There's a typo. $M'$ and $N'$ are the reflections of $M, N$ wrt $AC$ and $AB$ respectively

Let $O$ be the circumcenter of $\triangle ABC$, and construct point $N''$ with $\triangle OIN \stackrel{+}{\cong} \triangle ON''A$ (in particular $OI = ON''$). Construct $M''$ similarly. Claim. $N''$ is the reflection of $N'$ over $\overline{AI}$. Proof. Write \begin{align*} \measuredangle NAN'' & = \measuredangle NAO + \measuredangle OAN''\\ & = \measuredangle ONA + \measuredangle ONI\\ & = \measuredangle ONA + \measuredangle ONC\\ & = 2 \measuredangle ONM\\ & = \measuredangle ONM + \measuredangle NMO\\ & = \measuredangle NOM\\ & = 2 \measuredangle NAM\\ & = 2 \measuredangle BIC\\ & = \measuredangle BAC \end{align*}which is sufficient. $\square$ Similarly $M''$ is the reflection of $M'$ over $\overline{AI}$. Let $O'$ be the reflection of $O$ over $\overline{AI}$. Since $OI = OM'' = ON''$, $O$ is the circumcenter of $\triangle IM''N''$, so $O'$ is the circumcenter of $\triangle IM'N'$. Clearly $O'$ lies on the $A$-altitude, so we are done.

Original statement, equivalent to this one: Let $ABC$ be an acute-angled triangle whose circumcircle is $\Gamma$ and let $I$ be its incenter. In $\Gamma$ , $M$ is the midpoint of the arc $AB$ which does not contain $C$ and $N$ is the midpoint of the $AC$ which does not contain $B$. Let $K$ and $L$ be the reflections $M$ and $N$ in the lines $AB$ and $AC$, respectively. Suppose that the circle passing through the points $I$, $K$ and $L$ meets $BC$ at $P$ and $Q$. Prove that $AP=AQ$.

plagueis wrote: Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$. This problem is quite easy if one is familiar with the Fuhrmann Triangle and its circumcircle. Let $P$ be the midpoint of minor arc $BC$ in $\triangle ABC$, $P'$ its reflection about $BC$, $H$ the orthocenter, $Na$ the Nagel point, $N_9$ the nine-point center, $O$ the circumcenter, and $G$ the centroid (all w.r.t $\triangle ABC$.) Consider the anticomplementary triangle $A_1B_1C_1$ of $\triangle ABC$. Note that $A_1P', B_1M'$, and $C_1N'$ are angle bisectors in $\triangle A_1B_1C_1$, so they concur at the Nagel point of $\triangle ABC$. Furthermore, note that $\angle A_1P'H = \angle B_1M'H = \angle C_1N'H = 90^{\circ}$, so the circumcircle of $M'N'P'$ has diameter $HNa$. We then invoke two crucial claims which kill the original problem immediately. Claim 1: $I$ is the orthocenter of $\triangle M'N'P'$. This is a fairly simple job for complex numbers. Place $A, B, C$ on the unit circle, and note that there exist complex numbers $x, y, z$ such that $A = x^2, B = y^2, C = z^2, P = -yz, M = -xz, N = -xy$, and $I = -(xy + xz + yz)$. Then $P' = 2\cdot \frac{y^2 + z^2}{2} - (-yz) = y^2 + z^2 + yz$, and similar expressions hold for $M'$ and $N'$. Checking that $P'I \perp M'N'$ is just a boring exercise and is left to the reader. Claim 2: The nine-point centers of $\triangle M'N'P'$ and $\triangle ABC$ coincide. Consider the homothety centered at $G$ with factor $-2$, where $G$ is the centroid of $\triangle ABC$. By properties of the Nagel line and Euler line, it follows that $OI \parallel HNa$ and $HNa = 2OI$. Let $F$ be the midpoint of $HNa$ and thus the circumcenter of $\triangle M'N'P'$; then $OIHF$ is a parallelogram, so the midpoints of $OH$ and $IF$ coincide. But $I$ is the orthocenter of $\triangle M'N'P'$, so the common midpoint is in fact the nine-point center $N_9$ of $\triangle ABC$ and $\triangle M'N'P'$. Back to the original problem: The circumcenter of $IM'N'$ is the reflection of $F$ about $M'N'$ and thus the reflection of $P'$ about $N_9$, which must lie on the $A$-altitude as $N$ is the midpoint of $OH$.

This problem was a joint proposal by me and plagueis. Interestingly, the result remains true if we allow $I$ to be any arbitrary point and define $M$ and $N$ to be the intersections of $BI$ and $CI$ with the circumcircle. This was unknown to us and was discovered a week before the exam by one of the designers. It turns out that the official solutions we sent work for this general case as well, but I'm actually glad the weaker version was chosen as it seems to have facilitated some nice approaches.

Here is my solution for this problem Solution Let $P$ $\equiv$ $AI$ $\cap$ $\Omega$; $J$ be center of ($IM'N'$) We have: ($BI$; $BN'$) $\equiv$ ($BI$; $BA$) + ($BA$; $BN'$) $\equiv$ ($CM$; $CA$) + ($BN$; $BA$) $\equiv$ ($CA$; $CM'$) + ($CN$; $CA$) $\equiv$ ($CI$; $CM'$) (mod $\pi$) and $\dfrac{BI}{BN'}$ = $\dfrac{BI}{BN}$ = $\dfrac{CI}{CM}$ = $\dfrac{CI}{CM'}$ So: $\triangle$ $BIN'$ $\stackrel{+}{\sim}$ $\triangle$ $CIM'$ or $\triangle$ $IM'N'$ $\stackrel{+}{\sim}$ $\triangle$ $IBC$ But: $J$, $P$ are center of ($IM'N'$), ($IBC$) then: $\triangle$ $BIN'$ $\stackrel{+}{\sim}$ $\triangle$ $CIM'$ $\stackrel{+}{\sim}$ $\triangle$ $PIJ$ Hence: $\dfrac{PI}{PJ}$ = $\dfrac{BI}{BN'}$ and ($PI$; $PJ$) $\equiv$ ($BI$; $BN'$) (mod $\pi$) But: ($OB$; $OP$) $\equiv$ ($AB$; $AC$) $\equiv$ ($NB$; $NI$) (mod $\pi$) and ($PO$; $PB$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($BP$; $BC$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($AP$; $AC$) $\equiv$ ($IN$; $IB$) (mod $\pi$) so: $\triangle$ $BPO$ $\stackrel{+}{\sim}$ $\triangle$ $BIN$ or $\dfrac{PI}{PO}$ = $\dfrac{PB}{PO}$ = $\dfrac{IB}{IN}$ = $\dfrac{IB}{NB}$ Then: $PJ$ = $PO$ We also have: ($PI$; $PO$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($OP$; $OA$) $\equiv$ $\dfrac{\pi}{2}$ $-$ ($CP$; $CA$) $\equiv$ $\equiv$ $\dfrac{(BA; BC) - (CB; CA)}{2}$ $\equiv$ $-$ ($BI$; $BN'$) $\equiv$ $-$ ($PI$; $PJ$) (mod $\pi$) Hence: $J$ is reflection of $O$ through $AI$ or $AJ$ $\perp$ $BC$

My solution Pleasant wording wrote: In acute-angled scalene triangle $ABC$ with circumcircle $\Gamma$, incenter $I$, the lines $BI$ and $CI$ meet $\Gamma$ again at $M_B$ and $M_C$ respectively. Points $N_B$ and $N_C$ are the reflections of $M_B$ and $M_C$ in lines $AC$ and $AB$ respectively. Let $O$ be the circumcenter of triangle $IN_BN_C$. Prove that $AO \perp BC$. Let $H$ be the orthocenter of $\triangle ABC$ and define $M_A, N_A$ naturally. By Hagge's circle we see that $H$ lies on $\odot(N_AN_BN_C)$. Claim. $I$ is the orthocenter of $\triangle N_AN_BN_C$. (Proof) Too quick by complex numbers to even try synthetic. Indeed, set $A=x^2, B=y^2, C=z^2$ and $\Gamma$ as the unit circle. Then $N_A=(y^2+z^2+yz)$ and cyclic relations hold. In particular, $$IN_A \perp N_BN_C \iff \frac{(x+y+z)^2-x(x+y+z)}{z^2-y^2+x(z-y)} \in i\mathbb{R} \iff \frac{y+z}{z-y} \in i\mathbb{R}$$which is clearly true. $\blacksquare$ So, $\triangle IN_BN_C \sim \triangle ICB$. Denote by $J$ the orthocenter of $\triangle BIC$, by $K$ the reflection of $I$ in $BC$. By spiral similarity, we see that $\triangle IN_CB \sim \triangle IOM_A \sim \triangle IN_AJ$. Finally, we remark that $\frac{KI}{KJ}=\frac{AI}{AM_A}$ hence $\triangle IN_AK \sim \triangle IOA$. Thus, $\angle IAO=\angle IKN_A=\angle M_AIK$ so $AO \parallel IK$ and we're done.

By basic angle chasing, $\angle ICM' = \angle ICM- \angle M'CM = \frac{1}{2} (\angle C - \angle B )= \angle IBN'$. Moreover, \[ \frac{IC}{CM'} = \frac{IC}{CM} = \frac{IB}{BN} = \frac{IB}{BN'}. \]Thus $\triangle ICM' \sim \triangle IBN'$. Then by spiral similarity, $\triangle IM'N' \sim \triangle IBC$. Let $Q=NC \cap \odot (IM'N')$, then $N'Q \parallel BC$, there $O_1$ lies on the perpendicular bisector of $N'Q$, where $O_1$ is the circumcenter of $ \triangle IM'N'$. Now it is sufficient to prove that $AN' = AQ$. Now draw the circle of center of $A$ and radius $AN$, let $Q' = \odot A \cap CN$. Then by basic angle chasing, the angle between $NN'$ and BC is given by $\angle (NN', BC) = 90^\circ - \angle B$, and $180^\circ - \angle NN'Q' = \frac{1}{2} \angle NAQ' = 90^\circ - \angle ANQ' = 90^\circ - \angle B$. Therefore $N'Q' \parallel BC$, which means $Q' \equiv Q$. Hence $AO_1$ is the perpendicular bisector of $N'Q \parallel BC$. Remark: we can further prove that $O,M', O_1,N'$ is concyclic, but it is not very helpful in my proof.

Define points $P,Q$ on $CI$ and $BI$, respectively, such that $\angle APC=180^\circ-\angle B$, and $\angle AQB=180^\circ-\angle C$. Note that, $\angle APN=\angle B=\angle ANC$, so $AP=AN$ and similarly $AM=AQ$. Also, $AM'=AM$ and $AN'=AN$, hence $AP=AN', AQ=AM'$. In addition, obviously $\angle AN'B=\angle ANB=180^\circ-\angle C=\angle AQB$, so $N'$ and $Q$ belong to the circle $(A,H,B)$ where $H$ is the orthocenter. Similarly $P,M'$ belong to the circle $(A,H,C)$. Trivially the radii of these two circles are equal (and equal to the circumradius as well). Hence, we may write: $\dfrac{PH}{\sin \angle PAH}=2R_{BAH}=2R_{CAH}=\dfrac{N'H}{\sin \angle N'AH}$, and since $\angle PAH=\angle PCH=\angle N'BH=\angle N'AH$ (where the middle equality follows by an easy angle-chase), we obtain $PH=N'H$. Similarly, $HQ=HM'$. Hence, using $AP=AN'$ and $AM'=AQ$, we have that $AH$ is the perpendicular bisector of both $PN'$ and $QM'$. So $PN'QM'$ is an iscosceles trapezoid, hence cyclic. Now, we perform an angle-chase to obtain that $I$ belongs to the circumcircle of $PN'QM'$: (refer to the diagram attached) $\angle PN'Q=\angle AN'H-\angle AN'P+\angle HN'Q=90^\circ+\angle A-(90^\circ-\angle N'AH)+\angle QBH=\angle A+\angle N'AH+\angle QBH=\angle A+\angle N'BH+\angle B/2+\angle C-90^\circ=90^\circ-\angle B/2+\angle ABH-\angle ABN'=180^\circ-\angle A-\angle B/2-\angle C/2=90^\circ-\angle A/2=\angle PIQ$. Therefore, $\angle PN'Q=\angle PIQ$, so $I$ belongs to the circle $(P,Q,M',N')$. To end, the center of this circle is on the perpendicular bisector of $PN'$, $QM'$ which is $AH$. So, the circumcenter of $(P,I,Q)$ belongs on the $A-$ altitude, as desired.

It turns out that we don’t even need the fact that $I$ is the incenter. Let $H$ be the orthocenter of $\triangle ABC$. We claim that $\triangle IBN’ \sim \triangle ICM’$. To see this, just notice that $IB : BN’ = IB : BN = IC : CM = IC : CM’$ and $\angle IBN’ = \frac{1}{2}|\angle B - \angle C| = \angle ICM’$. Therefore, by spiral similarity, we also have $\triangle IN’M’ \sim \triangle IBC$. Also, by the orthocenter lemma, $N’$ lies on $(AHB)$. Now let $X$ be the reflection of $N’$ over $\overline{AH}$, and consider the Steiner line $\ell$ of $N’$ with respect to $\triangle AHB$. $\ell$ passes through the reflection of $N’$ over $\overline{AB}$, which is $N$, and it also passes through the orthocenter of $\triangle AHB$ which is $C$, so it follows that $l \equiv \overline{CN}$, and hence $X$ lies on $\overline{CN}$. Now angle chasing shows that \[ \measuredangle IXN’ = \measuredangle ICB = \measuredangle IM’N’ \]and so $X$ lies on $(IM’N’)$ and the conclusion follows.

Not sure why the above solution is so complicated As usual complex bash with $(ABC)$ as the unit circle, picking $A=x^2,B=y^2=C=z^2$ such that $M=-xz$ and $N=-xy$, so $I=-xy-xz-yz$. It is easy to calculate $M'=x^2+z^2+xz$ and $N'=x^2+y^2+xy$. Therefore, subtracting $A=x^2$, it suffices to show that the circumcenter of $xy+y^2=y(x+y)$, $xz+z^2=z(x+z)$, and $-x^2-xy-xz-yz=-(x+y)(x+z)$ is an imaginary multiple of $y^2-z^2$. Apply the following transformations: Divide by $(x+y)(x+z)$, so our points are now $\tfrac{y}{x+z},\tfrac{z}{x+y},-1$. Add $1$, so our points are now $\tfrac{x+y+z}{x+z},\tfrac{x+y+z}{x+y},0$. Multiply by $\tfrac{(x+y)(x+z)}{x+y+z}$, so our points are now $x+y,x+z,0$. Subtract $x$, so our points are now $y,z,-x$. In the last step, note that $y,z,-x$ all lie on the unit circle, hence their circumcenter is $0$. Undoing these operations implies that the original circumcenter was $-yz$. It is easy to see that $\tfrac{-yz}{y^2-z^2}$ is imaginary, since its conjugate is its negative, so we are done. $\blacksquare$

Let $A = x^2, B = y^2, C = z^2, M = -xz, N = -xy$ so that $M' = x^2 + z^2 + xz$, $N' = x^2 + y^2 + xy$, and $I = -xy - yz - xz$. Then we have $$\frac{M' - I}{C - I} = \frac{x^2 + z^2 + 2xz + xy + yz}{z^2 + xy + yz + xz} = \frac{(x+z)(x+z+y)}{(z+x)(z+y)} = \frac{x+y+z}{y+z} = \frac{N'-I}{B-I}$$ where the final equality follows from the fact that the LHS is symmetric in $y$ and $z$. The above implies that there is a spiral similarity mapping $\triangle ICB\to \triangle IM'N'$, which will map the midpoint $-yz$ of $\widehat{BC}$ to the circumcenter $O$ of $(IM'N')$: $$O = \frac{x+y+z}{y+z}(-yz - (-xy - yz - xz)) + (-xy - yz - xz) = x^2 + xy + xz - xy - yz - xz = x^2 + yz$$ as desired.

Letting $A = a^2$, $B = b^2$, and $C = c^2$ gives $I = -(ab+bc+ca)$ and \[N'= a^2+b^2+ab \text{ and } M' = a^2+c^2+ac\]Shift the complex system so $I = 0$, to get \[N'' = N'-I = (a+b+c)(a+b) \text{ and } M'' = M'-I = (a+b+c)(a+c)\]Now let $O_1$ be unshifted the center of $\Delta IM'N'$. We have \begin{align} O_1 &= I - \frac{(a+b+c)^2(a+b)(a+c)\left(\frac{(ab+bc+ca)(a+b)}{a^2b^2c} - \frac{(ab+bc+ca)(a+c)}{a^2bc^2}\right)}{(a+b+c)(a+b)(a+c)(ab+bc+ca)\left(\frac{1}{a^2bc^2}-\frac{1}{a^2b^2c}\right)} \\ &= I - \frac{(a+b+c)\left(c(a+b)-b(a+c)\right)}{b-c} \\ &= I + a^2+ab+ac \\ &= a^2-bc \end{align}It suffices to show that the foot of $A$ to $\overline{BC}$ is equal to the foot of $O_1$ to $\overline{BC}$. This is equivlent proving \[a^2-b^2c^2\frac{1}{a^2} = a^2-bc-b^2c^2\left(\frac{1}{a^2}-\frac{1}{bc}\right)\]\[\iff bc\left(1-\frac{bc}{a}\right) = b^2c^2\left(\frac{1}{bc}-\frac{1}{a^2}\right)\]which is clearly true, so we're done.

This problem is actually entirely and orthocentric configuration problem which is ingeniously hidden as an incentric configuration. To avoid notational nightmares we rename $M'$ and $N'$ to $M_1$ and $N_1$ respectively. We can start off with the following observation. Claim : Circles $(AM_1C)$ and $(AN_1B)$ intersect at $H$ - the orthocenter of $\triangle ABC$. Proof : Let $H' = (AM_1C) \cap (AN_1B) \neq A$. Then, it is easy to see that, \[\measuredangle AH'C = \measuredangle AM_1C = \measuredangle CMA = \measuredangle CBA \]and \[\measuredangle BH'A = \measuredangle BN_1A = \measuredangle AN_1B = \measuredangle ACB\]from which it is clear that $H'=H$, and the claim is proved. Now, let $M_2$ and $N_2$ be the reflections of $N_1$ and $N_2$ across the $A$-altitude respectively. We first locate the points $M_2$ and $N_2$, with the following claims. Claim : Points $M_2$ and $N_2$ lie on lines $\overline{BI}$ and $\overline{CI}$ respectively. Proof : Note that, \begin{align*} 2\measuredangle ANN_2 & = \measuredangle NAN_2\\ & = \measuredangle NAN_1 + \measuredangle N_1AN_2 \\ &= 2\measuredangle NAB + 2 \measuredangle N_1AH\\ &= 2\measuredangle NAB + 2 \measuredangle N_1AB + 2\measuredangle BAH\\ &= 2\measuredangle BAH\\ &= 2(90 + \measuredangle ABC)\\ &= 2\measuredangle ABC\\ &= 2\measuredangle ANC \end{align*}which implies that $N_2$ lies on $\overline{NC}$. Similarly we can show that $M_2$ lies on $\overline{MB}$ which proves the claim. Claim : Points $M_2$ and $N_2$ lie on circles $(AN_1C)$ and $(AM_1B)$ respectively. Proof : For this, simply note that \begin{align*} \measuredangle HAN_1 &= \measuredangle N_1AN + \measuredangle NAH \\ &= \measuredangle BCA + \measuredangle NCB + \measuredangle HCB\\ &= \measuredangle NCA + \measuredangle HCB\\ &= \measuredangle BCN + \measuredangle HCB\\ &= \measuredangle HCN \\ &= \measuredangle HCN_2 \end{align*}from which it follows that $N_2$ indeed lies on $(AHC)$ as desired. We can similrly show that $M_2$ lies on $(AHB)$. Now, note that $N,N_1$ and $N_2$ lie on a circle centered at $A$ and radius $AN$. Further, points $M$ , $M_1$ and $M_2$ also lie on a circle centered at $A$ with radius $AM$. Further, it is clear that due to the reflections, $N_2M_2M_1N_1$ must be an isosceles trapezoid, and hence it is cyclic. Using this we can finally note that, \[\measuredangle IN_2M_1 = \measuredangle CN_2M_1 = \measuredangle CAM_1 = \measuredangle MAC = \measuredangle MBC\]and \[2\measuredangle IM_2M_1 = 2\measuredangle MM_2M_1 = \measuredangle MAM_1 = 2\measuredangle MAC = 2\measuredangle MBC\]from which we have \[\measuredangle IM_2M_1 = \measuredangle MBC = \measuredangle IN_2M_1\]which implies that $I$ lies on $(M_2M_1N_1)$. Thus, points $M_2$ and $N_2$ must lie on the circle $(IMN)$. Thus, the center of this circle, $O$ must lie on the perpendicular bisector of $M_1M_2$ and $N_1N_2$ which is simply the $A$-altitude by the nature of construction. Thus, the center of the circle passing through $I$, $M_1$, $N_1$ lies on the altitude of triangle $ABC$ from $A$ as desired.

We will complex bash with $(ABC)$ as the unit circle, and $A=a^2$, $B=b^2$, $C=c^2$. Note that $N$ is the intersection of the altitude of the midpoint of major arc $AB$ with $(ABC)$, hence by orthocentre reflection lemma, we have $n'=a^2+b^2+ab$. Similarly, $m'=a^2+c^2+ac$. Now, let us shift $\displaystyle{\triangle IM'N'}$ to the origin. $I=-ab-ac-bc$, hence $m'$ goes to $a^2+c^2+2ac+ab+bc=(a+c)(a+b+c)$. Similarly, $n'$ goes to $(a+b)(a+b+c)$. Now, using the circumcentre formula, we have \begin{align*} \frac{(a+b+c)^2(a+c)(a+b)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{b}-\frac{1}{c}\right)}{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\left(\frac{1}{a}+\frac{1}{b}\right)(a+c)-(a+b)\left(\frac{1}{a}-\frac{1}{c}\right)\right)}&=\frac{(a+b+c)(a+c)(a+b)\left(\frac{1}{b}-\frac{1}{c}\right)}{\left(\frac{1}{a}+\frac{1}{b}\right)(a+c)-(a+b)\left(\frac{1}{a}-\frac{1}{c}\right)}\\ &=\frac{(a+b+c)(a+c)(a+b)\left(\frac{1}{b}-\frac{1}{c}\right))}{\frac{a}{b}-\frac{b}{a}+\frac{c}{b}-\frac{b}{c}+\frac{c}{a}-\frac{a}{c}}\\ &=\frac{a(a+b+c)(a+c)(a+b)(c-b)}{ca^2-cb^2+ac^2-ab^2+bc^2-ba^2}\\ &=\frac{a(a+b+c)(a+c)(a+b)(c-b)}{(c-b)(a+c)(a+b)}\\ &=a(a+b+c) \end{align*}Shifting back, we have $a(a+b+c)-ab-ac-bc=a^2-bc$. Thus, we have the centre of $\triangle IM'N'$. Now, we show that the line through $a^2$ and the centre is perpendicular to $b^2-c^2$. Indeed, \[\frac{bc}{b^2-c^2}=\frac{-bc}{c^2-b^2}=\frac{-\frac{1}{b^2c^2}}{\frac{1}{b^2}-\frac{1}{c^2}}\]As desired.