Here is a solution that I found after the test: We will prove the main lemma: Let $K,L$ be arbitrary points on segment $BC$. Let $\omega_{BK}$, $\omega_{CK}$, $\omega_{BL}$, $\omega_{CL}$ be the incircles of triangles $ABK, ACK, ABL, ACL$. Then, the pairs $(\omega_{BL}, \omega_{CK})$ and $(\omega_{CL}, \omega_{BK})$ have the same exsimilicenter. [asy][asy] import olympiad; size(10cm); pair A = dir(110), B = dir(210), C = dir(330), K = (B+C)/2, L = (K+C)/2, Ibk = incenter(A,B,K), Ick = incenter(A,C,K), Ibl = incenter(A,B,L), Icl = incenter(A,C,L), P = extension(Ibk,Icl, Ick,Ibl), I = incenter(A,B,C), U = foot(Ibl,B,C), V = foot(Icl,B,C); D(incircle(A,B,K),green); D(incircle(A,C,L),green); D(incircle(A,B,L),orange); D(incircle(A,C,K),orange); D(P--Ibl,orange); D(P--Ibk,green); D(L--A--B--C--A--K--P); D(B--I--C,dashed+blue); D("A",A,A); D("B",B,B); D("C",C,dir(270)); D("K",K,dir(270)); D("L",L,dir(270)); D("U",U,dir(270)); D("V",V,dir(270)); D("I_{BK}",Ibk,dir(140)); D("I_{CK}",Ick,dir(80)); D("I_{BL}",Ibl,dir(140)); D("I_{CL}",Icl,dir(80)); D("I",I,dir(110)); D("T",P,dir(0)); [/asy][/asy] Let $\omega_{BK}, \omega_{CK}, \omega_{BL}, \omega_{CL}$ have centers $I_{BK}, I_{CK}, I_{BL}, I_{CL}$, respectively. It suffices to show that $I_{BL}I_{CK}$ and $I_{BK}I_{CL}$ cocnur on $BC$. To prove this, we'll use moving points. Fix $K$ and move $L$. Then, we see that $I_{CK}$ and $I_{BK}$ are fixed. The key claim is: Let $U,V$ be the touchpoints of $\omega_{BL}$, $\omega_{CL}$ with $BC$. Then, $U\to V$ is projective. We proceed via Stewart's Theorem. Let $BU = u$, $BL = x$, $CL = y$, $CV = v$, $AL = d$, $BC = a$, $CA = b$, and $AC = c$. By Stewart's, we have that: \[ a(d^2+xy) = b^2x+c^2y. \]Now, we see that \[ u = \frac{c+x-d}{2}, v = \frac{b+y-d}{2}, \]so we can solve for \[ d = \frac{a+b+c}{2}-(u+v). \]Then, \[ x = 2u - c+d = u-v + \frac{a+b-c}{2} \]and \[ y = 2v - b + d = v-u+\frac{a-b+c}{2}. \]Plugging in, we have, for appropriate constants $A,B$: \[ \begin{aligned} \frac{b^2-c^2}{a}(u-v) + A &= \frac{b^2x+c^2y}{a} = d^2+xy \\ &= \left(\frac{a+b+c}{2} - (u+v)\right)^2 + \left(u-v+\frac{a+b-c}{2}\right)\left(v-u + \frac{a-b+c}{2}\right) \\ &= B - (a+b+c)(u+v) + (u+v)^2 - (u-v)^2 + (c-b)(u-v)\\ &= B - (a+b+c)(u+v) + (c-b)(u-v) + 4uv. \end{aligned} \]Thus in particular we have, for constants $P,Q, R$: \[ 4uv + uP + vQ + R = 0\implies v = - \frac{Pu+R}{4u+Q}. \]Thus $u\to v$ is a M\"obius function and so is projective if we parameterize $BC$ by $\mathbb R\mathbb P^1$; we can easily check that it is nonconstant. $\Box$ Now we are almost done. We note that $U\to I_{BL}$ is projective since we project through $\infty_{\perp BC}$ onto $BI$, and then we project through $I_{CK}$ to get $I_{BL}\cap I_{CK} = T'$. Then, $V\to I_{CL}$ is also projection through $\infty_{\perp BC}$, this time onto $CI$. Then, $I_{CL}\to I_{CL}I_{BK}\cap BC = T''$ is also projective. So, $U\to T', U\to T''$ are projective, and so it suffices to check $T' = T''$ for $3$ values of $U$. In order to do this, we just need to check $3$ values of $L$. At $L = K$, $T' = T'' = I_{BK}I_{CK}\cap BC$. At $L = B$, $I_{BL} = B$, $I_{CL} = I$,so $I_{BL}I_{CK}\cap BC = B = I_{BK}I_{CL}\cap BC$, i.e. $T' = T'' = B$. Similarly, at $L = C$, $T' = T'' = C$. Thus, $T' = T''$ in general, and so $I_{BL}I_{CK} \cap BC = I_{CL}I_{BK}\cap BC$. Thus the central lemma is proven. $\Box$ To finish from here, let $T$ be the common exsimilicenter in the original problem. Let $X\in AK$ be the insimilicenter of $\omega_{BK}$, $\omega_{CK}$, and let $Y\in AL$ be the insimilicenter of $\omega_{BL}, \omega_{CL}$. Then, $P,Q$ are the insimilicenters of $(\omega_{BK}, \omega_{BL})$ and $(\omega_{CK}, \omega_{CL})$, respectively, so by Monge's we have $T, X, P$ collinear, $T, Y, Q$ collinear, $T, X, Q$ collinear, and $T, Y, P$ collinear, so $T, X, Y, P, Q$ are collinear. Now let $TXYPQ$ meet $\omega_{BL}$ again at $Z\neq P$. The homothety at $T$ taking $\omega_{BL}$ to $\omega_{CK}$ takes $Z$ to $Q$, and so $I_{BL}Z\parallel I_{CK}Q = IC$. So, $\triangle PIQ\sim \triangle PI_{BL}Z$, and the latter is clearly $I_{BL}$-isosceles. $\blacksquare$ Remark: tastymath75025 notes that the main claim is equivalent to $I_{BL}\to I_{CL}$ being projective, which is true because $AI_{BL}\to AI_{CL}$ is a rotation.

Sadly, I did not solve this synthetically in the exam. Here is by length bash solution I found in the exam. Lemma : Let $ABC$ be a triangle with inradius $r$. Let $X$ be a variable point along $BC$. If $r_b,r_c$ are inradii of $\triangle ABX$, $\triangle ACX$. Then $$\left(\frac{r}{r_b}-1\right)\left(\frac{r}{r_c}-1\right)$$is not depending on $X$. Proof Let $I, I_B, I_C$ be the incenters of $\triangle ABC$, $\triangle ABX$, $\triangle ACX$. First, we claim that ratio $\tfrac{BI_B}{II_B}\cdot\tfrac{CI_C}{II_C}$ is constant. To prove this, denote another position of $X$ by $X'$ and let $I_B', I_C'$ be the corresponding incenters. Then as rotation by $0.5\angle A$ is projective map, we get $$(B, I; I_B, I_B') = (C, I; I_C, I_C') \implies \frac{BI_B}{II_B}\cdot\frac{CI_C}{II_C} = \frac{BI_B'}{II_B'}\cdot\frac{CI_C'}{II_C'}$$which is constant as claimed. To finish, note that $$\frac{II_B}{BI_B} = \frac{BI}{BI_B}-1 = \frac{\mathrm{dist}(I,BC)}{\mathrm{dist}(I_B,BC)}-1 = \frac{r}{r_b}-1$$so we are done. Back to the main problem. WLOG $B,K,L,C$ lie in this order. Let $r_1, kr_1, r_2, \ell r_2$ be the inradii of $\triangle ABK$, $\triangle ABL$, $\triangle ACL$, $\triangle ACK$ respectively. Let $I_B, I_C$ be the incenters of $\triangle ABK$, $\triangle ACL$ respectively. Moreover, let $b = 0.5\angle B$ and $c = 0.5\angle C$. First, note that $$k = \frac{BI_B - r}{BI_B+r} = \frac{1-\sin b}{1+\sin b}$$and similarly, $\ell = \tfrac{1-\sin c}{1+\sin c}$. By the lemma, \begin{align*} \left(\frac{r}{r_1}-1\right)\left(\frac{r}{\ell r_2}-1\right) &= \left(\frac{r}{kr_1}-1\right)\left(\frac{r}{r_2}-1\right) \\ k(r-r_1)(r-\ell r_2) &= \ell(r-kr_1)(r-r_2) \\ kr^2 - kr_1r - k\ell rr_2 + k\ell r_1r_2 &= \ell r^2 - \ell r_2r - k\ell rr_1 + k\ell r_1r_2 \\ (k-\ell)r + k(\ell - 1)r_1 &= \ell(k-1)r_2 \\ \left(\frac{1}{\ell-1} - \frac{1}{k-1}\right)r + \frac{k}{k-1}r_1 &= \frac{\ell}{\ell-1}r_2 \\ \end{align*}By plugging in $k,\ell$, it's easy to see that $\tfrac{2k}{k-1} = 1 + \tfrac{1}{\sin b}$ and $\tfrac{2}{k-1} = \tfrac{1}{\sin b}$, thus this is equivalent to $$\frac{r}{\sin b} - r_1\left(1+\frac{1}{\sin b}\right) = \frac{r}{\sin c} - r_2\left(1+\frac{1}{\sin c}\right)$$Finally, note that $BI = \tfrac{r}{\sin b}$, $BI_B = \tfrac{r_1}{\sin b}$ and $I_BP = r_1$. Hence the LHS is just $IP$. Similarly the RHS is $IQ$ so $IP=IQ$.

Here is a solution found during the test Let $I_1, I_2, I_3, I_4$ denote the incenters of $\triangle ABK, \triangle ABL, \triangle ACL, \triangle ACK$ respectively. Let $\omega_1, \omega_2, \omega_3, \omega_4$ denote their respective incircles. Lemma 1. $\frac{BI_1}{I_1I} \cdot \frac{II_4}{I_4C} = \frac{AB \cdot AC}{AI^2}.$ Proof. This is straightforward from Law of Sines in $\triangle ABI_1, \triangle AII_1, \triangle AII_4 \triangle AI_4C$. $\blacksquare$ Analogously, we get that $\frac{BI_2}{I_2I} \cdot \frac{II_3}{I_3C}$. From the two previous length relations, it's clear by Menelaus that $I_1I_3, I_2I_4$ concur at a point on $BC,$ say at $T.$ Observe that $(B, P; I_1, I_2) = (C, Q; I_3, I_4) = -1.$ This is true because $B, P$ are the exsimilicenter and insimilicenter of $\omega_1, \omega_2$, and $I_1, I_2$ their centers, and analogously for $C, Q, I_3, I_4$. Hence, since $BC, I_1I_3, I_2I_4$ concur at $T$, we must also have that $PQ$ goes through $T$ (simply project $(B, P; I_1, I_2)$ from $T$ onto $CI$). We are in the endgame now. Let $\omega$ be a circle centered at $I$ with radius $IP.$ Let $Q'$ be the insimilicenter of $\omega, \omega_3.$ By Monge's Theorem on $\omega, \omega_1, \omega_3$, we get that $P, Q', T$ are collinear. Hence, $Q' = PT \cap CI$, which we know is just $Q$. Therefore, we have that $Q$ is the insimilicenter of $\omega, \omega_3$. Since $Q \in \omega_3$, we therefore have that $Q \in \omega.$ Finally, from the definition of $\omega$, we have that $IQ = IP$, and so we're done . $\square$

[asy][asy] size(12cm); defaultpen(fontsize(10pt)); pen pri=royalblue; pen sec=deepcyan; pen tri=rgb(41, 207, 255); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; real r=1.7353; pair A, B, C, I, IB, P,Q, IC, K, L, JB, JC, T; A=dir(120); B=dir(210); C=dir(330); I=incenter(A,B,C); IB=(7B+r*I)/(7+r); P=IB+unit(I-B)*length(IB-foot(IB,B,C)); Q=I+unit(C-I)*length(P-I); IC=incenter(C,extension(A,C,Q,Q+A-foot(A,C,I)),extension(B,C,Q,Q+A-foot(A,C,I))); K=extension(B,C,A,2*foot(B,A,IB)-B); L=extension(B,C,A,2*foot(C,A,IC)-C); JB=incenter(A,B,L); JC=incenter(A,C,K); T=extension(IB,IC,B,C); draw(B--I--C,sec); draw(Q--T,sec); draw(JC--T--IC,sec); fill(A--B--C-- cycle,fil); draw(B--A--C--T,pri); draw(K--A--L,pri); filldraw(incircle(A,B,L),tfil,tri); filldraw(incircle(A,B,K),tfil,tri); filldraw(incircle(A,C,L),tfil,tri); filldraw(incircle(A,C,K),tfil,tri); dot("$A$",A,N); dot("$B$",B,S); dot("$C$",C,SE); dot("$T$",T,SW); dot("$I_B$",IB,dir(285)); dot("$I_C$",IC,dir(255)); dot("$J_B$",JB,NW); dot("$J_C$",JC,NE); dot("$P$",P,dir(85)); dot("$Q$",Q,dir(120)); dot("$K$",K,S); dot("$L$",L,S); dot("$I$",I,N); [/asy][/asy] Let $I_B$, $J_B$, $I_C$, and $J_C$ denote the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACL$, and $\triangle ACK$, respectively. Also let $r_{\triangle XYZ}$ denote the inradius of $\triangle XYZ$ for all $X,Y,Z$. Lemma. For any points $K$ and $L$ on $\overline{BC}$ of $\triangle ABC$, if $I_B$, $J_B$, $I_C$, and $J_C$ denote the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACL$, and $\triangle ACK$, respectively, then $\overline{I_BI_C}$, $\overline{J_BJ_C}$, and $\overline{BC}$ concur at a point $T$. Proof. We can easily determine that $$A(BI_B;IJ_B)=\frac{\sin\tfrac12\angle BAC\cdot\sin\tfrac12\angle KAL}{\sin\tfrac12\angle KAC\cdot\sin\tfrac12\angle BAL}=A(CI_C;IJ_C),$$and the result follows. $\blacksquare$ Considering the homothety centered at $B$ sending $(I_B)$ to $(J_B)$, we can check that the scale factor is $$\frac{PJ_B}{PI_B}=\frac{r_{\triangle ABL}}{r_{\triangle ABK}}=\frac{BJ_B}{BI_B}\implies -1=(BP;I_BJ_B),$$and similarly $-1=(CP;I_CJ_C)$. It is immediate that $\overline{PQ}$ also passes through $T$. By Menelaus on $\triangle II_BI_C$, $$-1=\frac{IP}{PI_B}\cdot\frac{I_BT}{TI_C}\cdot\frac{I_CQ}{QI}=\frac{IP}{QI}\cdot\frac{I_BT}{TI_C}\cdot\frac{I_CQ}{PI_B}=\frac{IP}{QI}\cdot\frac{r_{\triangle ABK}}{r_{\triangle ACL}}\cdot\frac{r_{\triangle ACL}}{r_{\triangle ABK}}=\frac{IP}{QI},$$whence $IP=IQ$, as desired. $\square$

This is my problem. It arose out of an experiment to create a geometry problem with a condition as absurd as possible. I think I did a good job; I’m still amazed that this statement is true. I thought this problem was quite difficult, but the results don’t agree (11/75 sevens). It seems the problem was quite simple to bash, unfortunately. My solution is essentially identical to the one in #6, so I won’t post it.

Let $w,w_1,w_2,w_3,w_4$ be the inscribed circles of $ABC,ABK,ACL,AKC,ABL$. $I,I_1,I_2,I_3,I_4$ the corresponding centers and $D,D_1,D_2,D_3,D_4$ the touching points with $BC$. It suffices to show that circles $w_1,w_2$ and $w_3,w_4$ have a common external homothetic center $H$ on $BC$. $AK$ ,$AL$ are the common internal tangents of $w_1,w_3$ and $w_2,w_4$ respectively well let $e_1, e_2$ be the other internal tangents and $H_1$ be the internal homothetic center of $w_1,w_4$ and $H_2$ that of $w_2,w_3$. We prove that $e_1,e_2$ concur at $D$. That is simple length chasing first prove that $DD_4=D_1K$ , $DD_3=D_2L$ and the result follows (you need a simple lemma that states that if $e_1$ intersects $BC$ at $D$ then $DD_4=D_1K$) Next we prove that $AH_1DH_2$ has an inscribed circle $C$. This requires more length chasing to prove that $H_1D+H_2A=H_1A+H_2D$ . After we prove the existence of such a circle $C$ we apply monge theorem for circles $w_1,w_2,C$ and prove that the line that connects the homothetic centers $H_1H_2$ intersects $BC$ at the external homothetic center of $w_1,w_2$ $H$. The same happens for $w_3,w_4$ and thus we proved that $w_1,w_2$ and $w_3,w_4$ have a common external homothetic center $H$ which is the intersection of $H_1H_2$ and $BC$.

Here are the two official solutions. The first one is already posted in #6 up to small edits, but it's referenced later so I included it anyways for completeness. First solution, mostly elementary (original) Let $I_B$, $J_B$, $I_C$, $J_C$ be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACK$, $\triangle ACL$ respectively. [asy][asy] unitsize(15); real r1, r2; path w1, w2; pair J1 = (-0.5, 2.7); pair J2 = (1, 3); real r1 = abs(J1.y); real r2 = abs(J2.y); path w1 = circle(J1, r1); path w2 = circle(J2, r2); pair X1 = intersectionpoints(w1, w2)[0]; pair X2 = intersectionpoints(w1, w2)[1]; pair H = extension((1.9, 0), (2, 1), X1, X2); pair P = intersectionpoints(circle(H, sqrt(abs(H-X1)*abs(H-X2))), w1)[1]; pair Q = intersectionpoints(circle(H, sqrt(abs(H-X1)*abs(H-X2))), w2)[0]; pair I = extension(J1, P, J2, Q); pair B = extension(J1, P, (0, 0), (1, 0)); pair C = extension(J2, Q, (0, 0), (1, 0)); pair A = extension(B, reflect(B, I) * C, C, reflect(C, I) * B); pair K = extension(B, C, A, reflect(A, J2) * C); pair L = extension(B, C, A, reflect(A, J1) * B); pair I_B = incenter(A, B, K); pair J_B = incenter(A, B, L); pair I_C = incenter(A, C, K); pair J_C = incenter(A, C, L); pair R = extension(P, Q, B, C); filldraw(A--B--C--cycle, invisible, blue); draw(incircle(A, B, K)^^incircle(A, C, L), deepcyan); draw(L--A--K, blue); draw(w1^^w2, deepcyan); draw(I--P^^I--Q, red); add(pathticks(I--P, 2, 0.5, 0, 10, red)); add(pathticks(I--Q, 2, 0.5, 0, 10, red)); draw(B--P^^C--Q, orange); draw(Q--R--B, deepgreen); draw(J_C--R, deepgreen+dashed); draw(I_C--R, deepgreen+dashed); dot("$A$", A, dir(A-I)); dot("$B$", B, dir(230)); dot("$C$", C, dir(310)); dot("$I$", I, dir(2*I-P-Q)); dot("$K$", K, dir(270)); dot("$L$", L, dir(270)); dot("$P$", P, dir(80)); dot("$Q$", Q, dir(20)); dot("$R$", R, dir(-90)); dot("$I_B$", I_B, dir(-90)); dot("$I_C$", I_C, dir(-90)); dot("$J_B$", J_B, dir(-90)); dot("$J_C$", J_C, dir(-90)); [/asy][/asy] We begin with the following claim which does not depend on the existence of tangency points $P$ and $Q$. Claim: Lines $BC$, $I_BJ_C$, $J_BI_C$ meet at a point $R$ (possibly at infinity). Proof. By rotating by $\frac{1}{2} \angle A$ we have the equality $A(BI; I_B J_B) = A(IC;I_C J_C)$. It follows $(BI; I_B J_B) = (IC; I_C J_C) = (CI; J_C I_C)$. Therefore, the concurrence follows from the so-called prism lemma on $\overline{I B I_B J_B}$ and $\overline{I C J_C I_C}$. $\blacksquare$ Remark: [Nikolai Beluhov] This result is known; it appears as {4.5.32} in Akopyan's Geometry in Figures. The cross ratio is not necessary to prove this claim: it can be proven by length chasing with circumscribed quadrilaterals. (The generalization mentioned later also admits a trig-free proof for the analogous step.) We now bring $P$ and $Q$ into the problem. Claim: Line $PQ$ also passes through $R$. Proof. Note $(BP; I_BJ_B) = -1 = (CQ; J_CI_C)$, so the conclusion again follows by prism lemma. $\blacksquare$ We are now ready to complete the proof. Point $R$ is the exsimilicenter of the incircles of $\triangle ABK$ and $\triangle ACL$, so $\tfrac{PI_B}{RI_B} = \tfrac{QJ_C}{RJ_C}$. Now by Menelaus, \[\frac{I_BP}{PI} \cdot \frac{IQ}{QJ_C} \cdot \frac{J_CR}{RI_B} = -1 \implies IP = IQ.\] Second solution, inversion (Nikolai Beluhov) As above, the lines $BC$, $I_BJ_C$, $J_BI_C$ meet at some point $R$ (possibly at infinity). Let $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$ be the incircles of $\triangle ABK$, $\triangle ACL$, $\triangle ABL$, and $\triangle ACK$. Claim: There exists an inversion $\iota$ at $R$ swapping $\{\omega_1, \omega_2\}$ and $\{\omega_3, \omega_4\}$. Proof. Consider the inversion at $R$ swapping $\omega_1$ and $\omega_2$. Since $\omega_1$ and $\omega_3$ are tangent, the image of $\omega_3$ is tangent to $\omega_2$ and is also tangent to $BC$. The circle $\omega_4$ is on the correct side of $\omega_3$ to be this image. $\blacksquare$ Claim: Circles $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$ share a common radical center. Proof. Let $\Omega$ be the circle with center $R$ fixed under $\iota$, and let $k$ be the circle through $P$ centered at the radical center of $\Omega$, $\omega_1$, $\omega_3$. Then $k$ is actually orthogonal to $\Omega$, $\omega_1$, $\omega_3$, so $k$ is fixed under $\iota$ and $k$ is also orthogonal to $\omega_2$ and $\omega_4$. Thus the center of $k$ is the desired radical center. $\blacksquare$ The desired statement immediately follows. Indeed, letting $S$ be the radical center, it follows that $\overline{SP}$ and $\overline{SQ}$ are the common internal tangents to $\{\omega_1, \omega_3\}$ and $\{\omega_2, \omega_4\}$. Since $S$ is the radical center, $SP = SQ$. In light of $\angle SPI = \angle SQI = 90^{\circ}$, it follows that $IP = IQ$, as desired. Remark: [Nikolai Beluhov] There exists a circle tangent to all four incircles, because circle $k$ is orthogonal to all four, and line $BC$ is tangent to all four; thus the inverse of line $BC$ in $k$ is a circle tangent to all four incircles. The amusing thing here is that Casey's theorem is completely unhelpful for proving this fact: all it can tell us is that there is a line or circle tangent to these incircles, and line $BC$ already satisfies this property. Remark: [Generalization by Nikolai Beluhov] The following generalization holds: \begin{quote} Let $ABCD$ be a quadrilateral circumscribed about a circle with center $I$. A line through $A$ meets $\overrightarrow{BC}$ and $\overrightarrow{DC}$ at $K$ and $L$; another line through $A$ meets $\overrightarrow{BC}$ and $\overrightarrow{DC}$ at $M$ and $N$. Suppose that the incircles of $\triangle ABK$ and $\triangle ABM$ are tangent at $P$, and the incircles of $\triangle ACL$ and $\triangle ACN$ are tangent at $Q$. Prove that $IP = IQ$. \end{quote} The first approach can be modified to the generalization. There is an extra initial step required: by Monge, the exsimilicenter of the incircles of $\triangle ABK$ and $\triangle ADN$ lies on line $BD$; likewise for the incircles of $\triangle ABL$ and $\triangle ADM$. Now one may prove using the same trig approach that these pairs of incircles have a common exsimilicenter, and the rest of the solution plays out similarly. The second approach can also be modified in the same way, once we obtain that a common exsimilicenter exists. (Thus in the generalization, it seems we also get there exists a circle tangent to all four incircles.) Remark: [Author's comments on drawing the diagram] Drawing the diagram directly is quite difficult. If one draws $\triangle ABC$ first, they must locate both $K$ and $L$, which likely involves some trial and error due to the complex interplay between the two points. There are alternative simpler ways. For example, one may draw $\triangle AKL$ first; then the remaining points $B$ and $C$ are not related and the task is much simpler (though some trial and error is still required). In fact, by breaking symmetry, we may only require one application of guesswork. Start by drawing $\triangle ABK$ and its incircle; then the incircle of $\triangle ABL$ may be constructed, and so point $L$ may be drawn. Thus only the location of point $C$ needs to be guessed. I would be interested in a method to create a general diagram without any trial and error.

Without LOG let $B,K,L,C$ be on a line in this order. Let $D$ be the projection of $I$ onto $BC$ and $I_b,I_c,J_b,J_c$ be the incenters of $ABK,ACK,ABL,ACL$. Claim. $\dfrac{BI_b \cdot CI_c}{II_b \cdot II_c}=\dfrac{BJ_b \cdot CJ_c}{IJ_b \cdot IJ_c}$. Proof. It’s well known that $\dfrac{BI_b}{II_b}=\dfrac{AB}{AI} \cdot \dfrac{\sin BAI_b}{\sin IAI_b}$ and $\dfrac{CI_c}{II_c}=\dfrac{AC}{AI} \cdot \dfrac{\sin CAI_c}{\sin IAI_c}$ (easily follows by two applications of sine law). Multiplying and taking into account that $\angle BAI_b=\angle IAI_c, \angle IAI_b=\angle CAI_c$ we get $\dfrac{BI_b \cdot CI_c}{II_b \cdot II_c}=\dfrac{AB \cdot AC}{AI^2}$. Let $IP=x$ and $IQ=y$. Obviously, $BI_b=\dfrac{BI-x}{1+\sin \frac\beta 2}, CI_c=\dfrac{CI-y}{1-\sin \frac \gamma 2}$. Now it’s easy to get $\dfrac{BI_b \cdot CI_c}{II_b \cdot II_c}=\dfrac{(BI-x)(CI-y)}{(x+DI)(y-DI)}$, so Claim rearranges to $\dfrac{x+DI}{x-DI}=\dfrac{y+DI}{y-DI}$ which implies $x=y$ as desired.

General problem. Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are any two points on segment $BC$. Let $\omega_{BK}$, $\omega_{CK}$, $\omega_{BL}$, and $\omega_{CL}$ be the incircles of triangles $ABK$, $ACK$, $ABL$, and $ACL$ respectively. Then, the pairs $(\omega_{BK}, \omega_{BL})$ and $(\omega_{CK}, \omega_{CL})$ have the insimilicenters $P$, $Q$ respectively. Let $R_{BK}$, $R_{CL}$ be the radius of circles $\omega_{BK}$, $\omega_{CL}$ respectively. Prove that $$\frac{IP}{IQ}=\frac{PI_{BK}}{R_{BK}}:\frac{QI_{CL}}{R_{CL}}.$$When $P$ lies on $\omega_{BK}$ and $Q$ lies on $\omega_{CL}$ then easily seen ${PI_{BK}}={R_{BK}}$ and ${QI_{CL}}={R_{CL}}$, so that $\frac{PI_{BK}}{R_{BK}}:\frac{QI_{CL}}{R_{CL}}=1$ or $IP=IQ$.

Let $M,N,R,S$ be the incenters of triangles $\triangle ABK, \triangle ACL, \triangle ABL, \triangle ACK$ respectively. We have $(B,M;R,I) = (C,N;S,I)$ (for example, by projecting from $A$), so by the "Prism lemma", lines $BC, MN, RS$ concur at some point $X$. Since $(B,P;M,R) = (C,Q;N,S) = -1$, $PQ$ also passes through $X$. Use Menelaus on $\triangle IMN$ and line $XPQ$, and noticing that $\frac{XM}{XN} = \frac{MP}{MQ}$, we have $IP=IQ$ as desired.

Idio-logy wrote: Let $M,N,R,S$ be the incenters of triangles $\triangle ABK, \triangle ACL, \triangle ABL, \triangle ACK$ respectively. We have $(B,M;R,I) = (C,N;S,I)$ (for example, by projecting from $A$), so by the "Prism lemma", lines $BC, MN, RS$ concur at some point $X$. Since $(B,P;M,R) = (C,Q;N,S) = -1$, $PQ$ also passes through $X$. Use Menelaus on $\triangle IMN$ and line $XPQ$, and noticing that $\frac{XM}{XN} = \frac{MP}{MQ}$, we have $IP=IQ$ as desired. excuse me, can you tell me about Prism lemma?

Idio-logy appears to be referencing the following lemma: Let $\ell_1, \ell_2$ be two lines and let $P = \ell_1 \cap \ell_2$. Let $A, B, C \in \ell_1$ and $X, Y, Z \in \ell_2$. Then $AX, BY, CZ$ concur iff $(A, B; C,P) = (X, Y; Z, P)$.

Solution with amar_04 and A-Thought-Of-God. The credit in making the diagram goes to A-Thought-Of-God . tastymath75025 wrote: Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are chosen on segment $BC$ such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$, and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$. Prove that $IP=IQ$. Ankan Bhattacharya [asy][asy] import olympiad; import math; import geometry; size(15cm); defaultpen(fontsize(10pt)); pair B = (0,0), C = (9,0), A = (2,6), K = (1.04,0) , L = (5.77,0), P=(1.04,0.76), Q=(5.37,1.35), Ibk = incenter(A,B,K), Ick = incenter(A,C,K), Ibl = incenter(A,B,L), Icl = incenter(A,C,L), I = incenter(A,B,C), U = extension(Q,P,B,C); D(incircle(A,B,K),green+blue); D(incircle(A,C,L),green+blue); D(incircle(A,B,L),red+blue); D(incircle(A,C,K),red+blue); D(A--B--C--A,red); D(A--K,blue); D(A--L,blue); D(I--B,cyan); D(I--C,cyan); D(U--Q,yellow); D(U--B,red); D(U--Ick,yellow); D(U--Icl,yellow); D("A",A,N); D("B",B,S); D("C",C,dir(270)); D("K",K,dir(270)); D("L",L,dir(270)); D("I_{1}",Ibk,dir(140)); D("I_{4}",Ick,S); D("I_{2}",Ibl,S); D("I_{3}",Icl,NE); D("I",I,dir(110)); D("P",P,N); D("Q",Q,SE); D("U",U,N); [/asy][/asy] Suppose $\overline{PQ}$ intersects $\overline{BC}$ at $U$ and $I_1, I_2, I_3, I_4$ denote the incenters of $\triangle ABK, \triangle ABL, \triangle ACL, \triangle ACK$ respectively. Now, we note that $$(B,P;I_1, I_2)=(C,Q;I_3,I_4) = -1\text{ (equal cross ratios). }$$Since $B, P$ are the exsimilicenter and insimilicenter of incircles of $\triangle ABK$, $\triangle ABL$, and $I_1, I_2$ their centers, and similar thing for $C, Q, I_3, I_4$, hence $BC, I_1I_3, I_2I_4$ concur at $U$; also, $PQ$ passes through $U$ due to Prism Lemma. Hence, $U$ is the exsimilicenter of incircles of $\triangle ABK$ and $\triangle ACL$, so $$\frac{PI_1}{UI_1} = \frac{QI_3}{UI_3}$$Now, from Menelaus, we get: $$\frac{I_1P}{PI} \cdot \frac{IQ}{QI_3} \cdot \frac{I_3U}{UI_1} = -1 \implies IP = IQ. \quad \blacksquare$$

Let $I_{BK},I_{BL},I_{CK},I_{CL}$ be the incenters of $\triangle ABK, \triangle ABL, \triangle ACK, \triangle ACL$, respectively. Let $\omega_{\triangle XYZ}$ denote the incircle of any $\triangle XYZ$, and $r_{\triangle XYZ}$ denote its radius. Now remark that a rotation around $A$ with angle $\angle A/2$ sends the lines $\overline{AB},\overline{AI},\overline{AI_{BK}}, \overline{AI_{BL}}$ to the lines $\overline{AI},\overline{AC},\overline{AI_{CK}}, \overline{AI_{CL}}$, respectively. This implies that $(BI;I_{BK}I_{BL}) = (IC;I_{CK}I_{CL})$, so by prism lemma we see that lines $BC, I_{BK}I_{CL}, I_{BL}I_{CK}$ concur at a point $T$. Let $k$ be the scale factor of the homothety sending $\omega_{\triangle ACL}$ to $\omega_{\triangle ACK}$. We may compute $k$ in two ways: note that $$k = \frac{CI_{CK}}{CI_{CL}} = \frac{r_{\triangle ACK}}{r_{\triangle ACL}} = \frac{QI_{CK}}{QI_{CL}}.$$ This implies that $-1 = (CQ;I_{CL}I_{CK})$. Similarly $-1 = (BP;I_{BK}I_{BL})$, so line $PQ$ also passes through $T$ by prism lemma. Note that $T$ is the exsimilicenter of $\omega_{\triangle ABK}$ and $\omega_{\triangle ACL}$, so $\tfrac{PI_{BK}}{TI_{BK}} = \tfrac{QI_{CL}}{TI_{CL}}$. Now note that by Menelaus on $\triangle II_{BL}I_{CK}$ we see that $IP=IQ$, as desired.

masadca wrote: As you may see, I am not an Olympiad. Ok sorry but this just made me laugh so hard As for your actual question: many of the people taking this test are 11th and 12th graders. You're in 8th grade right now: think about how good you are at math now, and also think about how good you were at it in 5th and 4th grade. 4 years is a long time for improvement.

tastymath75025 wrote: Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are chosen on segment $BC$ such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$, and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$. Prove that $IP=IQ$. Ankan Bhattacharya This wasnt too hard but yeah but quite interesting beauty.My solution pretty much simiar to others Let $I_B$, $J_B$, $I_C$, $J_C$ be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACK$, $\triangle ACL$ respectively. [asy][asy]size(12cm); defaultpen(fontsize(10pt)); pen pri=royalblue; pen sec=deepcyan; pen tri=rgb(41, 207, 255); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; real r=1.7353; pair A, B, C, I, IB, P,Q, IC, K, L, JB, JC, T; A=dir(120); B=dir(210); C=dir(330); I=incenter(A,B,C); IB=(7B+r*I)/(7+r); P=IB+unit(I-B)*length(IB-foot(IB,B,C)); Q=I+unit(C-I)*length(P-I); IC=incenter(C,extension(A,C,Q,Q+A-foot(A,C,I)),extension(B,C,Q,Q+A-foot(A,C,I))); K=extension(B,C,A,2*foot(B,A,IB)-B); L=extension(B,C,A,2*foot(C,A,IC)-C); JB=incenter(A,B,L); JC=incenter(A,C,K); T=extension(IB,IC,B,C); draw(B--I--C,sec); draw(Q--T,sec); draw(JC--T--IC,sec); fill(A--B--C-- cycle,fil); draw(B--A--C--T,pri); draw(K--A--L,pri); filldraw(incircle(A,B,L),tfil,tri); filldraw(incircle(A,B,K),tfil,tri); filldraw(incircle(A,C,L),tfil,tri); filldraw(incircle(A,C,K),tfil,tri); dot("$A$",A,N); dot("$B$",B,S); dot("$C$",C,SE); dot("$T$",T,SW); dot("$I_B$",IB,dir(285)); dot("$I_C$",IC,dir(255)); dot("$J_B$",JB,NW); dot("$J_C$",JC,NE); dot("$P$",P,dir(85)); dot("$Q$",Q,dir(120)); dot("$K$",K,S); dot("$L$",L,S); dot("$I$",I,N);[/asy][/asy] Claim: $I_BI_C$,$J_BJ_C$,$BC$ are concurrent say at a point $T$ Proof :Rotating $(BJ_B;I_BI)$ through $A$ by $\tfrac{A}{2}$ we get $(BJ_B;I_BI)=(CJ_C;I_CI)$ thus by the Prism Lemma $BC,I_BJ_B,I_BI_C$ are concurrent$\square$ Claim : $PQ$ passes through $T$ Proof :Note that $P,B$ are the Homothety centers of incircle of $\triangle ABK$ and $\triangle ABL$.Hence $-1=(BP;I_BJ_B)$ and similarly $-1=(CQ;I_CJ_C)$.Hence again by Prism Lemma $PQ$ passes through $T$ Now by Menelaus Theorm \[-1=\dfrac{IQ}{QI_C}\cdot\dfrac{IC_T}{TI_B}\cdot\dfrac{I_BP}{PI}=\dfrac{IQ}{PI}\left(\dfrac{TI_C}{QI_C}\cdot \dfrac{I_BP}{TI_B}\right)=\dfrac{IQ}{PI}\Longleftrightarrow IP=IQ\]and we are done.

$\textbf{LEMMA}$(anti-homologous points):In every two points which are collinear with respect to a similitude center but are not homologous points, The product of distances from a homothetic center to two anti-homologous points is a constant. The other property of this points is that in every pair of anti-homologous points there is a circle which is tangent to the two other circles at that points. The proof is easy just by thales theorem and homothety. now let's solve the problem: it is obvious that $(BP,I_1I_3)=-1=(CQ,I_2I_4)$ so it implies that $I_3I_4,PQ,I_1I_2,BC$ are conccurent at external homothetic center of $(I_1)$ and $(I_2)$ named $T$ now we just apply the $\textbf{LEMMA}$on anti-homologous points $P$ and $Q$ and also because $I_1P$ and $I_2Q$ intersect at I so $IP=IQ$ so we are done.$\blacksquare$ (the circle with radius $IP$ is tangent to $(I_1)$ and $(I_2)$ at points $P$ and $Q$) [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(22.15969431137711cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ real xmin = -29.88720851620066, xmax = 14.43218010655356, ymin = -15.439174074706292, ymax = 22.773450222129547; /* image dimensions */ pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); 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filldraw(circle((5.671787825599821,-0.42356159229146023), 2.1500307304818906), invisible, linewidth(2.) + linetype("2 2") + sexdts); /* draw figures *//* special point *//* special point *//* special point */ draw((-3.79365102990033,12.53635641196014)--(-9.586493200442964,-2.530740465116278), linewidth(2.) + dbwrru); draw((-9.586493200442964,-2.530740465116278)--(10.73125864894795,-2.5878128017718707), linewidth(2.) + dbwrru); draw((10.73125864894795,-2.5878128017718707)--(-3.79365102990033,12.53635641196014), linewidth(2.) + dbwrru); draw((-3.79365102990033,12.53635641196014)--K, linewidth(2.)); draw((-3.79365102990033,12.53635641196014)--L, linewidth(2.)); /* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw((5.67178782559982,-0.4235615922914604)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + linetype("2 2") + qqzzcc); draw((3.6954595639694174,0.42305446253944107)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + linetype("2 2") + qqzzcc); draw((-9.586493200442964,-2.530740465116278)--(-25.73875160010579,-2.4542508876070626), linewidth(2.) + qqzzcc); draw((-25.73875160010579,-2.4542508876070626)--(-0.8118343963566137,2.349887988824497), linewidth(2.) + qqzzcc); draw((-1.8261325606399366,2.7837665712829005)--(3.6954595639694174,0.42305446253944107), linewidth(2.) + dtsfsf); draw((1.2198937175515965,2.2705502786417124)--(0.6494332857778794,0.9362707551806287), linewidth(2.) + dtsfsf); draw((-1.8261325606399366,2.7837665712829005)--(-6.779651039942243,-0.6009271762414112), linewidth(2.) + dtsfsf); draw((-3.8935572632802167,0.49235628305876683)--(-4.7122263373019635,1.6904831119827222), linewidth(2.) + dtsfsf); draw((-6.779651039942243,-0.6009271762414112)--(-9.586493200442964,-2.530740465116278), linewidth(2.) + zzffqq); draw((3.6954595639694174,0.42305446253944107)--(10.73125864894795,-2.5878128017718707), linewidth(2.) + zzffqq); /* dots and labels */ label("$A$", (-3.5253347924341663,13.15983113479268), NE * labelscalefactor,uququq); 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Spoonfed the solution by jj_ca888, Psyduck909, and Spacesam Let the centers of the incircles of $ABK, ACK, ACL, ABL$ be $K_1, K_2, L_1, L_2$, respectively. Note that by a bit of angle chasing, we can find\[\angle BAK_1 = \angle IAK_2 = \tfrac12 \angle BAK \text{ and } \angle K_1AL_2 = \angle K_2AL_1 = \tfrac12 \angle KAL \text{ and } \angle L_2AI = \angle L_1AC = \tfrac12 \angle CAL.\]The angles formed by the sets of lines are the same, hence $(BL_2;K_1I) = (IL_1; K_2C) = (CK_2; L_1I)$. By Prism Lemma, it follows that $K_1L_1, K_2L_2, BC$ concur at $T$. Furthermore, note that $P$ and $Q$ are insimilicenters, hence it follows that $(BP; K_1P_2) = (CQ; L_1K_2) = -1$. Projecting $(BP; K_1P_2)$ through $T$ yields $(CQ'; L_1K_2) = -1$, where we let $Q' = TP \cap IC$. It follows that $Q' = Q$, hence $T \in PQ$. Lastly, let $\omega$ be the circle with center $I$ and radius $IP$, and $\omega_K, \omega_L$ be the incircles with centers $K_1, L_1$, respectively. By Monge on the three circles, we get that $T, P,$ and the insimilicenter $Z$ of $\omega$ and $\omega_L$ are collinear. $Z$ lies on $L_1I$, and must lie on $TP$ as well, hence $Z = Q$. Since $Q \in \omega_L$, it follows that $Q \in \omega$, hence $IP = IQ$ and we done. $\blacksquare$

I can see why Ankan likes this one so much Sketch: Let the incenters be $I_B$ ($ABK$), $J_B$ ($ABL$), $I_C$ ($AKC$), $J_C$ ($ALC$). 1. Deduce that $(I_B,J_B;B,I)=(J_C,I_C;C,I)$ by simply observing angles, and hence $BC,I_BJ_C,J_BI_C$ concur at some point $X$ by the Prism Lemma. 2. Deduce that $(B,P;I_B,J_B)=(C,Q;J_C,I_C)=-1$ by exsimilicenters/insimilicenters, and hence $PQ$ also passes through $X$ by Perspectives. 3. Finish with Menelaus.

Solved with MathJams. Let the incircles of $ABK, ABL, ACK, ACL$ be $\omega_1, \omega_2, \omega_3, \omega_4$ with centers $O_1, O_2, O_3, O_4$ respectively. Furthermore, let $r_1, r_2, r_3, r_4$ denote the radii of these $4$ circles, and let $r_i$ be the inradius of $ABC$. It's easy to see $O_1, O_2 \in BI$ and $O_3, O_4 \in CI$. Notice $$\angle O_1AO_3 = \angle O_1AK + \angle KAO_3 = \frac{\angle BAK}{2} + \frac{\angle KAC}{2} = \frac{\angle A}{2}.$$Analogously, we conclude $\angle O_2AO_4 = \frac{\angle A}{2}$. Now, consider a counterclockwise rotation of measure $\frac{\angle A}{2}$ about point $A$, and let the images of $B, O_1, O_2, I$ under this rotation be $B', O_1', O_2', I'$ respectively. It's easy to see $B', O_1', O_2', I'$ are collinear and $$B' \in AI, O_1' \in AO_3, O_2' \in AO_4, I' \in AC$$via previous angle chasing. Thus, $$(B, O_2; O_1, I) = (B', O_2'; O_1', I') \overset{A}{=} (I, O_4; O_3, C) = (O_3, C; I, O_4) = (C, O_3; O_4, I)$$by basic projections and cross ratio manipulations. Now, the Prism Lemma implies $BC, O_2O_3, O_1O_4$ are concurrent at some point $T$. Claim: $(O_1, O_2; B, P)$ and $(O_4, O_3; C, Q)$ are harmonic bundles. Proof. By properties of homotheties, we know $$\frac{BO_1}{BO_2} = \frac{r_1}{r_2} = - \frac{PO_1}{PO_2}$$where lengths are directed, implying the the first result by the definition of the cross ratio. The second result follows analogously. $\square$ Claim: $T \in PQ$. Proof. Let $TP \cap CI = Q'$. Then, $$-1 = (O_1, O_2; B, P) \overset{T}{=} (O_4, O_3; C, Q').$$But $(O_4, O_3; C, Q) = -1$, so $Q' = Q$ as desired. $\square$ As a result, we know $$(I, O_1; P, B) \overset{T}{=} (I, O_4; Q, C)$$so $$\frac{IP}{PO_1} \div \frac{BI}{BO_1} = \frac{IQ}{QO_4} \div \frac{CI}{CO_4} \implies \frac{IP}{IQ} = \frac{CO_4 \cdot PO_1 \cdot BI}{BO_1 \cdot QO_4 \cdot CI}$$(where the direction of lengths is irrelevant). To finish, we observe that homotheties imply $$\frac{IP}{IQ} = \frac{BI}{BO_1} \cdot \frac{CO_4}{CI} \cdot \frac{PO_1}{QO_4} = \frac{r_i}{r_1} \cdot \frac{r_4}{r_i} \cdot \frac{r_1}{r_4} = 1$$as desired. $\blacksquare$ Remark: $A$ is the Miquel Point of $O_1IO_3K$ and $O_2IO_4L$. (This can be verified with basic angle chasing.)

USA TSTST 2019 P9 wrote: Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are chosen on segment $BC$ such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$, and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$. Prove that $IP=IQ$. Denote the incenters of $\triangle ABK, \triangle ABL, \triangle ACK,$ and $\triangle ACK$ as $I_B, J_B, I_C,$ and $J_C,$ respectively. We begin with a crucial claim: Claim- $BC$, $I_BI_C$, $J_BJ_C$, and $PQ$ concur at some point $X$. Proof. We can see that $BC$, $I_BI_C$, $J_BJ_C$ concur at some point $X$ by Prism Lemma since $$A(B, I; I_B, J_B)=A(I, C; I_C, J_C).$$Since $P, B$ are insimilicenter, and exsimilicenters of incenters of $\triangle ABK$, $\triangle ABL$ $$-1=(B, P; I_B, J_B)=(C, Q; I_C, J_C)$$So by Prism Lemma again implies that $PQ$ passes through $X.\square$ [asy][asy] import graph; size(17cm); 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 = -30.53, xmax = 7.65, ymin = -2.46, ymax = 14.78; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ubqqys = rgb(0.29411764705882354,0.,0.5098039215686274); pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((-6.37,9.94)--(-11.,2.)--(3.,2.)--cycle, linewidth(0.5) + zzttqq); /* draw figures */ draw((-6.37,9.94)--(-11.,2.), linewidth(0.5) + zzttqq); draw((-11.,2.)--(3.,2.), linewidth(0.5) + zzttqq); draw((3.,2.)--(-6.37,9.94), linewidth(0.5) + zzttqq); draw((-11.,2.)--(-5.545193307851359,5.133647281153756), linewidth(0.5) + xdxdff); draw((-5.545193307851359,5.133647281153756)--(3.,2.), linewidth(0.5) + xdxdff); draw(circle((-9.42594737269626,2.9042530770234087), 0.9042530770234085), linewidth(0.5) + red); draw((-6.37,9.94)--(-8.753257409829024,2.), linewidth(0.5) + ubqqys); draw(circle((-6.329313718559153,4.683190109747784), 2.6669871773326093), linewidth(0.5) + red); draw((-6.37,9.94)--(-1.612508515592935,2.), linewidth(0.5) + ubqqys); draw(circle((-0.8190089387275968,3.4004863958489855), 1.4004863958489857), linewidth(0.5) + red); draw(circle((-4.872503344984124,4.886962039842536), 2.8869620398425346), linewidth(0.5) + red); draw((-4.872503344984124,4.886962039842536)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + qqwuqq); draw((-0.8190089387275968,3.4004863958489855)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + qqwuqq); draw((-11.,2.)--(-25.794084103041374,1.9605476148542396), linewidth(0.5) + zzttqq); /* dots and labels */ dot((-6.37,9.94),linewidth(3.pt) + dotstyle); label("$A$", (-6.29,10.1), NE * labelscalefactor); dot((-11.,2.),linewidth(3.pt) + dotstyle); label("$B$", (-11.5,1.35), NE * labelscalefactor); dot((3.,2.),linewidth(3.pt) + dotstyle); label("$C$", (3.09,1.4), NE * labelscalefactor); dot((-5.545193307851359,5.133647281153756),linewidth(3.pt) + dotstyle); label("$I$", (-5.7,5.3), NE * labelscalefactor); dot((-9.42594737269626,2.9042530770234087),linewidth(3.pt) + dotstyle); label("$I_B$", (-10.2,2.9), NE * labelscalefactor); dot((-8.753257409829024,2.),linewidth(3.pt) + dotstyle); label("$K$", (-8.81,1.4), NE * labelscalefactor); dot((-8.64186668623104,3.3546874278654695),linewidth(3.pt) + dotstyle); label("$P$", (-8.8,3.66), NE * labelscalefactor); dot((-6.329313718559153,4.683190109747784),linewidth(3.pt) + dotstyle); label("$J_B$", (-7.2,4.8), NE * labelscalefactor); dot((-1.612508515592935,2.),linewidth(3.pt) + dotstyle); label("$L$", (-1.79,1.4), NE * labelscalefactor); dot((-4.872503344984124,4.886962039842536),linewidth(3.pt) + dotstyle); label("$I_C$", (-4.83,4.9), NE * labelscalefactor); dot((-0.8190089387275968,3.4004863958489855),linewidth(3.pt) + dotstyle); label("$J_C$", (-0.9,3.56), NE * labelscalefactor); dot((-25.794084103041374,1.9605476148542396),linewidth(3.pt) + dotstyle); label("$X$", (-26.03,1.3), NE * labelscalefactor); dot((-2.147226471973991,3.8875631783032922),linewidth(3.pt) + dotstyle); label("$Q$", (-2.7,4.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now by Menelaus' Theorem on $\triangle II_BJ_C$, using that $\frac{PI_B}{I_BX}=\frac{QJ_C}{J_CX}$ we get that$$\frac{IP}{PI_B}\cdot\frac{IB_X}{XJ_C}\cdot\frac{J_CQ}{IQ}=-1\implies IP=IQ$$

Denote $I_1, I_2, I_3, I_4$ the incenters of $\triangle ABK, \triangle ABL, \triangle ACK, \triangle ACL$ respectively. First we present the following lemma: Lemma. $I_1I_4, I_2I_3$, and $BC$ concur. Use the prism lemma. In particular, $$(BI_1;I_2I) = A(BI_1;I_2I) = \frac{\sin \angle BAI_2}{\sin \angle BAI} \div \frac{\sin \angle I_1AI_2}{\sin \angle I_1AI} = \frac{\sin \frac 12 \angle BAL}{\sin \frac 12 \angle BAC} \div \frac{\sin \frac 12 \angle KAL}{\sin \frac 12 \angle KAC},$$which is symmetric and thus also equals $(CI_4; I_3I)$. Thus the lines concur by prism. $\blacksquare$ Now $(BP;I_1I_2)=-1$ and similarly $(CQ;I_4I_3)=-1$, which implies that $\overline{PQ}$ passes through the concurrency point $X$ as well. Therefore, we can conclude that $$(BP;I_1I) = \frac{BI_1}{BI} \cdot \frac{PI}{PI_1} = \frac{\frac{r_1}{\sin \frac B2}}{BI} \cdot \frac{IP}{r_1} = \frac{IP}{BI \sin \frac B2}.$$Similarly, $(CQ; I_4I) = \frac{IQ}{CI \sin \frac C2}.$ But $BI \sin \frac B2 = CI \sin \frac C2$, so $IP=IQ$ as required.

sup nerds WLOG suppose that the inradius of $\triangle ABK$, which we call $r_1$, is greater than the inradius of $\triangle ABL$, and that the inradius of $\triangle ACK$, which we call $r_2$, is less than the inradius of $\triangle ACL$. Now, if we let $\angle ABC=2\angle B$ (to avoid fractions), and we let $\angle AKB=2\theta$, then we eventually get that $r_1=b\sin C(\cos C-\sin C\cdot \cot \theta)$ and $r_2=c\sin B(\cos B - \sin B\cdot \tan \theta)$ so we see that they must satisfy the equation $$cr_1\sin 2B + br_2\sin 2C-2r_1r_2=2bc\sin B\sin C\cos (B+C).$$To show that $IP=IQ$ we require $$\frac{a\sin C}{\sin(B+C)}-\left(\frac{1-\sin B}{1+\sin B}r_1\right)\left(1+\frac{a\sin C}{r\sin (B+C)}\right)=\frac{a\sin B}{\sin(B+C)}-r_2\left(1+\frac{a\sin B}{r\sin(B+C)}\right)$$where $r$ is the inradius of $\triangle ABC$. Repeating the same logic from earlier we can also see that $$c\sin 2B\cdot \frac{1-\sin B}{1+\sin B}r_1+b\sin 2C\cdot \frac{1+\sin C}{1-\sin C}r_2-\frac{2(1-\sin B)(1+\sin C)}{(1+\sin B)(1-\sin C)}r_1r_2=cr_1\sin 2B+br_2\sin 2C-2r_1r_2.$$Note that from here we're basically done since we combine this with the other equation to get some line which should match the $IP=IQ$ condition. We begin the calculation. The equation is equivalent to $$\frac{4(\sin B-\sin C)}{(1+\sin B)(1-\sin C)}r_1r_2-c\sin 2B\frac{2\sin B}{1+\sin B}r_1+b\sin 2C\frac{2\sin C}{1-\sin C}r_2=0$$so if we combine this with the first expression to remove the $r_1r_2$ term we end up with the equation $$2b\sin 2C\sin B(1+\sin C)r_2-2c\sin 2B\sin C(1-\sin B)r_1=4bc\sin B\sin C\cos(B+C)(\sin B-\sin C).$$Note that $r=\frac{a\sin B\sin C}{\sin(B+C)}$ so the original equation is actually $$\frac{1+\sin C}{\sin C}r_2-\frac{1-\sin B}{\sin B}r_1=\frac{a\sin B}{\sin(B+C)}-\frac{a\sin C}{\sin(B+C)}$$so it would be enough that $$2c\sin 2B\sin B\sin C(\sin B-\sin C)\left(\frac{a}{\sin(B+C)}\right)=4bc\sin B\sin C\cos(B+C)(\sin B-\sin C)$$or that $$\sin 2B\cdot \frac{a}{\sin(B+C)}=2b\cos(B+C)$$or that $$\frac{\sin 2B}{b}=\frac{\sin(2B+2C)}{a}=\frac{\sin 2A}{a}$$which is just the Law of Sines. We are done. $\blacksquare$

Let the incenters of $\triangle ABK, \triangle ABL, \triangle ACL, \triangle ACK$ be $I_B,J_B.I_C,J_C$ respectivily and note that $$(B, I; I_B, J_B) \overset{\text{rotation by} \; \frac{\angle BAC}{2}}{=} (I, C; J_C, I_C)=(C, I; I_C, J_C)$$Now using projections its easy to know that $J_BJ_C, I_BI_C, BC$ are concurrent at $T$ and since $P$ is insimilicenter of the incircles of $\triangle ABK, \triangle ABL$ and $Q$ is insimilicenter of the incircles $\triangle ACK, \triangle ACL$ we have. $$(C, Q; I_C, J_C)=-1=(B, P; I_B, J_B)$$Which means that $PQ$ also passes through $T$, now consider an inversion with center $T$ that sends the incircle of $\triangle ABL$ to the incircle of $\triangle ACK$, note that $T$ is the exsimilicenter of the incircles of $\triangle ABK, \triangle ACL$ and of the incircles of $\triangle ABL, \triangle ACK$ so we have that the inverse of the incircle of $\triangle ABK$ is a circle tangent to the incircle of $\triangle ACK$ and this circle eith the incircle of $\triangle ABK$ share same exsimilicenter $T$ hence the inverse of the incircle of $\triangle ABK$ is the incircle of $\triangle ACL$ which means that the inverse of $P$ is $Q$, to the final step, build a circle $\omega$ that passes through $P,Q$ and its tangent to the incircles of $\triangle ABK, \triangle ABL$ after inversion $\omega$ is fixed but that means that $\omega$ is also tangent to the incircles of $\triangle ACK, \triangle ACL$ meaning that the center of this circle $\omega$ is $I$, hence $IP=IQ$ thus we are done

Great problem! Here is a slightly different solution to the above ones. Let $\omega_1$, $\omega_2$, $\omega_3$ and $\omega_4$ be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle AKC$ and $\triangle ALC$, with centers $I_1$, $I_2$, $I_3$ and $I_4$, respectively. I will prove that these four circles have a radical center. To that end, let $U$ and $V$ be the insimilicenters of $\omega_1$ and $\omega_3$, and of $\omega_2$ and $\omega_4$. Claim: $P-Q-U-V$ are colinear. Pf: As in above solutions, we use the prism lemma to prove that $I_1I_4$, $I_2I_3$ and $BC$ concur (for completeness this can also be achieved by Monge), as well as $PQ$; the concurrence point $S$ is the exsimilicenter of $\omega_2$ and $\omega_3$. Now, by Monge's Theorem on $\omega_1$, $\omega_2$ and $\omega_3$ we known that the insimilicentre of $\omega_1$ and $\omega_2$ ($P$), the insimilicentre of $\omega_1$ and $\omega_3$ ($U$) and the exsimilicentre of $\omega_2$ and $\omega_3$ ($S$) are colinear. Hence $P-Q-S$. Similarly, we also find that $Q-V-S$ are also colinear. $\blacksquare$ Using the claim, we now apply the Simson Theorem on $\triangle II_1I_3$, and points $P\in II_1$, $U\in I_1I_3$ and $Q\in II_3$. Since $P-U-V$ are colinear, the perpendiculars through each of these points to $II_1$, $I_1I_3$ and $II_3$, respectively, concur at a point $X$. However, these lines are in fact the pairwise radical axes of $\omega _1$ and $\omega_2$, $\omega_1$ and $\omega_3$ and $\omega_3$ and $\omega_4$. Thus, $X$ is the radical center of the four incircles. However, that means that $$PX^2=P_{\omega_1}(X)=P_{\omega_3}(X)=QX^2\implies PX=QX\implies IP=IQ,$$as desired. $\square$

Let $I_1$, $I_2$, $I_3$, $I_4$ be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle AKC$, $\triangle ALC$ respectively. Claim: Lines $BC$, $I_1I_4$, $I_2I_3$, and $PQ$ concur and a point $X$. Proof. We first prove that $(BI_2;I_1I)=(CI_3;I_4I)$. More precisely, we prove $(AB,AI_2;AI_1,AI)=(AC,AI_3;AI_4,AI)$. To do so, notice that \[\measuredangle BAI_1= \frac 12 \measuredangle BAK =\frac 12\left( \measuredangle BAC-\measuredangle KAC\right) =\measuredangle IAC-\measuredangle I_3AC=\measuredangle IAI_3\]and similarly $\angle I_2AI=\measuredangle I_4AC$. Thus pencils $AB,AI_2,AI_1,AI$ and $AC,AI_3,AI_4,AI$ are rotations by half of $\angle BAC$ and the cross ratios are equal. The concurrence of $I_2I_3$, $I_1I_4$, $BC$ follows by prism lemma. Also, from in/ex similicenter harmonic config, $-1=(BQ;I_1I_2)=(CP;I_4I_3)$ and prism lemma again implies the result. $\blacksquare$ Now, let $\omega$ be the circle with center $I$ through $P.$ Notice that the insimilcenter of $\omega$ and $\omega_4$ is on line $IC.$ But also, by Monge on $\omega_1,\omega_4,\omega$ we have that this insimilicenter line on line $XQ$. Thus the insimilicenter is $P=HQ\cap IC$ implying that $\omega$ and $\omega_4$ are tangent at $P$ so $P$ lies on $\omega$ and we are done.

Nice!

Claim 1: $I_BI_C,J_BJ_C,BC$ concur.

Now, Clearly $(IP;I_BJ_B)=(IQ,I_CJ_C)$ therfore $P-Q-T$; now by menelaus we may finish easily.

Let $I_{BK}$, $I_{BL}$, $I_{CL}$, and $I_{CK}$, be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACL$, and $\triangle ACK$ respectively. Observe that $(I_{BK}I_{BL}; BI) \stackrel{\text{rot} \tfrac{1}{2}\angle BAC}= (I_{CL} I_{CK}; CI)$. Now prism lemma implies $\overline{I_{BK}I_{CL}}$, $\overline{I_{BL}I_{CK}}$, and $\overline{BC}$ concur, call this concurrency point $N$. Note that due to insimilar/exsimilar centers $-1=(CQ; I_{CK} I_{CL})= (BP; I_{BL} I_{BK})$. So $N \in \overline{PQ}$. Consider the circle centered at $I$ with radius $PI$, then let the insimilar center of this circle and the incircle of $\triangle AKC$ be $Q'$. Note that by monge $Q'$, $P$, and $N$ are collinear also note that $Q' \in \overline{IQ}$ therefore $Q'=Q$ and we can conclude.

Let the centers of the incircles of $\triangle ABK$, $\triangle ABL$, $\triangle AKC$ and $\triangle ALC$ be $M$, $N$, $U$, $V$ respectively. Notice that $\angle MAU = \angle NAV = \frac{A}{2}$ so $M \rightarrow AM \rightarrow AU \rightarrow U$ is projective and hence we have $(I,B;N,M) = (I,C;V,U)$. Let $BC$ intersect $MV$ at $T$ (possibly at infinity) and let $TU$ intersect $IB$ at $N'$. Projecting through $T$ we get: $(I,C;V,U) = (I,B;N',M) = (I,B;N,M)$ so $N' = N$ meaning that $NU$, $MV$ and $BC$ are concurrent. Now let $r_{1}$ and $r_{2}$ be the inradii of the incircles of $\triangle ABK$ and $\triangle ABL$. Since $B$ is the intersection of the common tangents to these circles from homothety we have that $\frac{BM}{BN} = \frac{r_{1}}{r_{2}} = \frac{PM}{PN}$ hence $(B,P;M,N)= -1$. Similarly we get $(C,Q;V,U) = -1$ so proceeding like before we get that $PQ$, $MV$ and $BC$ meet at $T$. Finally, $T$ is the intersection of the exterior common tangents to the incircles of $\triangle ABK$ and $\triangle ALC$. Denote by $r_{1}$ and $r_{4}$ they're inradii respectively. By homothety again we get $\frac{TM}{TV} = \frac{r_{1}}{r_{4}} = \frac{PM}{QV}$. And from Menelaus in $\triangle IMV$ we get $$\frac{IQ}{QV}\cdot \frac{VT}{TM}\cdot \frac{MP}{PI}= 1$$ So by the above we get $IP = IQ$. Done!

Let the incenters of $\triangle ABK, \triangle ABL, \triangle ACL, \triangle ACK$ be $B_1$, $B_2$, $C_1$, and $C_2$ respectively. Then $B$ and $P$ are the exsimilicenter and insimilicenters of the incircles of $\triangle AKK$ and $\triangle ABL$. So this implies that $\frac{BB_1}{BB_2} = \frac{PB_1}{PB_2} \implies (B, P; B_1, B_2) = (C, Q; C_1, C_2)= -1$. Then notice that $\angle B_1AC_2 = C_1AB_2 = \frac{\angle A}{2}$ which implies that $(AB, AB_2; AB_1, AI) = (AI, AC_1; AC_2, AC)$. Then Prism Lemma implies that $BC$, $B_1C_1$, and $B_2C_2$ concur at some point $K$. Then applying it again with the knowledge that $(B, P; B_1, B_2) = (C, Q; C_1, C_2)= -1$ we get that $PQ$ also passes through this point. Notice that since $K$ lies on the external tangent of the circles with centers $B_1$ and $C_1$ and lies on the line passing through the centers it is the exsimilicenter of both circles, and similarly for the circles with center $B_2$ and $C_2$. Let $r(XYZ)$ be the inradius of some triangle with vertices $\triangle XYZ$. Then Menelaus's weak theorem implies that \[\frac{KC_1}{KB_1} \cdot \frac{B_1P}{PI} \cdot \frac{IQ}{QC_1} = 1.\]However, $\frac{KC_1}{KB_1} = \frac{QC_1}{B_1P} = \frac{r(\triangle ACL)}{r(\triangle ABK)} \implies \frac{IP}{IQ} = 1$ as desired.

Let $\beta_1, \beta_2, \gamma_1, \gamma_2$ be the incircles of $\triangle ABK$, $\triangle ABL$, $\triangle ACL$, and $\triangle ACK$, respectively. Let $B_1, B_2, C_1, C_2$ be the centers of these incircles, respectively. We have \begin{align*} \angle BAB_1 &= \frac{\angle BAK}2 = \angle IAC_2\\ \angle BAB_2 &= \frac{\angle BAL}2 = \angle IAC_1\\ \angle BAI &= \frac{\angle BAC}2 = \angle IAC.\\ \end{align*}Therefore, \begin{align*} (AB, AI; AB_1, AB_2) &= (AI, AC; AC_2, AC_1)\\ &= (AC, AI; AC_1, AC_2). \end{align*}Because $B$, $B_1$, $B_2$, and $I$ lie on the bisector of $\angle B$, they are collinear. The same is true of $C$, $C_1$, $C_2$, and $I$. So, \[(BI;B_1B_2) = (CI;C_1C_2).\]By the prism lemma, $BC$, $B_1C_1$, and $B_2C_2$ concur at some point $E$. Because \[(BP;B_1B_2)=(CQ;C_1C_2)=-1,\]by the prism lemma, $PQ$ passes through $E$. Therefore, $PQ$ forms an equal angle with $\beta_1$ and $\gamma_1$. So, if the tangent to $\beta_1, \beta_2$ at $P$ and the tangent to $\gamma_1, \gamma_2$ at $Q$ intersect at $X$, then $\triangle XPQ$ is isosceles. Since $I$ is the antipode of $X$ on $(XPQ)$, $IP=IQ$.

Solved with hints a while back and posting for storage.

Let $X_K, X_L, Y_K, Y_L$ be the incenters of triangles $ABK, ABL, ACK, ACL$, which lie on angle bisectors $BI, CI$. First notice an $\angle A/2$ rotation gives \[(AB, AX_K; AX_L, AI) = (AI, AY_K; AY_L, AC) \implies (BX_K; X_LI) = (CY_L; Y_KI),\] so Prism Lemma tells us $BC$, $X_KY_L$, $X_LY_K$ concur at point, say $T$. Then length ratios give \[-1 = (BP; X_KX_L) = (CQ; Y_LY_K) \implies T \in PQ\] again by Prism Lemma. We finish by using Menalaus on $\triangle IX_KY_L$ with transversal $PQT$, as \[1 = \frac{IP}{PX_K} \cdot \frac{X_KT}{TY_L} \cdot \frac{Y_LQ}{QI} = \frac{IP}{r_{XK}} \cdot \frac{r_{XK}}{r_{YL}} \cdot \frac{r_{YL}}{QI} \implies IP = IQ. \quad \blacksquare\]