I was really really rude Sorry @Vickmath

VicKmath7 wrote: Yes, of course I know, but I believe that everyone who knows what IGO is will understand what A means in this case. Anyways, I will correct it. Yes but even I got confused at first sight,it's reccomendable to write the full form cuz there are a lot of people who don't know abt IGO

Not the best solution, but I guess it works. Ignore config issues (I don't think there are any but who knows).

Let $O,H,G$ denote the circumcenter, orthocenter, and centroid of $\triangle IKL$, respectively. We will show that $A$, $O$, $H$ are collinear $A$, $G$, $D$ are collinear which imply the conclusion.

Additional fact: $\overline{KL}$, $\overline{OX}$, $\overline{BC}$ seem to concur, although I didn't use/prove it.

Let $M,N$ denote the reflections of $A$ with respect to $BI,CI$ respectively, and let $G,H$ be the centroid and orthocenter of $\triangle AMN$. First, notice that by symmetry $IM=IA=IN$, so $I$ is the circumcenter of $\triangle AMN$ and $D$ is the midpoint of $MN$.

to know that $A$ lies on the Euler line of $\triangle IKL$. Since $G \in AD$, we have that $AD$ is the Euler line of $\triangle IKL$. $\blacksquare$

Here is my simple proof using barycentric coordinates.

Here's a solution me and my friends found at our 3 A.M Claim 1.In $\triangle ABC$ ,let $P$ be a point such that $\angle PBA=\angle PCA=180-A$. Then $P$ is on the Euler line of $\triangle ABC$. Proof.It's well-known that the locus of points $X$ such that $\angle XBA=\angle XCA$ is a hyperbola passing through $A,B,C,H,A'$ where $H$ is the orthocenter of $ABC$ and $A'$ is the antipod of $A$ in $(ABC)$ . So $P$ is on this hyperbola. Also, by the definition of $P$, we have $PC \cap AB$ lies on the perpendicular bisector of $AC$ . Now , applying Pascal on the hyperbola $AA'CPHBA$ we'll get that a line through $PC \cap AB$ and $HP \cap AA'$ is perpendicular to $AC$ and since $PC \cap AB$ lies on the perpendicular bisector of $AC$, $HP \cap AA'$ is in fact the circumcenter of $(ABC)$ . So $H,O,P$ are collinear and we're done. Claim2 .In $\triangle ABC$ with incenter $I$ , If a circle with center $I$ intersects $AB,BC,CA$ at $X,{Y,Z},T$ , then $XY,ZT,AD$ are concurrent. Proof. Let the lines through $D$parallel to $AB,AC$ intersects $XY,ZT$ at $E,F$ respectively. Since $CT=CZ$ and $DF||AC$ , we have $DZ=DF$, and similarly , $DE=DF$ . On the other hand , since $IY=IZ$ ,it follows that $DY=DZ$ . So $DE=DF$ and $\triangle DEF$ is an isosceles triangle. Also , with a little angle chasing , $AXY$ is an isosceles to , so since $DE||AB$ and $DF||AC$ , $\triangle AXY$ and $\triangle DEF$ are homothetic. the conclusion follows. Back to the problem. By Claim 1 , Since $\angle AKI = \angle ALI = 90 - \angle A /2 = 180 - \angle KIL$ , $A$ lies on the Euler line of $\triangle KIL$. Now , let the perpendicular bisector of $AI$ intersects $AC,AB$ at $M,N$ respectively. Since $MI=MA$ and $\angle AMI = 2\angle ALI$ , $M$ is the circumcenter of $\triangle ALI$. Similarly , $N$ is the circumcenter of $\triangle AKI$ . Now , let $M',N'$ be the reflection of $M,N$ to $IL,IK$ respectively . Since $MI=MK$ , $M'I = M'L$ and similar for $N$ . But $M',N'$ lie on $BC$ . So since $I$ is the center of $(MNM'N')$ , and since $MM',NN'$ are the perpendicular bisectors of $IL,IK$ respectively , by claim 2 , the circumcenter of $\triangle IKL$ lies on $AD$. So $AD$ is the Euler line of $\triangle IKL$ and we're done.

Proposed by Le Viet An -Vietnam

It seems that this one is not hard to bash.

We have that $APB$ is opositive similar to $CAB,$ hence ${KB}{/IB} = {AB}{/BC}.$ Hence, if we define $X$ as the reflection of $A$ over line $BI,$ we will have $KX \parallel IC.$ Similarly, $Y$ is the reflection of $A$ with respect to $CI,$ and $L$ is the intersection of $CI$ and the parallel of $BI$ through $Y.$ We are going to take this redefinition later on. [asy][asy] import olympiad; import graph; import geometry; size(250); pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt; A = dir(-240.54); B = dir(-170.37); C = dir(-10.43); O = (0,0); I = incenter(A,B,C); D = foot(I,B,C); X= 2*foot(A,B,I) - A; Y = 2*foot(A,C,I) - A; Bt= B*0.5 + I*0.5; Mt = extension(I,Y,C,Bt); L =extension(B,Mt,C,I); Ct= C*0.5 + I*0.5; Nt = extension(I,X,B,Ct); K = extension(C,Nt,B,I); dot("$K$",K,dir(120)); dot("$L$",L, dir(60)); dot("$Y$", Y, dir(-90)); dot("$X$", X, dir(X)); dot("$B$",B, dir(-90)); dot("$C$",C, dir(-90)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); draw(circle(I,distance(I,D)),dotted + green); draw(A--B--C--cycle,gray); draw(B--I,blue); draw(C--I,blue); draw(K--X, dotted+magenta); draw(Y--L,dotted+magenta); dot("$A$",A, dir(A)); [/asy][/asy] Note now, that on triangle $\triangle AXY,$ we have that $BI$ and $CI$ are the perpendicular bissectors of $AX$ and $AY,$ respectively. Furthermore, $YL$ and $XK$ are altitudes of $\triangle AXY.$ So the new problem reads as follows Let $\triangle AXY$ be an triangle with circumcenter $I$. Point $K$ is the intersection of the $X$-altitude with the perpendicular bissector of $AX;$ similarly point $L$ is the intersection of the $Y$-altitude with the perpendicular bissector of $AY.$ If $D$ is the midpoint of side $XY,$ show that $AD$ is the euler line of thriangle $IKL.$ So let $H$ be the orthocenter of $\triangle AXY.$ Also let $M$ and $N$ be the midpoints of sides $AX$ and $AY,$ respectively, and let $E$ and $F$ be the foots of $X$ and $Y$ respectively. Also let $T$ be the orthocenter of $IKL$ [asy][asy] import olympiad; import graph; import geometry; size(250); pair A,B,C,O,I,D,X,Y,K,L,H,I,Bt,Mt,Ct,Nt; A = dir(-240.54); B = dir(-170.37); C = dir(-10.43); O = (0,0); I = incenter(A,B,C); D = foot(I,B,C); X= 2*foot(A,B,I) - A; Y = 2*foot(A,C,I) - A; Bt= B*0.5 + I*0.5; Mt = extension(I,Y,C,Bt); L =extension(B,Mt,C,I); Ct= C*0.5 + I*0.5; Nt = extension(I,X,B,Ct); K = extension(C,Nt,B,I); pair M,N,E,F,H,T; M = A*0.5 + X*0.5; N = A*0.5 + Y*0.5; E = foot(X,A,Y); F = foot(Y,A,X); H = extension(E,X,F,Y); T = extension(A,D,K,A+K-Y); draw(A--D, dotted); draw(K--T--L,magenta); draw(circumcircle(A,F,N), dotted + green); draw(circumcircle(A,E,M),dotted + blue); draw(A--X--Y--cycle,gray); draw(M--I--K--cycle,orange); draw(N--I--L--cycle,orange); draw(X--E,red); draw(F--Y,red); dot("$A$", A, dir(A)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$D$",D, dir(-90)); dot("$M$",M, dir(30)); dot("$N$", N, dir(N)); dot("$E$",E, dir(E)); dot("$F$",F, dir(-10)); dot("$I$",I, dir(90)); dot("$K$",K,dir(260)); dot("$L$",L,-dir(L)); dot("$H$",H,dir(-90)); dot("$T$",T, dir(45)); [/asy][/asy] First, note that the centroid of $\triangle IKL$ lies on $AD.$ Indeed, by definition $IKHL$ is an parallelogram, so the centroid of $\triangle IKL$ divides the segment $IH$ in the ratio $1{/3};$ hence the centroid of $IKL$ coincides with the centroid of $AXY$ which indeed lies on the $A$-median, $AD$. Now it is sufficient to show that $T$ lies on $AD.$ On one hand, $AD$ is the $A$-median, so $T$ lies on $AD$ if and only if $[ATY]=[ATX].$ On the other hand, by construction $TK \parallel AY$ and $TL \parallel AX$ and hence $[ATY] = [AKY]$ and $[ATX] =[ALX]. $ Then it reduces to $$[AKY] = [ALX] \iff KE \cdot AY = LF \cdot AX.$$ So let $b =AX, c =AY, p = XE, q = YF.$ Since $AEKM$ is cyclic by construction, it follows by power of a point that $$ XH \cdot XE = XM \cdot XA = \frac{b^2}{2} \Rightarrow HE = \frac{2p^2 -b^2}{2p} \Rightarrow KE\cdot AY = \frac{(2p^2-b^2)c}{2p}$$Similarly, $LF \cdot AX = \frac{(2q^2-c^2)b}{2q} $ and it is sufficient to show $$(2p^2 -b^2)cq =(2q^2-c^2)bp $$which is obviously since $p = b\cdot \sin \angle XAY$ and $q = c \cdot \sin \angle XAY.$

official solution !

Consider point $R, S$ on side $AB, AC$ s.t. $RIA=SIA=90$, then i claim that $(RKIA)$ cyclic: $$\angle AKI=\beta/ 2+\gamma /2=90-\alpha /2 = \angle ARI$$similarly the other one.now said $O_b, O_c$ the midpoint of $AR, AS$ we have that the axis of $KI, LI$ pass through this points. Now consider point $X, Y$ as the second intersection of $(RKIA),(SLIA)$ with $AC, AB$. I claim that the height from $K$ and $L$ in triangle $\triangle KIL$ intersect side $AC, AB$ on $X,Y$: $$\angle IKX=\angle AIX=\alpha/2=\angle KIL-90$$Now consider the point $E, F$ as the projection on the side $AB, AC$ of $I$, then we have that the bisector $BI\perp ED$ and so $ED$ is parallel to the axis of $KI$, and the height of $L$ wrt. $IK$. Now the problem can be finished by noting that $AX/AY=AO_b/AO_c=AE/AF=1$ and so by homothety in $A$ the point $O,H,D$ are aligned.

Let $G, O, H$ be the centroid, circumcenter, orthocenter of $\triangle IKL$. Claim 1: $A \in OH$.

Claim 2: $G \in AD$.

Combining these two claims clearly solves the problem.

I will share my proof of showing that centroid of $\Delta IKL$ lies on $AD$ This proof is very interesting and beautiful and as i have seen no one else posted it before Also i found this without geogebra Points $P, Q$ are basically irrelevant, $K$ and $L$ are just points on $BI$ and $CI$ such that $\angle KAB=\frac{\angle C}{2}$ and $\angle LAC=\frac{\angle B}{2}$ As $\angle KAB=\angle ACL$ and $\angle KBA=\angle LAC$ we see that $\Delta KBA\sim \Delta LAC$ Because these triangles are oriented the same way we can now apply the gliding principle Let $M, N, T$ be midpoints of $AC, AB, KL$ By the gliding principle $\Delta TNM\sim \Delta KBA\sim \Delta LAC$ Let $X=MT\cap AK$ and $Y=NT\cap AL$ $\angle YNM=\angle TNM=\angle LAC=\angle YAM$ so $ANYM$ cyclic similarly $ANXM$ cyclic Hence $ANXYM$ cyclic Consider a homothety at $A$ with scaling ratio $2$ $N\mapsto B, M\mapsto C$ and hence $(ANXYM)\mapsto (ABC)$ Now let $AX$ and $AY$ intersect $(ABC)$ at $R$ and $S$ $X\mapsto R, Y\mapsto S$ Let $J=BS\cap CR$ $T\mapsto J$ so $A-T-J$ and $T$ is midpoint of $AJ$ $\angle JBC=\angle SBC=\angle SAC=\angle ABI=\angle IBC$ similarly $\angle JCB=\angle ICB$ This means that $J$ is the reflection of $I$ in $BC$, hence $I-D-J$ and $D$ is midpoint of $IJ$ Let $G=IT\cap AD$ Consider triangle $\Delta IJA$ $T$ is midpoint of $AJ$, $D$ is midpoint of $IJ$ hence $G$ is centroid of $\Delta IJA$ Now $\frac{GT}{GI}=\frac{1}{2}$ As $T$ is midpoint of $KL$ as well we see that $G$ is centroid $\Delta IKL$ as needed

Let $K' = \overline{CI} \cap \overline{AK}$, and $L' = \overline{BI} \cap \overline{AL}$. Note that $\triangle ABC \sim \triangle PBA \sim \triangle QAC$, so $\triangle BIC \sim \triangle BKA \sim \triangle ALC$. In particular, $\triangle IKK'$ and $\triangle ILL'$ are isosceles, and so $K$, $L$, $K'$ and $L'$ are concyclic. By EGMO Theorem 10.5, the Euler line of $\triangle IKL$ must then be the radical axis $\ell$ of circles with diameters $KK'$ and $LL'$, call them $\omega_K$ and $\omega_L$. Obviously $A$ must lie on $\ell$ since $KK'L'L$ is cyclic, so we just need to show that $D$ has equal powers to both circles. Let $\omega_K$ intersect $\overline{BI}$ and $\overline{CI}$ at $W$ and $X$. Define $Y$ and $Z$ similarly for $\omega_L$. We claim that $D$ lies on $\overline{WX}$. Let $M_B$ be the midpoint of minor arc $AC$. By the incenter lemma, $M_BI = M_BC$, so $\triangle IWX \sim \triangle IK'K \sim \triangle IM_BC$, so $\overline{WX} \parallel \overline{M_BC}$. Let $I'$ be the reflection of $I$ over $\overline{BC}$. Then $\triangle BI'C$ and $\triangle BKA$ are spirally similar, so $\triangle BKI'$ and $\triangle BAC$ are spirally similar as well. In particular, $\angle BKI' = \angle BAC = \angle BM_BC$, so $\overline{DW} \parallel \overline{I'K} \parallel \overline{M_BC}$. This shows the collinearity. Since $WXYZ$ is cyclic, it then follows that $D$ has equal power to both $\omega_K$ and $\omega_L$. Therefore, $D$ lies on $\ell$ as desired.

Let $P,Q$ be the $A-$mixtilinear touch points to $AB,AC$ respectively. Let $E,F$ be the tangency points of the incircle with $AC,AB$. Let $O,H$ be the circumcenter and orthocenter of $\triangle IKL$. Claim: $K=BI\cap (API)$ and $L=CI\cap (AQI)$. Proof: \[\measuredangle KAP=\measuredangle KAB=\frac{\measuredangle C}{2}=\measuredangle BIP=\measuredangle KIP\]\[\measuredangle QAL=\measuredangle CAL=\frac{\measuredangle B}{2}=\measuredangle QIC=\measuredangle QIL\]These yield the desired result.$\square$ Now we present a lemma. Lemma: $ABC$ is a triangle whose circumcenter is $O$. Let $O'$ be the reflection of $O$ with respect to $BC$. The tangent to $(ABC)$ at $A$ intersect $BC$ at $D$. Points $E,F$ are taken on $O'C,O'B$ such that $EA=EC$ and $FA=FB$. Then, $D,E,F$ are collinear. Proof: Let $X,Y$ be the intersection of the tangent to $(ABC)$ at $A$ with the tangent at $C,B$ respectively. By appyling Desargues theorem on $YOX$ and $BO'C,$ since $YB,OO',XC$ are concurrent, we get that these triangles are perpective. Hence $YO\cap BO'=F,OX\cap O'C=E,XY\cap BC=D$ are collinear.$\square$ Claim: $A,O,D$ are collinear. Proof: Invert the diagram from $I$ with radius $ID$. Note that $K^*=IB\cap A^*P^*$ and $L^*=IC\cap A^*Q^*$. $O^*$ is the reflection of $I$ according to $K^*L^*$. Let $EF$ intersect $BC$ at $T$. $M,N,W$ are the midpoints of $IE,IF,IT$ respectively. $M,N,W$ are collinear. Let $K_1,L_1$ be the reflections of $I$ to $K^*,L^*$. By applying the lemma on $\triangle DEF,$ we get that $T,K_1,L_1$ are collinear. Take the homothety centered at $I$ with ratio $\frac{1}{2}$. Then, $W,K^*,L^*$ are collinear. Since $K^*L^*$ is the perpendicular bisector of $IO^*,$ we have $WT=WI=WO^*$. Hence $O^*$ is on $(TIDA^*)$ which proves the claim.$\square$ Claim: $A,H,D$ are collinear. Proof: Let $(AIP)\cap AC=R$ and $(AIQ)\cap AB=S$. \[\measuredangle HKI=90-(180-\measuredangle CIB)=\frac{\measuredangle A}{2}=\measuredangle RKI\]Thus, $H$ lies on $KR$. Similarily, $H$ lies on $LS$. So $H=KR\cap LS$. Applying Desargues theorem on $RHS$ and $EDF,$ since $RH\cap ED,RS\cap EF,HS\cap DF$ is the line at infinity $A,H,D$ are collinear as desired.$\blacksquare$