Let the midpoint of arc $ACB$ be $R$. Let $I_A,I_B,I_C$ be the $A,B,C$ excentres of $\triangle ABC$ Observe that $\odot ABC$ is the nine point circle of $\triangle ABC$ with $I_AA,I_BB,I_CC$ altitudes. $\implies R$ is the midpoint of $I_BI_C$ Also $I,P,I_A$ and $I,Q,I_B$ are lines. $\angle CDP = \frac{B-A}{2}$ and $\angle CDI = A+\frac{C}{2}$ That is $CI \perp DPQ$ $\implies PQ || I_BI_C$ Now consider the homothety centred at $I$ that takes $P$ to $I_A$ and $Q$ to $I_B$ Then it takes midpoint of $PQ$ to midpoint of $I_AI_B$ That is $M \to R$ That is $I,M,R$ are collinear. QED

My solution: Let $X=BI\cap \omega$ and $Y=AI\cap \omega$ and $Z$ is the midpoint of arc $ACB$. By angle-chasing is easy to see that: $IXYZ$ is parallelogram and $\angle DQA=\angle IBA$ and $\angle QPB=\angle IAB$ $\Longrightarrow$ $PQ\parallel XY$ and let $N$ is the midpoint of $XY$ $\Longrightarrow$ $I,N,M$ are collinear and $I,N,Z$ are collinear since $IXYZ$ is parallelogram $\Longrightarrow$ $I,Z,M$ are collinear...

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posted before http://www.artofproblemsolving.com/community/c6h1159531p5513985

My solution: Let the $C$-mixtilinear incircle touch the circumcircle at $T$, $M_1$ be the midpoint of arc $ACB$, $M_2$ be the antipode of it in the circumcircle. Denote by $l$ the line parallel to $CM_1$ thru $I$. Let $R$ be the point at infinity on $PQ$. Note that $PQ \parallel CM_1$ so $$ -1 = ( P, Q; M, R ) = ( B, A; MI \cap AB, l \cap AB) $$Also, $$ -1 = ( B, A; M_1, M_2 ) = (B, A; TM_1 \cap AB, TM_2 \cap AB)$$Since $l, AB, TM_2$ are concurrent, from the above 2 cross ratios, we get $TM_1 \cap AB \equiv MI \cap AB$. But we know that $T,I, M_1$ are collinear, so the conclusion follows. Motivation: The line joining $I$ and $M_1$ immediately reminded me of the mixtilinear incircle configuration. After adding in the mixtilinear incircle, the rest was a 5-minute cross-ratio chase. And I like ad-hoc proofs of different problems, so I thought it would be nice to post it here!

$MB=MC$ and $\angle MBI=\angle IAB$ $\implies$ M is the intersection of tangents on $\odot IAB$ at A,B $\implies$ IM is the symmedian in $\triangle IAB$.$\angle AQP+\angle ABI=\pi$$\implies$ $APQB$ is cyclic so PQ and AB are anti-parallel and we're done.

Either I've misread something or this is too easy for it's position. Let $X, Y$ be the midpoint of arc $AB$ containing and not containing $C$, respectively. Let $M^*$ be the dracula (or phantom whatever) point on $PQ$ such that $I, M^*, X$ are colinear. Let $\Gamma = \omega_{BIC}$, $Z = XI \cap \Gamma$ (second intersection). Note that by some very easy angle chasing, we get $CX || PQ || \text{Tangent to I at } \Gamma$. So, $$ 90 = \angle XBY = \angle XAY \Rightarrow \text{XA, XB are tangent to} \Gamma$$$$ \Rightarrow -1 = (I, Z; A, B) \overset{I}{=} (P_{\infty}^{PQ}, M^*; P, Q) \Rightarrow M^* = M$$$\blacksquare$

Kayak wrote: Either I've misread something or this is too easy for it's position.

Solution. Without loss of generality, we assume $\angle A>\angle B.$ Let $I_A$ and $I_B$ be the $A$ and $B-$excenters of $\triangle ABC$ respectively. Let $S_C'$ be the midpoint of arc $\overarc{ACB}$ of $\omega.$ Now, it is well-known that $S_C'$ is the midpoint of $I_A,I_B,$ and that the quadrilateral $ABI_BI_A$ is cyclic. Claim. $I_AI_B$ is parallel to $PQ.$ Proof of Claim. We observe that \begin{align*}&\angle ACD=\angle ABC=B \\\implies &\angle CDB=A-B \\\implies &\angle CDP=\angle PDA=\frac{A-B}{2} \\\implies &\angle APD=\angle PAB-\angle PDA=\frac{A}{2}-\frac{A-B}{2}=\frac{B}{2}~.\end{align*}Using the above equality and the fact that $ABI_BI_A$ is cyclic, we obtain - $$\angle ABI_B=\angle AI_AI_C=\frac{B}{2}=\angle APD\implies PQ\parallel I_AI_B.$$So the proof to our Claim is complete. $\blacksquare$ Now, consider the triangles $II_AI_B$ and $IQP.$ By the Claim above, we know that $I_AI_B$ is parallel to $PQ.$ Using this and the fact that $Q$ lies on $II_A,P$ lies on $II_B,$ we can say there exists a homothety $\mathcal{H}(I,k)$ with factor $k$ which sends $QP$ to $I_AI_B.$ Thus $\mathcal{H}:M\mapsto S_C,$ implying $I,M,S_C$ are collinear, as desired. $\square$

utkarshgupta wrote: A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$. Let $M,N$ respectively be the midpoints of $\overline{PQ}$ and $\overarc{ACB}$. Now, note that $\angle NBI = \frac{\alpha}{2} = \angle IAB \implies IN$ is the symmedian of $\triangle IAB$. Now also observe the $\angle QAB = \angle IAB = \angle IPQ = \angle QPB \implies APQB$ cyclic. Thus $PQ$ is anti-parallel to $AB$ in the angular region $\angle AIB \implies IN$ bisects $PQ$

Let $N$ denote the midpoint of $AB$ and $T$ the tangency point of $AB$ with the incircle. Let $X$ and $Y$ represent the midpoints of major arc $AB$ and minor arc $BC$, respectively. Our goal is to show that $\angle XIY = \angle MIY$. Lemma 1. The triangles $ITN$ and $ICX$ are similar. Proof. I don't have a nice way to show this, but it is not difficult to bash with trigonometry or complex numbers. Lemma 2. The triangles $AIB$ and $QIP$ are cyclic. Proof. We have that \[\angle IQP = 180^\circ - \angle QDA - \angle DAQ = 180^\circ - \frac{1}{2} CDA - (180 - \angle A/2) = \angle A/2 - \frac{1}{2}(\angle A - \angle B) = \angle B/2. \]Likewise $\angle IPQ = \angle A/2$, so we are done. Obviously, $A, I, Y$ collinear, so \[ \angle XIY= 180^\circ - \angle XIA = 180^{\circ} - (\angle XIC + \angle CIA) = 180^\circ - (90^\circ - \angle CXI) - (90^\circ + \angle B/2) = \angle CXI - \angle B/2 \]On the other hand, by Lemma 2, $\angle MIQ = \angle NIB$. We angle-chase: \[ \angle NIB = 180^\circ - \angle INB - \angle B/2 = \angle INT - \angle B/2. \]By Lemma 1, $\angle INT = \angle CXI$, so $\angle MIQ = \angle XIQ$. Done.

Let $N$ be the midpoint of arc $ACB$. Claim: $ABQP$ is cyclic. Note that $Q,P$ are the incenter and $D$-excenter of $\triangle ACD$ and $\triangle BDC$, respectively. Thus, $$\measuredangle AQP=\measuredangle ADQ+\measuredangle QAD=\frac{\measuredangle ABC}{2}=\measuredangle ABP.\quad \square$$ Let $K$ be the midpoint of $AB$. It is well-known that $\measuredangle IKB=\measuredangle CNI$. As $\triangle IAB\sim\triangle IPQ$, we also have $\measuredangle QMI=\measuredangle IKB$. Also, $$\measuredangle (PQ,CI)=\measuredangle QDC +\measuredangle DCB+\measuredangle BCI=\frac{\measuredangle ABC-\measuredangle CAB}{2}+\measuredangle CAB+\frac{\measuredangle BCA}{2}=90^\circ.$$ Therefore, $QP\perp CI\perp CN$, thus $I,M,N$ are collinear, we are done.

Let $J$ be the incenter of $\triangle DAC$. Claim: $ABPQ$ is cyclic. Proof. Notice that $P, Q$ are the excenter and incenter of $\triangle ADC$ and $\triangle BDC$ respectively. So by fact 5, we have $CJAP$ is cyclic. And since $\triangle DCA\sim\triangle DCB$ and $J, Q$ are the corresponding incenters therefore $\triangle ACJ\sim\triangle BCQ$. So \begin{align*} \angle QPA=\angle JPA=\angle JCA=\angle CBQ=\angle ABQ \end{align*}which means $ABPQ$ is cyclic. $\blacksquare$ Let $E$ be the midpoint of $AB$ and $F$ be the midpoint of the arc $\widehat{ACB}$. Notice that $\angle IAB=\angle IBF$ thererfore $BF$ is tangent to $(AIB)$ and $FA$ is also tangent because $FA=FB$. So $IF$ is $I-$symmedian of $\triangle AIB$. Since $AB$ and $PQ$ are antiparallel and $\triangle IAB\sim\triangle IPQ$ therefore $IF$ is the $I-$median of $\triangle IPQ$.

Let $\omega$ meet rays $AI,BI$ again at $K,L.$ Denote the midpoint of arc $ACB$ by $N.$ Claim. $IKNL$ is a parallelogram. Proof. Notice that $CN\parallel LK$ as both lines are perpendicular to $CI.$ Then by the Trident theorem $$|IL|=|CL|=|KN|,\text{ } |IK|=|CK|=|LN|\text{ } \Box$$ Now it's suffice to prove that $PQ\parallel KL.$ Indeed (angles and arcs are oriented): $$\angle (PQ,AB)=\frac{\angle (CD,AB)}{2}=\frac{\widehat{CB}+\widehat{CA}}{2}=\widehat{KB}+\widehat{LA}=\angle (KL,AB)\text{ } \blacksquare$$

Let $I_B$ and $I_C$ denote the $B$ and $C$ excenters of $\triangle ABC$. Note that $\angle PQA = \angle QAB - \angle QDA = \frac{A}{2} - \frac{A-B}{2} = \angle PBA$, so $ABQP$ is cyclic. But note that $\angle BI_CI_B = \angle BAI = \angle BAQ = \angle BPQ$ so lines $PQ$ and $I_BI_C$ are parallel. Therefore, $IM$ intersects $I_BI_C$ at its midpoint, which is indeed the midpoint of arc $ACB$, as desired. $\blacksquare$

overcomplicated

Let $I_B$ and $I_C$ denote the excenters

And hence done by homothety.

WLOG let $AC<BC$. Note that $P$ is the incenter of $\triangle DBC$ and $Q$ is the $D$-excenter of $\triangle DAC$. Thus we have: $\angle DCI=\angle DCA+\angle ACI=\angle B+\frac{1}{2}\angle C$. $\angle DCP=\frac{1}{2}\angle DCB=\frac{1}{2}(180^\circ-\angle A)=90^\circ-\frac{1}{2}\angle A$. $\angle DCQ=90^\circ+\frac{1}{2}\angle DCA=90^\circ+\frac{1}{2}\angle B$. Then $\angle PCQ=\frac{1}{2}(\angle A+\angle B)$, and since $\angle PIQ=\angle AIB=90^\circ+\frac{1}{2}\angle C$ it follows that $CPQI$ is cyclic. We now have $\angle PQA=\angle PCI=\angle B+\frac{1}{2}(\angle A+\angle C)-90^\circ=\frac{1}{2}\angle B$, so $ABPQ$ is cyclic as well. Letting $N$ be the midpoint of arc $ACB$, it suffices to show that $\overline{IN}$ is a symmedian in $\triangle BIC$. Since $\angle IAN=\angle BAN-\angle BAI=90^\circ-\frac{1}{2}\angle C-\frac{1}{2}\angle A=\frac{1}{2}\angle B$, $\overline{AN}$ is tangent to $(ABI)$. Likewise, $\overline{BN}$ is also tangent to $(ABI)$, so the conclusion follows. $\blacksquare$

Whaaaaaa????? Why is 2015 Russia full of not-nice geo Relabel $C\to A$, $A\to B$, $B\to C$ because why not. So by USAJMO 2014/6 if $N_a$ is the arc midpoint of $BAC$ on $\omega$ then $IN_a$ is the $I$-symmedian in $\triangle IBC$, since if $M_a$ is the point diametrically opposite $N_a$ on $\omega$ $M_a$ circumcentre of $\triangle IBC$ by Incentre-Excentre and then $BN_a\perp BM_a$ and $CN_a\perp CM_a$ obviously so $IN$ symmedian. Thus $IN$ is the $I$-median of $\triangle IPQ$ if we can show $BCPQ$ cyclic. But $\angle QPC=\angle PCB+\angle PDC=\frac{1}{2}(\angle ADB+\angle ACB)=\frac{1}{2}(\angle ABC-\angle DAB+\angle ACB)=\frac{1}{2}\angle ABC=\angle QBC$ as desired.