Just the details: 1. Let $X$ be on $\overline{BC}$ such that $IX\perp IO$ and let $A_1$ be the reflection of $A$ in $OI.$ 2.Consider $B_1$ the second intersection of $AB$ with $(BA_1X)$ and $C_1$ the second intersection of $AC$ with $(CA_1X).$ 3.By this we know that $\triangle{IXA_1}\sim\triangle{AIA_1}.$ 4. Easy angle chasing now yields that $l=\overline{B_1-X-C_1}$, so $K=B_1,L=C_1.$ 5. Since $AA_1(\perp OI)$ is the radical axis of $(AKL)$ and $(ABC)$, the problem is now clear.

Let the incircle of $ABC$, $\omega$, be tangent to $BC,CA,AB$ at $X,Y,Z$ respectively. Observe that $KL$ is tangent to $\omega$ at, say, $T$. Under the inversion wrt $\omega$, the circumcircles of $ABC$ and $AKL$ go to nine-point circles of $XYZ$ and $TYZ$. Therefore $I,O$ and $H$, the orthocenter of $XYZ$, are collinear and we want to prove that $Q$, the orthocenter of $TYZ$, lies on $IH$. Since $\vec{XH}=\vec{TQ}$, $HQ||XT||IH$, as desired.

What a nice problem! Let $X$ be the point on $BC$ such that $\angle OIX = 90^{\circ}$ and let $X'$ be the reflection of $X$ across $I$. Let $M_B, M_C$ be the intersection of rays $BI, CI$ and $\odot(ABC)$. By Butterfly theorem, $X'\in M_BM_C$. But $\overline{M_BM_C}$ is the perpendicular bisector of $AI$. Thus $AX' = X'I$. Let $P$ be the point on $\odot(ABC)$ such that $AP\parallel KL$ and let $A', P'$ be the reflections of $A,P$ across $OI$. By reflection, $A'P'\parallel BC$ thus $AA', AP'$ are isogonal w.r.t. $\angle BAC$. However, as $\measuredangle IAX' = \measuredangle X'IA = \measuredangle A'IA$, we get $A, X', P'$ are colinear. Reflecting back across $OI$ gives $A', X, P$ are colinear. Finally, notice that $$\measuredangle XLC = \measuredangle PAC = \measuredangle PA'C = \measuredangle XA'C$$thus $A'\in\odot(XLC)$. Similarly $A'\in\odot(XKB)$ so by Miquel's theorem, $A'\in\odot(AKL)$. As perpendicular bisector of $AA'$ is $OI$, we are done.

Let $M$ be the reflection of $A$ over $OI.$ By ARMO 2017 we have that $TM=TI,$ where $T$ is the intersection of $s$ with line $BC.$ Hence, $MI$ bisects angle $\angle{TMA}.$ Hence, by Serbia 2019 we have that $M$ lies on the circumcenter of triangle $AKL,$ which finishes the problem as the line determined by the two circumcenters is perpendicular on the radical axis $AM.$

HKIS200543 wrote: In a non-equilateral triangle $ABC$ point $I$ is the incenter and point $O$ is the circumcenter. A line $s$ through $I$ is perpendicular to $IO$. Line $\ell$ symmetric to like $BC$ with respect to $s$ meets the segments $AB$ and $AC$ at points $K$ and $L$, respectively ($K$ and $L$ are different from $A$). Prove that the circumcenter of triangles $AKL$ lies on the line $IO$. Let $A’$ be the reflection of $A$ over $BC$. Let $I_aI_bI_c$ be the excentral triangle of $\triangle ABC$. Let $M = BC \cap \text{s}$ and $M’ = I_bI_c \cap \text{s}$. Note that it is enough to prove that $A, K, L, A’$ are concyclic. Note that by butterfly theorem in $(I_aI_bI_c)$, we have that $M’I = 2 \cdot MI$. This implies that $MA’ = MI$. So, we have that $A’I$ bisects $\measuredangle{MA’A}$. Now let $T = (AKL) \cap (ABC)$. We shall prove that $T \equiv A’$. Let $AI \cap (AKL) = N’$ and $AI \cap (ABC) = N$. Note that $T$ is the centre of spiral similarity mapping $KL \mapsto BC$. So, we have $$\frac{NI}{N’I} = \frac{BI}{KI} = \frac{TN}{TN’}$$Thus, by converse of angle bisector theorem, we have that $TI$ bisects $\measuredangle{N’A’N}$. This means that $T \equiv A’$ and so $A, K, L, A’$ are concyclic. $\blacksquare$

I will use this interesting Lemma: Given a triangle $ABC$ with circuimcenter $O$, incenter $I$ and inradius $r$. Then, locus of points $P$ such $dist(P;AB)+dist(P;BC)+dist(P;CA)=3r$ (all distances are directed) is line which pass through $I$ perpendicular to $OI$. Proof of original problem: Let $s \cap BC = X$. It's easy to see that line $l$ tangents incircle of $\triangle ABC$. According with lemma wrt $\triangle ABC$, $3r = dist(X;AB)+dist(X; AC)+0 =dist(X;AK)+dist(X; AL)+0$ since $IX \perp OI$ . If $O'$ is circuimcenter of $\triangle AKL$, then (using our lemma wrt $\triangle AKL$) $O'I \perp IX$ $\implies O' \in OI$. We are done!

inversion at incircle, trivial

MarkBcc168 wrote: What a nice problem! Let $X$ be the point on $BC$ such that $\angle OIX = 90^{\circ}$ and let $X'$ be the reflection of $X$ across $I$. Let $M_B, M_C$ be the intersection of rays $BI, CI$ and $\odot(ABC)$. By Butterfly theorem, $X'\in M_BM_C$. But $\overline{M_BM_C}$ is the perpendicular bisector of $AI$. Thus $AX' = X'I$. Let $P$ be the point on $\odot(ABC)$ such that $AP\parallel KL$ and let $A', P'$ be the reflections of $A,P$ across $OI$. By reflection, $A'P'\parallel BC$ thus $AA', AP'$ are isogonal w.r.t. $\angle BAC$. However, as $\measuredangle IAX' = \measuredangle X'IA = \measuredangle A'IA$, we get $A, X', P'$ are colinear. Reflecting back across $OI$ gives $A', X, P$ are colinear. Finally, notice that $$\measuredangle XLC = \measuredangle PAC = \measuredangle PA'C = \measuredangle XA'C$$thus $A'\in\odot(XLC)$. Similarly $A'\in\odot(XKB)$ so by Miquel's theorem, $A'\in\odot(AKL)$. As perpendicular bisector of $AA'$ is $OI$, we are done. Sorry I must ask, how does reflection imply $A’P’$$//$$BC$?

A really nice problem here's my solution: let $s \cap BC =x$ and let $\omega ,D,E,F$ be the incircle, and the in-touch points with $BC,AC,AB$ note that $LK$ is tangent to $\omega$ let the tangency point be $G$ let $H_1$ be the orthocenter of $\triangle DEF$ and $ O'$ be the the circumcenter of $\triangle AKL$ $H_2$ is the orthocenter of $\triangle GEF$ well-known that $O,I,H_1$ are collinear since $\omega$ is the excircle of $\triangle AKL$ we have also $O',H_2,I$ are collinear so it suffices to show that $O,H_1,H_2$ are collinear let $E',F'$ are the reflection of $E,F$ in $s$ since we have $D$ is the reflection of $G$ in $s$ easy to see that $EE' || FF' ||OI$ and $E',F' \in \omega$ so $\triangle F'E'D$ is the reflection of $\triangle EFG$ and the following claim will end the problem claim: $\triangle DEF$ and $\triangle DE'F'$ share the same euler line proof: let $M,N$ be the midpoints of $EF,E'F'$ we have $MN || OI $ let $G,G' $ be the centroids of $\triangle DEF , \triangle DE'F'$ we have $\frac{DG}{DM}=\frac{DG'}{DN}=\frac{2}{3} \implies GG' || OI$ $\blacksquare$ and we win

Let $\overline{KL}$ intersect $\overline{BC}$ at $D$. It is easy to see that $\overline{KL}$ is tangent to the incircle of $\triangle ABC$, and so $I$ is also the incenter of $\triangle AKL$. Thus \[ \measuredangle BKI = \measuredangle AIL - 90^\circ = \measuredangle BID - 90^\circ = \measuredangle BIO, \]so $(BIK)$ is tangent to $\overline{OI}$. Similarly, $(CIL)$ is also tangent to $\overline{OI}$. Now consider an inversion at $I$ that preserves $(ABC)$. The problem now becomes the following: Inverted Version wrote: $\triangle ABC$ has orthocenter $I$ and Euler line $\ell$, and $K$ and $L$ are points on $(IAB)$ and $(IAC)$ such that $BK \parallel CL \parallel \ell$. Show that the center of $(AKL)$ lies on $\ell$. Let the circumcenter of $\triangle ABC$ be $S$, which we will consider to be the origin. Notice that $(IAC)$ is the image of $(IAB)$ translated by $\overrightarrow{BC}$. As $BK \parallel CL$, this means that $L$ is the image of $K$ under the same transformation, so $\overrightarrow{BK} = \overrightarrow{CL} = \mathbf{x}$ for some vector $\mathbf{x}$. Now let $A'$ be the point defined by $\overrightarrow{A'} = \overrightarrow{A} - \mathbf{x}$. Then obviously, $AA'BK$ is a parallelogram, so $A'$ lies on $(ABC)$. Consider the translation $\mathcal{T}$ of the whole diagram by $\mathbf{x}$. This sends $(A'BC)$ to $(AKL)$, so the circumcenter of $\triangle AKL$ is $\overrightarrow{S} + \mathbf{x} = \mathbf{x}$, which lies on $\ell$ as $\mathbf{x}$ is parallel to $\ell$.

HKIS200543 wrote: In a non-equilateral triangle $ABC$ point $I$ is the incenter and point $O$ is the circumcenter. A line $s$ through $I$ is perpendicular to $IO$. Line $\ell$ symmetric to like $BC$ with respect to $s$ meets the segments $AB$ and $AC$ at points $K$ and $L$, respectively ($K$ and $L$ are different from $A$). Prove that the circumcenter of triangle $AKL$ lies on the line $IO$. Dušan Djukić Consider the inversion at $I$ with an arbitrary radius and denote the inverse of the points the same as the original ones. Then $I, O$ are the orthocenter and circumcenter of $\triangle ABC$, respectively. Line $s$ is fixed, $l$ maps to the circle $\omega$ that is symmetric to $(BIC)$ wrt $s$. $\omega$ cuts $(AIB)$ and $(AIC)$ at $K, L$, respectively. It suffices to prove the circumcenter $J$ of $\triangle AKL$ lies on $IO$. Let $O_1,O_2$ be the center of $(AIB),(AIC)$, respectively, then $O_1,O_2$ are symmetric to $O$ wrt $AB, AC$. Let $S, T$ be the center of $(BIC)$ and $\omega$, respectively. Note that clearly the radius of $(O),(BIC), (AIB), (AIC),\omega$ are equal, so $\overrightarrow{TL}=\overrightarrow{IO_2}=\overrightarrow{O_1A}\implies AL\parallel TO_1\perp KI$. Similarly, we imply $I$ is the orthocenter of $\triangle AKL$, thus $\overrightarrow{JT}=\overrightarrow{AI}=\overrightarrow{OS}\implies OJ\parallel ST\perp s\perp IO$. The conclusion follows.

Apparently yesterday I wrote a geo problem, only to discover a few hours later that it's actually this. Here is a (sketch of) new solution that led me to discover this problem. Let $M$ be the center of $\odot(BIC)$; $J$ be the intersection of $IO$ and $\odot(BIC)$; and $D$ is the second intersection of $MJ$ and $\odot(ABC)$. Observe that $\triangle ABC$ and $\triangle DBC$ share the same $OI$-line. Moreover, a simple angle chasing (omitted) gives $\triangle AKL\stackrel{-}\sim \triangle DCB$. Thus, if $S$ is the circumcenter of $\triangle AKL$, then $\triangle AKL\cup S\cup I \stackrel{-}\sim\triangle DCB\cup O\cup J$. Hence, $\measuredangle(KL, SI) = -\measuredangle(BC, OJ)$. This implies that $S\in OI$.