How do you solve this?

I think complex bashing is possible. And when you try to calculate the point, you find the similarity and then done.

Just intersect the tangents with the A-bisector. You'll get that they're concurrent in A'. Then with angle chasing you get two cyclic quadrilaterals, which implies that the angles belong to triangle BCA'.

I think it is too easy for P4. Sketch: construct the tangent parallel to BC to the incircle, see everything in the triangle enclosed by the tangent and AB, AC, then it is almost obvious using homothety.

Funny. Solved pretty quickly yippeee!!!! Take the tangents at $X$ and $Y$ to the incircle other than $BC$, have them meet at $Z$. Then note that \[ \measuredangle ZXB = \measuredangle ZXC = \measuredangle ACX = \measuredangle ACB = \measuredangle APB = \measuredangle ZPB, \]so $BXZP$ cyclic, similarly $CYZP$ cyclic. Now note that $\measuredangle BZC = \measuredangle LIK$ by homothety (note the parallelogram) so \[ \measuredangle BZC + \measuredangle XPY = \measuredangle BZP + \measuredangle PZC + \measuredangle XPY = \measuredangle BXP + \measuredangle PYC + \measuredangle XPY = \measuredangle YXP + \measuredangle PYX + \measuredangle XPY = 0. \] We are done.

LoloChen wrote: I think it is too easy for P4. Sketch: construct the tangent parallel to BC to the incircle, see everything in the triangle enclosed by the tangent and AB, AC, then it is almost obvious using homothety. I did the same thing. I feel its too easy for P4.

Solution Sketch: Reflect A over I to get A', we get a parallelogram, so $\angle KIL = \angle BA'C$. Now, an angle chase gives $BXA'P$ cyclic. If we draw in the tangent line $\ell$ to the circumcircle at $P$, then we show that $\angle BA'A = \angle(\ell, PX)$. Now adding all the angles together yields the desired result.

Cross-ratio bashing We prove a stronger claim that $\angle IPX = \angle IKL$ (and similarly, $\angle IPY = \angle ILK$). Let $D=AI\cap BC$. Let the line through $P$ parallel to $BI$ meet $BC$ at $T$ and meet $AB$ at $U$. Notice that $\angle PTX = \tfrac{\angle B}2 = 90^\circ - \angle PIC = \angle PIX$, so $P,I,T,X$ is cyclic. Thus, the claim is equivalent to $I,K,T$ are collinear. To do this, we first note that $$\triangle PDB\sim\triangle PBA\implies \frac{AB}{BD} = \frac{PB}{PD} \implies \frac{AI}{ID} = \frac{PI}{PD},$$so we have \begin{align*} (AB;KU) &= -\frac{UB}{UA} = -\frac{PI}{PA} = \frac{AI}{ID}\cdot\frac{PD}{PA} = (AD;IP), \end{align*}and by projection from $T$, it follows that $I, T, K$ are collinear, as desired.

Let $T$ be the reflection of $A$ over $I$, the most important point to add since it gets rid of $K$ and $L$ as follows. Claim: We have $\angle KIL = \angle BTC$, and lines $TX$ and $TY$ are tangent to the incircle. Proof. The first part is true since $\triangle BTC$ is the image of $\triangle KIL$ under a homothety of ratio $2$. The second part is true because lines $AB$, $AC$, $TX$, $TY$ determine a rhombus with center $I$. $\blacksquare$ We thus delete $K$ and $L$ from the picture altogether; they aren't needed anymore. [asy][asy] size(11cm); pair A = dir(105); pair B = dir(200); pair C = dir(340); pair P = dir(270); filldraw(unitcircle, invisible, blue); pair I = incenter(A, B, C); filldraw(incircle(A, B, C), invisible, blue); draw(A--B--C--cycle, blue); pair E = foot(I, C, A); pair F = foot(I, A, B); pair T = 2*I-A; pair U = 2*I-E; pair V = 2*I-F; draw(U--T--V, deepgreen); draw(B--T--C, red); pair K = midpoint(A--B); pair L = midpoint(A--C); draw(K--I--L, red+dashed); pair X = extension(U, T, B, C); pair Y = extension(V, T, B, C); draw(A--P, grey); draw(circumcircle(B, X, P), grey+dashed); draw(circumcircle(C, Y, P), grey+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(160)); dot("$C$", C, dir(20)); dot("$P$", P, dir(45)); dot("$I$", I, dir(250)); dot("$T$", T, dir(225)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(310)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(11cm); A = dir 105 B 160 = dir 200 C 20 = dir 340 P 45 = dir 270 unitcircle / 0.1 lightcyan / blue I 250 = incenter A B C incircle A B C / 0.1 lightcyan / blue A--B--C--cycle / blue E := foot I C A F := foot I A B T 225 = 2*I-A U := 2*I-E V := 2*I-F U--T--V / deepgreen B--T--C / red K = midpoint A--B L = midpoint A--C K--I--L / red dashed X = extension U T B C Y 310 = extension V T B C A--P / grey circumcircle B X P / grey dashed circumcircle C Y P / grey dashed */ [/asy][/asy] Claim: We have $BXPT$ and $CYPT$ are cyclic. Proof. $\measuredangle TYC = \measuredangle TYB = \measuredangle ABC = \measuredangle APC = \measuredangle TPC$ and similarly. (Some people call this Reim's theorem.) $\blacksquare$ To finish, observe that \[ \measuredangle CTB = \measuredangle CTP + \measuredangle PTB = \measuredangle CYP + \measuredangle PXB = \measuredangle XYP + \measuredangle XYP = \measuredangle XPY \]as desired. (The length conditions $AC > AB > BC$ ensure that $B$, $X$, $Y$, $C$ are collinear in that order, and that $T$ lies on the opposite side of $\overline{BC}$ as $A$. Hence the directed equality $\measuredangle CTB = \measuredangle XPY$ translates to the undirected $\angle BTC + \angle XPY = 180^{\circ}$.)

EthanWYX2009 wrote: I think complex bashing is possible. And when you try to calculate the point, you find the similarity and then done. what similarity Let $DEF$ be the intouch triangle and use complex numbers with $\omega$ as the unit circle. We calculate $A=\tfrac{2ef}{e+f}$ etc., so $L=\tfrac{ef}{e+f}+\tfrac{df}{d+f}$ and $K=\tfrac{ef}{e+f}+\tfrac{de}{d+e}$. Now $X$ is the intersection of the tangents at $D$ and the $E$-antipode, so it equals $\tfrac{2de}{e-d}$ and $Y$ equals $\tfrac{2df}{f-d}$. To calculate $P$ note that it is the center of $(BIC)$, so it equals $\tfrac{bc(\overline{c}-\overline{b})}{b\overline{c}-c\overline{b}}$ since $I=0$. By substitution this cancels nicely to $\tfrac{2def}{(d+e)(d+f)}$. Now it's sufficient to show that $(\tfrac{X-P}{Y-P})\div \overline{(\tfrac{L}{K})}$ is real. We compute these expressions separately. \begin{align*} \frac{X-P}{Y-P}=\frac{\frac{e}{e-d}-\frac{ef}{(d+e)(d+f)}}{\frac{f}{f-d}-\frac{ef}{(d+e)(d+f)}}&=\frac{e(d+e+2f)(d-f)}{f(d+2e+f)(d-e)}\\ \overline{\left(\frac{L}{K}\right)}=\frac{\frac{1}{e+f}+\frac{1}{d+f}}{\frac{1}{e+f}+\frac{1}{d+e}}&=\frac{(d+e+2f)(d+e)}{(d+2e+f)(d+f)}. \end{align*}Hence the desired expression equals $\tfrac{e(d^2-f^2)}{f(d^2-e^2)}$, which equals its conjugate. $\blacksquare$

Notice that such a point $X$ and $Y$ are unique (For $X$, let $J$ be the intersection of the tangent with $(AB)$. Then since $\omega$ is inscribed inside $AJXC$ we have $AC+JX = AJ+XC \Longrightarrow AC+ k AC = (1-k) AC + (1-k) BC$, by Thales where $k = \frac{JX}{AC}$. Solving this equation, we have $k$ unique hence $X$ is unique. ). A construction of such points $X$ and $Y$ is easy : take $S$ the symmetric of $A$ about $I$. Draw the tangents of $S$ to $\omega$ and their intersections $X$ and $Y$ with $(BC)$. The parallelism is obvious since we have a rhombus. Notice that $A, S, P$ (the south pole) belong to the bisector of $A$. Now the following lemma : $B,X,S,P$ are concyclic (as well as $P,S,Y,C$). Sketch of the proof : Angle chase with $\angle BXS =_{\text{parallelism}} = \pi - \angle BCA$ and $\angle BPS = \angle BPA =_{\text{A,B,P,C concyclic}} \angle BCA$. Remark with Thales theorem we have $\angle LIK = \angle BSC$. Finally $\angle BSC + \angle YPX = \angle BSC + (\angle BXP + \angle PYC) - \pi = \angle BSC + \angle BSP + \angle PSC = 2 \pi - \pi = \pi$, as desired.

Let $\ell$ denote line tangent to the incircle parallel to $BC$, let $\ell \cap AB,AC = U,V$ it is well known in config that $UY,VX$ concur at $I$, let the tangents through $X,Y$ meet at $Z$, then $ZXY,AUV$ are homothetic with ratio $-1$ at $I$ implying $Z$ lies on $AI$, now $\measuredangle XZP = \measuredangle PAC= \measuredangle CBP$ so $BXZP,CYZP$ are cyclic, and now by homothety $\measuredangle BZC = \measuredangle LIK$ so we're done.

OMG same solution as Evan Chen

This problem was proposed by Burii.

Let $D$ be the intersection of the tangent parallel to AC with $\omega$. Let $E$ be the intersection of the other tangent with $\omega$. Let $R = DX \cap EY$. Let the tangent parallel to $AB$ intersect $AC$ at H and the tangent parallel to $AC$ intersect AB at $J$ and let . Since we have two pairs of parallel sides, $AHRJ$ is at least a parallelogram with the circle $\omega$ inscribed inside. We will show that $R,I,A$ are collinear. Let $\angle HAJ = \angle HRJ = 2t$. $AI$ bisects $\angle HAJ$ and $RI$ bisects $\angle HRJ$, each to two angles of size $t$. Now $HRJ = 180^{\circ} -2t$ by angles in a parallelogram yielding $\angle RIA = 180^{\circ}$, so the points are collinear. We can also conclude that $AHRJ$ is a rhombus. $\angle BXD = \angle BCA = \alpha = \angle APB$, the latter holding due to $ABPC$ being cyclic. It follows that $RXB = 180^{\circ}-\alpha = 180^{\circ}-\angle RPB$. So $(PRXB)$ is cyclic. With a similar reasoning $(RYCP)$ is cyclic. Now let $\angle YPR = s$ and $\angle RPX = t$. This implies $\angle YCR = s$ and $\angle RBX = t$. Thus $\angle YPX = s+t$. $\angle CRB = 180^{\circ} - (s+t)$. If $\angle KIL + \angle YPX = 180^{\circ}$, then $\angle KIL = 180^{\circ} - \angle YPX = 180^{\circ} - (s+t) = \angle CRB$. It thus suffices to show that $CR || KI$ and $ RB || IL$. Because $AHRJ$ is a rhombus, $I$ bisects $AR$ and thus $\frac{AI}{IR} = 1 = \frac{AK}{KC}$, implying $\triangle AKI \sim \triangle ACR$ and that $KI || CR$. A similar argument can be done to show $LI || JR$ and we are done.

The shortest solution I found, unfortunately. Let $T_B$ and $T_C$ be the $B$ and $C$-mixtilinear touchpoints, and let $B_E, C_E$ be the $B$, $C$-excircle touchpoints, respectively. Observe that $P, Y, T_B$ and $P, X, T_C$ are collinear, so it suffices to prove $\angle{T_BYT_C} + \angle{KIL} = 180^\circ$. But $\angle{T_BPA}=\angle{CBB_E}$ is the angle between $KI$ and $BC$, which is also just $\angle{LKI}$ so we are done.

Let $A'$ be the reflection of $A$ about $I$. Notice $A'X$ and $A'Y$ are tangent to $w$. Thus we must have that $PA'XB$ is cyclic as $\angle A'PB=\angle ACB$ $=\angle CXA'$. Also $PA'YC$ is cyclic. Thus $$180^{\circ}-\angle YPX=180^{\circ}-\angle YCA'-\angle XBA'=\angle CA'B=\angle KIL$$

Let the line through $Y$ parallel to $AB$ meet $AC$ at $N$, and the line through $X$ parallel to $AC$ meet $AB$ at $M$. Also, let the intersection of these two tangent lines be point $Z$. We can show by equal tangents that since quadrilateral $AMZN$ has an incircle, it is a rhombus. This implies that $Z$ lies on line $AI$, so $A, I, Z, P$ are collinear in that order. Now, if we let $\angle BAC$ be equal to $2x$, by inscribed angles $\angle BCP$ and $\angle CBP$ are both equal to $x$. Since $\angle AZX = \angle AZY = x$, $BXZP$ and $CYZP$ are both cyclic. Since $AC = 2AK, AZ = 2AI, AB = 2AL$, there exists a homothethy centered at $A$ that sends triangle $KIL$ to triangle $CZB$, meaning $\angle KIL = \angle CZB$. Finally, $\angle KIL + \angle XYP = \angle CZB + \angle XYP =$ $2x + \angle BXZ + \angle CZY + \angle XPY = \angle BPC + 2x$$ = 180^{\circ}$.

Call $D$ the reflection of $A$ over $I.$ Claim: $BXDP,CYDP$ are cyclic. Proof: $180-\angle BXD=\angle CXD=\angle XCA=\angle BCA=\angle BPA=\angle BPD,$ and similarly for $CYDP.$ Thus $\angle YPX=\angle YPD+\angle DPX=\angle YCD+\angle DBX=\angle BCD+\angle DBC=180-\angle CDB.$ But homothety at $A$ with scale factor $2$ takes $\angle KIL$ to $\angle CDB,$ so $\angle KIL+\angle YPX=180$ as desired.

Here is an interesting approach using Iran's lemma found by AlphaGeo, with some modifications made by me. Let $T = BI\cap KL$, so by Iran's lemma, $\angle ATB = 90^\circ$ and $LA=LB=LT$. Claim. Quadrilaterals $ALIT$ and $CPXI$ are similar. Proof. Since $XI$ bisects the angle between both tangents from $X$ to the incircle, it follows that $\angle XIC=90^\circ$. Moreover, $\angle IAT = 90^\circ - \tfrac B2 - \tfrac A2 = \tfrac C2$, so $\triangle CXI\sim\triangle AIT$. Finally, $IC=IP$ and $LA=LT$. Moreover, $\angle IPC = \angle ABC = \angle ALT$, so $\triangle ALT\sim\triangle CPI$. Combining both similarities gives the result. $\blacksquare$ The claim gives $\angle IPX = \angle ILT = \angle ILK$, and similarly, $\angle IPY = \angle IKL$, so we are done.

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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\color{red}\textbf{Claim:-}$ $X,Q,P,C$ are cyclic. $\color{blue}\textbf{Proof:-}$ We have $MX\parallel AC$ and $NY\parallel AB$ $\implies$ $\triangle BMX\sim \triangle NCY\sim \triangle ABC.$ Also, since $L,K$ are the midpoints of $AB$ and $AC$ of triangle $ABC.$ Therefore we get, $LK\parallel BC.$ Now in $\triangle$ $LIK$ we get, $\angle KIL+\angle LKI+\angle ILK=180^o$ also $\angle YPX=\angle APX+\angle APY.$ Also we get, $\angle CQB=\angle KIL.$ Therefore we have to prove that $\boxed{\angle CQB+\angle YOX=180^o}$ Now, Let the line parallel to $LI$ be $BQ$ and the line parallel to $IN$ be $QC.$ You can easily prove that point $Q$ lies on $AP$ by converse of midpoint theorem we get, $AI=IQ.$ Now produce $MX$ to a point on the line $AP$ let's call that point be $F$ Now, since $MX\parallel AC$ we get, $\angle FAC=\angle MAF=\angle MFA.$ From this we have to prove that point $Q$ and point $F$ are coincides each other. Now for proving this we draw the inradius of the incircle of $\triangle ABC$ Therefore we get, $\triangle AVI\cong WIF$ where $IW$ and $IV$ are the inradius. From this we easily say that $AI=FI$ Therefore $Q$ and $F$ coincides each other. Now Since $A,B,P,C$ are concyclic therefore $\angle AQM=\angle PBC\implies$ $X,P,B,Q$ are concyclic. By similar logic we say that $Y,Q,P,C$ are cyclic. Therefore, $\boxed{BQC+XPY=180^o}$ $\implies$ $\boxed{\angle KIL+\angle XPY=180^o.}$

$ $

Synthetic length chase

We uploaded our solution https://calimath.org/pdf/IMO2024-4.pdf on youtube https://youtu.be/DNZw9wMdYH4.

Cool problem

Let the parallels from X and Y intersect AB and AC at M and N respectively, and let the parallels intersect at Z. Then this problem can be solved in 2 main claims. Claim 1. AMZN is a rhombus. (This implies Z is on line AI) Claim 2. The quadrilaterals XZPB and YZPC are cyclic. Also note that triangle BZC and LIK are similar. Therefore 180 = angle BZC + ZCB + CBZ = angle BZC + angle ZPY + angle ZPX = angle LIK + angle XPY. How long do I have to wait till aops allows me to post tatex stuff?? It's been three weeks now.

Let those paralel lines intersect with $AB$ and $AC$ at $R$ and $S$, respectively. Let $RX\cap SY=D$ Claim: $D$ is the reflection of the point $A$ wrt $I$. Since $ARSD$ is both tangent quadrilateral and paralellogram, $AI=ID$. Claim: $BXDP$ and $CYDP$ are cyclic quadrilaterals. Using Reim's Theorem, since $AC$ and $BP$ are antiparallel and $DX$ and $BP$ are paralel, the points $B$, $X$, $D$ and $P$ are concyclic. Similarly points $C$, $Y$, $D$ and $P$ are concyclic. Claim: $\angle KIL+\angle XYP=\pi$. Observe that cyclic quadrilaterals implies $$\angle XYP=\angle XPD+\angle YPD=\angle XBD+\angle YCD=\pi-\angle BDC=\pi-\angle KIL$$as desired.

Edit in sol later

ok this is neat stuff Let $a,b,c$ be $BC,CA,AB$. Let $s$ be the semi perimeter, $\Delta $ be the area, $h_{a},h_{b},h_{c}$ be the $A,B,C$ altitudes of$\triangle ABC$. Let the line through $X$ parallel to $AC$ meet $AB$ in $R$ and the line through $Y$ parallel to $AB$ meet $AC$ in $Q$. Let a line through $L$ parallel to $IL$ meet $AI$ in $M$ and the line through $Q$ parallel to $KI$ meet $AI$ in $N$. CLAIM 1: $XP,YQ,BI$ concur. PROOF: Indeed, let $XR \cap YQ=Z$ then $ARZQ$ is a parallelogram. But $\omega $ is an inscribed circle of a parallelogram, hence $ARZQ$ is a rhombus, so $AZ$ bisects $\angle A$ and consequently, $A-I-Z$. A result of this is that $\triangle ARZ \sim \triangle BPC \sim \triangle AQZ$ and that $I$ is the intersection of the angle bisectors of the opposite angles and hence the diagonals. CLAIM 2: $\frac{CX}{CB}= \frac{b}{s}$ PROOF: Note that $\frac{CX}{CB}=\frac{2r}{h_{b}}=\frac{\frac{2 \Delta}{s}}{\frac{2 \Delta}{b}}=\frac{b}{s}$ Similarly, $\frac{BY}{BC}=\frac{c}{s}$ CLAIM 3: $\frac{AM}{AZ}=\frac{CX}{CB}$ PROOF: Indeed, note that $\frac{CX}{CB}= \frac{b}{s}$ $\implies \frac{BX}{BC}=\frac{s-b}{s}=\frac{BR}{BA}$ $\implies \frac{AI}{AM}=\frac{AL}{AR}=\frac{\frac{AB}{2}}{AR}=\frac{1}{2} \cdot \frac{AB}{AR}=\frac{1}{2} \cdot \frac{1}{1-\frac{BR}{BA}}=\frac{1}{2} \cdot \frac{s}{b}=\frac{s}{2b}$ $\implies \frac{AM}{AZ}=\frac{1}{\frac{2AI}{AM}}=\frac{b}{s}$ As a consequence (using $\triangle ARZ \sim \triangle CPB$), $A \mapsto C, R \mapsto P, Z \mapsto B$ and $M \mapsto X \implies \angle ZRM=\angle BPX$. Similarly, $\angle ZQN =\angle YPC$ Now, finally, $\angle LIK=\angle AIK+\angle LIA=\angle QNA+\angle RMA=\angle AZQ+\angle ZQN+\angle AZR+\angle ZRM$ $=\angle A+\angle CPY+\angle BPX=180-\angle A-\angle XPY+\angle A=180-\angle XPY$ and we are done. $\blacksquare$

This is not an IMO level geometry problem. We set the incircle of $ABC$ as the unit circle and let $D,E$ and $F$ be the touching points of incircle and $ABC$ we get that $cyc: A=\frac{2ef}{e+f}$ and so on. Then $K=\frac{ef}{e+f}+\frac{ed}{e+d}$ and $L=\frac{ef}{e+f}+\frac{fd}{f+d}$ notice that by rotating $E$ wrt $I$ (call it $E'$) we get that $X$ is the intersection of tangents drawn from those points so::: $X=\frac{-2ed}{d-e}$ and similarly $Y=\frac{-2df}{d-f}$ Notice that $P=O_{BIC}$ by evaluating the determinat gives us that $P=\frac{2def}{(e+d)(f+d)}$ It is left to prove that $(\frac{p-y}{p-x}) \times \overline{(\frac{K}{L})}$ which is trivial...$\lambda $