reciprocal pairs }$$Hence, $D\in AP$. On the other hand, again by DDIT to the quadrilateral $FCABE$ from the point $A_0$, we have $$ (A_0B,A_0C),(A_0A,A_0A_2) \text{ and } (A_0B_0,A_0C_0) \text{ are reciprocal pairs }$$Hence, the points $P,A_2,X$ and $A_0$ are collinear. Now, let $R$ be the antipod of $A_0$ wrt $(ABC)$. Since the line $IO$ goes through the mid-points of both $A_0A$ and $A_0R$, $AR$ is parallel to the line $IO$. Let the miquel point of the lines $AB, BC, CA$ and $IO$ be $M$. Then , $$\angle MDF=\angle MBF=\angle MBA=\angle ARM \implies R\in MD.$$Now by pascal for the points $(P,A_0,R;M,AO\cap (ABC)=A',A)$, we get $A_2\in MA'$ since $PA\cap MR=D$. Which means $\angle AMA_2=90$ and finishes the problem.

Well this is kinda annoying that the exam had 3 geos which two has not so hard complex bash solution and third is nearly impossible. (The longest process which have achieved in the exam by contestants was guessing the concurrency point.) Cool problems though.

Let's complex bash this one too. Identify the incircle of $\triangle ABC$ with the unit circle, so that $$|x|=|y|=|z|=1$$$$a=\frac{2yz}{y+z}$$$$b=\frac{2xz}{x+z}$$$$c=\frac{2xy}{x+y}$$$$o=\frac{2xyz(x+y+z)}{(x+y)(x+z)(y+z)}$$We can quickly find the coordinate of $D$ by writing $$\overline{d} = \frac{d(xy+xz+yz)}{xyz(x+y+z)} = \frac{2x-d}{x^2}$$and so $$d = \frac{2xyz(x+y+z)}{x^2y+x^2z+2xyz+y^2z+yz^2}$$Similarly, $$e = \frac{2xyz(x+y+z)}{xy^2+y^2z+2xyz+x^2z+xz^2}$$$$f = \frac{2xyz(x+y+z)}{xz^2+yz^2+2xyz+x^2y+xy^2}$$Now we find the coordinate of $A_1$. Since $A_1$ lies on line $BE$, we have $$\frac{b-a_1}{b-e} \in \mathbb{R}$$Now $$b-e = \frac{2xz\left[(xy^2+y^2z+2xyz+x^2z+xz^2) - y(x+z)(x+y+z)\right]}{(x+z)(xy^2+y^2z+2xyz+x^2z+xz^2)} = \frac{2xz(x^2z+xz^2-x^2y-yz^2)}{(x+z)(xy^2+y^2z+2xyz+x^2z+xz^2)}$$So $$\frac{(xy^2+y^2z+2xyz+x^2z+xz^2)\left[2xz - a_1(x+z)\right]}{2xz(x^2z+xz^2-x^2y-yz^2)} = \frac{(xy^2+y^2z+2xyz+x^2z+xz^2)\left[2-\overline{a_1}(x+z)\right]}{2y(xy+yz-x^2-z^2)}$$$$y(xy+yz-x^2-z^2)\left[2xz - a_1(x+z)\right] = xz(x^2z+xz^2-x^2y-yz^2)\left[2-\overline{a_1}(x+z)\right]$$$$\overline{a_1}xz(x+z)(x^2z+xz^2-x^2y-yz^2) = a_1y(x+z)(xy+yz-x^2-z^2) - 2xz(x+z)(y^2-xz)$$$$\overline{a_1} = \frac{a_1y(xy+yz-x^2-z^2) - 2xz(y^2-xz)}{xz(x^2z+xz^2-x^2y-yz^2)}$$Similarly, since $A_1$ lies on line $CF$, we have $$\overline{a_1} = \frac{a_1z(xz+yz-x^2-y^2) - 2xy(z^2-xy)}{xy(x^2y+xy^2-x^2z-y^2z)}$$Setting these equal gives $$\frac{a_1y(xy+yz-x^2-z^2) - 2xz(y^2-xz)}{xz(x^2z+xz^2-x^2y-yz^2)} = \frac{a_1z(xz+yz-x^2-y^2) - 2xy(z^2-xy)}{xy(x^2y+xy^2-x^2z-y^2z)}$$Then solving for $a_1$ gives $$a_1 = \frac{2xyz\left[(y^2-xz)(x^2y+xy^2-x^2z-y^2z) - (z^2-xy)(x^2z+xz^2-x^2y-yz^2)\right]}{y^2(xy+yz-x^2-z^2)(x^2y+xy^2-x^2z-y^2z) - z^2(xz+yz-x^2-y^2)(x^2z+xz^2-x^2y-yz^2)}$$Factoring out $z-y$ from the numerator and denominator gives $$a_1 = \frac{2xyz(x^3y+x^3z-x^2y^2+x^2yz-x^2z^2-xy^3-xy^2z-xyz^2-xz^3+y^3z+y^2z^2+yz^3)}{x^4y^2+x^4z^2+x^3y^2z+x^3yz^2-x^2y^4-3x^2y^3z-3x^2yz^3-x^2z^4+xy^3z^2+xy^2z^3+y^4z^2+y^2z^4}$$Now we find the coordinate of $A_2$. First, $A_2$ lies on line $IO$, so $$\overline{a_2} = \frac{a_2(xy+xz+yz)}{xyz(x+y+z)}$$Also, $A_2$ lies on line $XA_1$. So we find the vector $$a_1-x = \frac{x(-x^4y^2-x^4z^2+x^3y^2z+x^3yz^2+x^2y^4+x^2y^3z+2x^2y^2z^2+x^2yz^3+x^2z^4-2xy^4z-3xy^3z^2-3xy^2z^3-2xyz^4+y^4z^2+2y^3z^3+y^2z^4)}{x^4y^2+x^4z^2+x^3y^2z+x^3yz^2-x^2y^4-3x^2y^3z-3x^2yz^3-x^2z^4+xy^3z^2+xy^2z^3+y^4z^2+y^2z^4}$$We factor the numerator to get $$a_1-x = -\frac{x(x-y)(x-z)(x+y+z)(xy^2+xz^2-y^2z-yz^2)}{x^4y^2+x^4z^2+x^3y^2z+x^3yz^2-x^2y^4-3x^2y^3z-3x^2yz^3-x^2z^4+xy^3z^2+xy^2z^3+y^4z^2+y^2z^4}$$Then $$\frac{a_1-x}{\overline{a_1}-\overline{x}} = -\frac{x^2(x+y+z)(xy^2+xz^2-y^2z-yz^2)}{(xy+xz+yz)(xy+xz-y^2-z^2)}$$and $$\frac{a_2-x}{\overline{a_2}-\overline{x}} = \frac{a_2-x}{\frac{a_2(xy+xz+yz)}{xyz(x+y+z)}-\frac1x} = \frac{yz(a_2-x)(x+y+z)}{a_2(xy+xz+yz) - yz(x+y+z)}$$Setting these equal gives $$-\frac{x^2(x+y+z)(xy^2+xz^2-y^2z-yz^2)}{(xy+xz+yz)(xy+xz-y^2-z^2)} = \frac{xyz(a_2-x)(x+y+z)}{a_2(xy+xz+yz) - yz(x+y+z)}$$We cancel out $x+y+z$ (nonzero since the triangle is scalene) and clear denominators to get $$-a_2x^2(xy+xz+yz)(xy^2+xz^2-y^2z-yz^2) + x^2yz(x+y+z)(xy^2+xz^2-y^2z-yz^2) = a_2xyz(xy+xz+yz)(xy+xz-y^2-z^2) - x^2yz(xy+xz+yz)(xy+xz-y^2-z^2)$$Solving for $a_2$ gives $$a_2 = \frac{xyz\left[(x+y+z)(xy^2+xz^2-y^2z-yz^2) + (xy+xz+yz)(xy+xz-y^2-z^2)\right]}{(xy+xz+yz)\left[x(xy^2+xz^2-y^2z-yz^2) + yz(xy+xz-y^2-z^2)\right]}$$This expands to give $$a_2 = \frac{xyz(2x^2y^2+2x^2yz+2x^2z^2-2y^3z-2y^2z^2-2yz^3)}{(xy+xz+yz)(x^2y^2+x^2z^2-y^3z-yz^3)}$$We factor out $x^2-yz$ (again, nonzero since the triangle is scalene) to get $$a_2 = \frac{2xyz(y^2+yz+z^2)}{(y^2+z^2)(xy+xz+yz)}$$Similarly, $$b_2 = \frac{2xyz(x^2+xz+z^2)}{(x^2+z^2)(xy+xz+yz)}$$$$c_2 = \frac{2xyz(x^2+xy+y^2)}{(x^2+y^2)(xy+xz+yz)}$$ Now for a complex number $t$, denote $$t^* = \frac{t(xy+xz+yz)}{2xyz} - 1$$Then $$a^* = \frac{yz}{x(y+z)}$$$$b^* = \frac{xz}{y(x+z)}$$$$c^* = \frac{xy}{z(x+y)}$$$$a_2^* = \frac{yz}{y^2+z^2}$$$$b_2^* = \frac{xz}{x^2+z^2}$$$$c_2^* = \frac{xy}{x^2+y^2}$$Now the equation of the circle with diameter $AA_2$ is $$\frac{a^*-t^*}{a_2^*-t^*} \in i\mathbb{R}$$$$\frac{(y^2+z^2)\left[t^*x(y+z) - yz\right]}{x(y+z)\left[t^*(y^2+z^2) - yz\right]} = -\frac{(y^2+z^2)\left[\overline{t}^*(y+z) - x\right]}{(y+z)\left[\overline{t}^*(y^2+z^2) - yz\right]}$$$$\left[t^*x(y+z) - yz\right]\left[\overline{t^*}(y^2+z^2) - yz\right] = -x\left[t^*(y^2+z^2)-yz\right]\left[\overline{t^*}(y+z)-x\right]$$$$t^*\overline{t^*}x(y+z)(y^2+z^2) - t^*xyz(y+z) - \overline{t^*}yz(y^2+z^2) + y^2z^2 = -t^*\overline{t^*}x(y+z)(y^2+z^2) + t^*x^2(y^2+z^2) + \overline{t^*}xyz(y+z) - x^2yz$$$$2t^*\overline{t^*}x(y+z)(y^2+z^2) - t^*x(xy^2+xz^2+y^2z+yz^2) - \overline{t^*}yz(xy+xz+y^2+z^2) + yz(x^2+yz)$$$$\overline{t^*} = \frac{t^*x(xy^2+xz^2+y^2z+yz^2) - yz(x^2+yz)}{2t^*x(y+z)(y^2+z^2) - yz(xy+xz+y^2+z^2)}$$The circles with diameters $BB_2$ and $CC_2$ are defined analogously. Intersecting them gives $$\frac{t^*y(x^2y+yz^2+x^2z+xz^2) - xz(y^2+xz)}{2t^*y(x+z)(x^2+z^2) - xz(xy+yz+x^2+z^2)} = \frac{t^*z(x^2z+y^2z+x^2y+xy^2) - xy(z^2+xy)}{2t^*z(x+y)(x^2+y^2) - xy(xz+yz+x^2+y^2)}$$Clearing denominators gives $$2(t^*)^2yz(x+y)(x^2+y^2)(x^2y+yz^2+x^2z+xz^2) - 2t^*xz^2(x+y)(x^2+y^2)(y^2+xz) - t^*xy^2(xz+yz+x^2+y^2)(x^2y+yz^2+x^2z+xz^2) + x^2yz(y^2+xz)(xz+yz+x^2+y^2)$$on one side, and on the other side is the same expression but with $y$ and $z$ interchanged. Subtracting and dividing out $y-z$ gives $$2(t^*)^2yz(x^3y^2+x^3z^2+x^2y^3+2x^2y^2z+2x^2yz^2+x^2z^3+2xy^2z^2+y^3z^2+y^2z^3)$$$$ + 2t^*x(x^4y^2+x^4yz+x^4z^2+x^3y^2z+x^3yz^2-xy^3z^2-xy^2z^3-y^4z^2-y^3z^3-y^2z^4)$$$$ - t^*x(x^4y^2+2x^4yz+x^4z^2+x^3y^2z+x^3yz^2+x^2y^4+3x^2y^3z+4x^2y^2z^2+3x^2yz^3+x^2z^4+xy^3z^2+xy^2z^3+y^4z^2+2y^3z^3+y^2z^4)$$$$ + x^2yz(-x^3+xyz+y^3+2y^2z+2yz^2+z^3) = 0$$Combining the coefficients of $t^*$ gives $$2(t^*)^2yz(x^3y^2+x^3z^2+x^2y^3+x^2z^3+y^3z^2+y^2z^3+2x^2y^2z+2x^2yz^2+2xy^2z^2)$$$$ - t^*x(-x^4y^2-x^4z^2-x^3y^2z-x^3yz^2+x^2y^4+3x^2y^3z+4x^2y^2z^2+3x^2yz^3+x^2z^4+3xy^3z^2+3xy^2z^3+3y^4z^2+4y^3z^3+3y^2z^4)$$$$ + x^2yz(-x+y+z)(x^2+y^2+z^2+xy+xz+yz) = 0$$We factor this as $$\left[t^*(x^3y^2+x^3z^2+x^2y^3+x^2z^3+y^3z^2+y^2z^3+2x^2y^2z+2x^2yz^2+2xy^2z^2) - xyz(x^2+y^2+z^2+xy+xz+yz)\right]$$$$\times \left[2t^*yz - x(-x+y+z)\right] = 0$$Thus we see that $$t^* = \frac{xyz(x^2+y^2+z^2+xy+xz+yz)}{x^3y^2+x^3z^2+x^2y^3+x^2z^3+y^3z^2+y^2z^3+2x^2y^2z+2x^2yz^2+2xy^2z^2}$$is a root of this quadratic, so it lies on both circles. Since its coordinate is symmetric in $x, y, z$, it lies on all three circles as desired. If we want, we can backsolve to see that the coordinate of the common point $T$ is $$\frac{2xyz(t^*+1)}{xy+xz+yz} = \frac{2xyz(x+y+z)(x^2y^2+x^2z^2+y^2z^2+x^2yz+xy^2z+xyz^2)}{(xy+xz+yz)(x^3y^2+x^3z^2+x^2y^3+x^2z^3+y^3z^2+y^2z^3+2x^2y^2z+2x^2yz^2+2xy^2z^2)}$$$\blacksquare$ (Yes, I did this entirely by hand. It took about three hours.)

AlperenINAN wrote: Well this is kinda annoying that the exam had 3 geos which two has not so hard complex bash solution and third is nearly impossible. (The longest process which have achieved in the exam by contestants was guessing the concurrency point.) Cool problems though. Your are right about the fifth problem which has an easier solution with complex and not so selective when it comes to denote the students with great geometry knowledge. But I think the first problem was pretty good for a tst p1 because it has a few and motivational steps and nearly all of the students solve it without using complex. And also tst p9 is supposed to be the hardest problem in the exam and it was asked to have the diffuculty level like an imo p6 geo. I think, when it comes to geometry, the only pitfall of the exam was p5

very hard and nice problem

Nice and cute problem!

Define $G$ as the miquel point of $\overline{ABCDEF}$ Define $A_0,B_0,C_0$ as the reflections of $A,B,C$ over the line $OI$. Clearly all of $A_0,B_0,C_0$ lie on the circumcircle. Denote by $A'_0$ the antipode of $A_0$ in the circumcircle Let $A'$ be the antipode of $A$ in $(ABC)$. Denote $Q$ as the intersection of $A_0X$ with $(ABC)$ We claim that $G$ is the desired point of concurrence of the three circles. Clearly, it suffices to prove that $\measuredangle BGB_2 = 90^{\circ}$ Claim 1: $A_0,X,A_1$ are collinear

Claim 2: $G,D,A'_0$ are collinear

Now, the problem follows from pascal on $QA_0A'_0GA'A$. $\blacksquare$

Let $M$ be the Miquel point of complete quadrilateral $ABCDEF$, which we claim is the desired intersection point. It suffices to show that $M$ lies on the circle with diameter $AA_2$. Let $l$ denote line $IO$, reflect $ABC$ about $l$ to $A'B'C'$, let $AD$ meet $(ABC)$ again at $G$, and let $A'B$ and $A'C$ meet $l$ at $S$ and $T$. Re-define $A_2$ as the intersection of $A'A_1$ with $l$. Claim: $A_2$ is the intersection of $BC'$ with $B'C$

Claim: $A'$, $X$, and $A_2$ are collinear

Claim: The circle with diameter $(AA_2)$, $(ABC)$, and $(AEF)$ concur

We claim that the desired concurrency point is the Miquel point of the four lines $\{AB, BC, CA, IO\}$. Let $P$, $Q$, $R$ be the reflections of $A$, $B$, $C$ across $IO$ respectively. Note that the line $QR$ is the reflection of line $BC$ across $IO$, so it must be tangent to the incircle. Doing this for all three sides, we see that the triangles $\triangle ABC$ and $\triangle PQR$ share the same incircle $\omega$. Moreover, the point $D$, $E$ and $F$ lie on $QR$, $PR$ and $PQ$ respectively. [asy][asy] import geometry; import olympiad; size(14cm); pair foot(pair P, pair A, pair B) { return foot(triangle(A, B, P).VC); } filldraw(unitcircle, cyan+white+white, blue); pair A = dir(120); pair B = dir(215); pair C = dir(325); point I = incenter(A, B, C); point O = (0,0); point D = intersectionpoint(line(I,O),line(B,C)); point E = intersectionpoint(line(I,O),line(A,C)); point F = intersectionpoint(line(I,O),line(B,A)); point X = foot(I, B, C); point A_1 = intersectionpoint(line(B, E), line(C, F)); transform reflect=reflect(line(I,O)); point P = reflect*A; point Q = reflect*B; point R = reflect*C; transform reflect1=reflect(perpendicular(O, line(P,X))); point P_1 = reflect1*P; point A_2 = intersectionpoint(line(P,X),line(I,O)); transform reflect2=reflect(perpendicular(O, line(P,O))); point P_2 = reflect2*P; transform reflect3=reflect(perpendicular(O, line(A,O))); point A_3 = reflect3*A; point S = intersectionpoint(line(A_3, A_2), line(D,P_2)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, N); dot("$O$", O, NE); dot("$D$", D, SW); dot("$E$", E, NE); dot("$F$", F, NW); dot("$X$", X, SE); dot("$P$", P, SE); dot("$Q$", Q, SW); dot("$R$", R, N); dot("$A_1$", A_1, SE); dot("$P_1$", P_1, NW); dot("$A_2$", A_2, N); dot("$P_2$", P_2, N); dot("$A_3$", A_3, SE); dot("$S$", S, W); draw(A -- B -- C -- cycle, red+0.2); draw(D -- B, red+0.2); draw(incircle(A, B, C), purple); draw(line(D, false, E), lightgreen+1); draw(line(P, false, Q, false), red); draw(line(P, false, R, false), red); draw(line(D, false, R, false), red); draw(line(B, false, E, false)); draw(line(C, false, F, false)); draw(segment(B, R)); draw(segment(C,Q)); draw(segment(A, A_3)); draw(segment(P, P_2)); draw(segment(D, A)); draw(segment(D,P_2)); draw(segment(P, P_1), heavygreen); draw(segment(S, A_3), dashed); draw(circumcircle(S, Q, F), opacity(0.05)+magenta); draw(S -- E -- F -- cycle, darkblue); draw(S -- B -- C -- cycle, darkblue); [/asy][/asy] Claim 1: Points $P$, $A_1$ and $X$ are collinear. Proof: By DDIT from $P$ to the degenerate quadrilateral $ABXC$, there's an involution swapping the pairs \[(PA, PX), (PB, PC), (PF, PE)\]where the last pair is due to the fact that $PF$ and $PE$ are tangent to $\omega$. Similarly, by DDIT from $P$ to the quadrilateral $BECF$ there's an involution swapping \[(PA_1, PA), (PB, PC), (PF, PE)\]The last two pairs of swaps are the same, which means that these involutions are actually the same. This implies that lines $PX$ and $PA_1$ are the same, which proves the claim. \qed Claim 2: $A_2 = BR \cap CQ$. Proof: Let $PX$ hit $(ABC)$ for the second time at $P_1$. If we project the involution from the previous claim onto $(ABC)$ we deduce that lines $QR$, $BC$ and $AP_1$ are concurrent. Then, as $IO$ is the perpendicular bisector of $AP$, we see that $A_2O$ is the external angle bisector of $\angle AA_2P_1$. Combined with the fact that $OA = OP_1$, we see that the points $A$, $P_1$, $A_2$ and $O$ are cyclic. Thus by PoP, \[DB\cdot DC = DP_1\cdot DA = DA_2\cdot DO\]so we deduce that $B$, $C$, $O$ and $A_2$ are cyclic. On the other since $BQRC$ is an isosceles trapezoid, we can easily show that $BR \cap CQ$ lies on line $IO$ and that it lies on circle $(BOC)$. This is enough to imply the claim. \qed Next up, let $P_2$ and $A_3$ be the reflection of $P$ and $A$ across $O$ respectively and redefine $S = DP_2 \cap (ABC) \ne P_2$ Claim 3: Points $S$, $A_2$ and $A_3$ are collinear. Proof: By Pascal on $SA_3AP_1PP_2$ the point $SA_3 \cap P_1P$ lies on line $IO$. But $P_1P \cap IO = A_2$ so this proves the claim. \qed The last claim shows that $\angle ASA_2 = ASA_3 = 90$ which means that it suffices to show that $S$ is the Miquel point of $\{AB, BC, CA, IO\}$. The following claim will show that $S$ is the center of the similarity sending $FE$ to $BC$. This will certainly give us our desired result. Claim 4: $S$ is the center of the similarity sending $FE$ to $BC$. Proof: We first show that $BFOP$ and $CEOP$ are cyclic. Indeed notice that $FO$ is the exterior angle bisector of $\angle BFP$ and combining this with the fact that $OB = OP$ we get that $BFOP$ is cyclic. Using a completely symmetric argument, we get that $CEOP$ is cyclic as well. Next, note that $P_2$ and $A_3$ are reflections of each other around $IO$. Thus, \[\angle QSA_2 = \angle QSA_3 = \angle P_2PB = \angle OPB = \angle OFB = \angle PFO = \angle QFA_2\]so $SQFA_2$ is cyclic. Finally, this implies that \[\angle SFE = \angle SFA_2 = \angle SQA_2 = \angle SQC = \angle SBC\]Using similar reasoning, we deduce that \[\angle SEF = \angle SCB\]so triangles $\triangle SFE$ and $\triangle SBC$ are similar, which proves the claim, and thus finishes the proof. \qed

25 mins solve, absolutely lovely problem. "MathLuis went Double Down and IT took Turkey TST 2024 P9's life down" Now reflect $A,B,C$ over $IO$ and let those points be $A',B',C'$ respectively, let $DA' \cap (ABC)=A_0$, also let $A'A_2 \cap (ABC)=A_3$ and $AO \cap (ABC)=A_4$. Now from DDIT spam on $ABXC, BFEC$ with perspector $A'$ and reflecting over $IO$ and projecting onto $(ABC)$ we get that $A_3, A_2, A_1, X, A'$ are colinear and $A,A_3,D$ are also colinear, so now let $Q$ the miquel point of complete quad $\{AB,AC,BC, IO \}$, then notice that $\angle QDE=\angle QBA=\angle QA_4O$ so $DQOA_4$ is cyclic but from $2\angle AA_0D=\angle AOA'=2\angle AOD$ we get that $AOA_0D$ is also cyclic so by radax we have that $Q,A_2,A_4$ are colinear but this means $\angle AQA_2=90$ so $Q$ is on $(AA_2)$, and analogously it also lies on $(BB_2), (CC_2)$ thus we are done .