Nice problem Let $t_{P, \omega}$ be the length of the tangent from $P$ to circle $\omega$. Let $\omega_1,\omega_2,\omega_3$ be the incircles of $AMD,CMD,BMC$. First remark $t_{M, \omega_1} + t_{M, \omega_3} = 0.5((AM+MD-AD) + (BM+CM-BC))$, which by Pitot is just $0.5(AB + CM+MD - (AD + BC)) = 0.5(CM+MD - BC) = t_{M, \omega_2}=t$. Now let $\alpha,\beta,\gamma$ be half of $\angle AMD, \angle DMC, \angle CMB$, respectively. Let $\omega_1, \omega_2$ meet $MD$ at $R,Q$, let $\omega_3$ meet $MC$ at $S$, and let $P=I_2Q\cap MI_1$. First, the previous paragraph tells us that $QR = MS=x$, hence $\dfrac{PI_1}{MI_3} = \dfrac{x}{\cos \alpha} \dfrac{\cos \gamma}{x} = \dfrac{\cos \gamma}{\cos \alpha}$. Meanwhile, $\dfrac{PI_2}{MI_2} = \dfrac{PQ+QI_2}{MI_2}=\dfrac{t (\tan \alpha +\tan \beta)}{\frac{t}{\cos \beta}} = \dfrac{ \sin (\alpha + \beta) \cos \beta}{\cos \alpha \cos \beta} = \dfrac{\cos \gamma}{\cos \alpha}$, hence $\dfrac{PI_1}{MI_3} = \dfrac{PI_2}{MI_2}$. It's also clear that $\angle I_1PI_2 = 90^{\circ}-\alpha = \angle I_2MI_3$, hence $PI_1I_2\sim MI_3I_2$, so $\angle PI_1I_2=\angle MI_3I_2$, implying $I_1I_2I_3M$ is cyclic as desired.

This one was AZE JBMO TST 2016 .Its original source is BMO 2016 SL G1 .lol

Lamp909 wrote: Let $ABCD$ be a quadrilateral ,circumscribed about a circle. Let $M$ be a point on the side $AB$. Let $I_{1}$,$I_{2}$ and $I_{3}$ be the incentres of triangles $AMD$, $CMD$ and $BMC$ respectively. Prove that $I_{1}I_{2}I_{3}M$ is circumscribed. Put another way, this problem states that in $\triangle CMD$ with incenter $I_3$, points $I_1, I_2$ slide along fixed lines so that $\angle I_1MI_2=90^{\circ}+\tfrac{1}{2}\angle CMD$ such that if $X, Y$ are projections of $I_1, I_2$ on $\overline{MC}, \overline{MD}$ respectively then $MX+MY=\tfrac{1}{2}{MC+MD-CD}$ is constant; then $I_3$ lies on $\odot(MI_1I_2)$. Now to solve it, phantom $I_2'$ to lie on $\overline{MI_2}$ and $\odot(MI_1I_3)$. Then $I_1 \mapsto I_2'$ is a linear map by spiral similarity pivoted at $I_3$. Also $I_1 \mapsto I_2$ is linear. Thus, it suffices to solve the problem for precisely two cases. It is obvious for the case when either of $I_1, I_2$ coincides with $M$, so we're done. EDIT: In my notation $I_3, I_2$ have been swapped so $I_3$ is the incenter of $\triangle CMD$. Oops.

$\angle AMI_1=\angle I_1MD=\alpha$, $\angle DMI_2=\angle I_2MC=\beta$, $\angle CMI_3=\angle I_3MB=\gamma$ $\alpha+\beta+\gamma=90^{\circ}$ $\angle DI_1O =180^{\circ}-\angle DI_1A=90^{\circ}-\alpha$, $\angle CI_3O =180^{\circ}-\angle CI_3B=90^{\circ}-\gamma$ $\angle DI_2C =90^{\circ}+\beta =90^{\circ}-\alpha+90^{\circ}-\gamma=\angle DI_1O+\angle CI_3O $ $\angle I_1DO=\angle I_2DC, \angle I_3CO=\angle I_2CD$ Point $O'$ on $\overline{DC}$, $\angle DI_2O' = \angle DI_1O$ Then $\triangle DI_1O \sim \triangle DI_2O'$ & $\triangle CI_3O \sim \triangle CI_2O'$ By spiral similarity $\triangle DI_1I_2 \sim \triangle DOO'$ & $\triangle CI_3I_2 \sim \triangle COO'$ $\angle I_1MI_3+\angle I_1I_2I_3=90^{\circ}+\beta+(360^{\circ}-\angle DI_2I_1-\angle CI_2I_3-\angle DI_2C)$ $=90^{\circ}+\beta+(360^{\circ}-\angle DO'I_2-\angle CO'I_2-(90^{\circ}+\beta))=180^{\circ}$ Hence $MI_1I_2I_3$ is cyclic

Can someone explain anantmudgal09's solution to me? It's the first time I'm seeing such a solution.

Let $ \angle AMI_1 = I_1MD = x$, $ \angle DMI_2 = I_2MC = y$, $ \angle CMI_3 = I_3MA = z$ and $ \angle ADI_1 = I_1DM = k$. Claim1 : $\angle I_1DO = I_2DC$ and $\angle I_3CO = I_2CD$. Proof : $\angle I_1DO = \angle 180 - \angle DI_1O - \angle I_1OD = \angle 180 - \angle (90 - x) - \angle (180 - A/2 - B/2) = \angle A/2 + \angle B/2 + x - \angle 90$ and $\angle I_2DC = (\angle B - 2k)/2$ so it's easy to see $\angle I_1DO = I_2DC$. we prove the other part with same approach. Claim2 : $\angle DI_1O + \angle CI_2O = \angle BI_2C$. Proof : $\angle DI_1O = \angle 90 - x$ , $\angle CI_2O = \angle 90 - z$ and $\angle BI_2C = \angle 90 + y$ so it's easy to see our claim is true. Let $S$ be a point on $CD$ such that $\angle DI_2S = \angle DI_1O$, then it's obvious $DI_1O$ and $DI_2S$ are similar so $\angle I_1I_2D = \angle OSD$. with same approach we have $\angle I_3I_2C = \angle OSC$ so $\angle I_1I_2I_3 = \angle180 - \angle BI_2C = \angle 90 - y = \angle 180 - \angle I_1MI_3$ so $I_1MI_3I_2$ is cyclic. we're Done.

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Let $\angle AMI_1=\angle DMI_1=\alpha$, $\angle DMI_2=\angle CMI_2=\beta$, and $\angle CMI_3=\angle BMI_3=\gamma$. We have $$MI_2=\frac{MD+MC-DC}{2\cos\beta},$$and similarly $$MI_{1}=\frac{MA+MD-AD}{2\cos\alpha},MI_{3}=\frac{MB+MC-BC}{2\cos\gamma}.$$However, note that $\sin\angle I_1MI_3=\sin\alpha+2\beta+\gamma=\sin90+\beta=\cos\beta,$ so $$MI_2\sin\angle I_1MI_3=\frac{MD+MC-DC}{2}.$$Similarly, $$MI_1\sin\angle I_2MI_3=\frac{MA+MD-AD}{2},MI_3\sin\angle I_1MI_2=\frac{MB+MC-BC}{2}.$$However, since $MA+MB+DC=AD+BC$, the latter two add to the former, so $MI_1I_2I_3$ is cyclic and we are done. remark: For this problem, the single-vertex form of Ptolemy is very useful since the cyclic quadrilateral is "centered" around vertex $M$. Furthermore, $MI_1,MI_2,MI_3$ are relatively well-understood, so the advantage of this form is that we do not have to deal with less-well-understood lines between incenters of different triangle, which many other approaches must do.

Lemma: If a circle divides a bisected angle as shown, then $AX+AY=2\cos \tfrac{\alpha}{2} \cdot AZ$, or twice the length of the projection of $AZ$. Simply drop altitudes and use congruent triangles. A key takeaway is that we have a similar result when the circle intersects $AY$ on its extension; in that case, we would have $AX-AY=2\cos \tfrac{\alpha}{2} \cdot AZ$. $\Box$ [asy][asy] size(125); pair A, X, Y, Z; A = dir(230); X = dir(90); Y = dir(310); Z = dir(20); draw(A--(Y*1.5-A*.5)--A--(X*1.2-A*.2)--A--(Z*1.2-A*.2)); draw(circle((0,0),1)); dot("$A$", A, dir(225)); dot("$X$", X, dir(110)); dot("$Y$", Y, dir(280)); dot("$Z$", Z, dir(350)); [/asy][/asy] Define the intersection of $(MI_2I_3)$ with $AB$ the bisector of $\angle AMD$ as $N$ (WLOG on segment $MB$) and $J$. Using the lemma, we compute \begin{align*} MP+MQ &= 2\cos \tfrac{\angle PMQ}{2} \cdot MI_2 \\ &= MC+MD-CD \\ &= (AM+MD-AD) + (BM+MC-BC) \\ &= 2\left(s_{\triangle AMD} - AD + s_{\triangle BMC} - BC\right) \\ &= 2\cos \tfrac{\angle AMD}{2} \cdot MI_1 + 2\cos \tfrac{\angle BMC}{2} \cdot MI_3 \end{align*} using Pitot's on $ABCD$. However, we also have \begin{align*} MP+MQ &= (MP-MN) + (MQ+MN) \\ &= 2\cos \tfrac{\angle AMD}{2} \cdot MJ + 2\cos \tfrac{\angle BMC}{2} \cdot MI_3, \end{align*} and hence we have $J = I_1$, so $MI_1I_2I_3$ is cyclic. $\blacksquare$

We want to show that $I_1I_2I_3M$ is cyclic. We will show this by Ptolemy's. It is needed to prove that $MI_1 \cdot \sin \angle I_2MI_3 + MI_3 \cdot \sin \angle I_1MI_2 = MI_2 \cdot \sin \angle I_1MI_3$. Now let $\triangle ABC$ be a triangle with angles $\alpha$, $\beta$ and $\gamma$ and incenter I. We will show that $AI = \frac{AC + AB - BC}{2\cos \frac{\alpha}{2}}$. Now we have to show that $\frac{AC + AB - BC}{2\cos \frac{\alpha}{2}} = \frac{2R(\sin \beta + \sin \gamma - \sin \alpha)}{2\cos \frac{\alpha}{2}} = 4R\sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}$ $\Leftrightarrow$ $\sin \beta + \sin \gamma - \sin \alpha = 4\sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2} \cdot \cos \frac{\alpha}{2}$ $\Leftrightarrow$ $\sin (\alpha + \beta) + \sin \beta - \sin \alpha = \sin (\alpha + \beta) + 2\cos \frac{(\alpha + \beta)}{2} \cdot \sin \frac{(\beta - \alpha)}{2} = 4\sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2} \cdot \cos \frac{\alpha}{2}$ $\Leftrightarrow$ $\sin (\alpha + \beta) + 2\cos \frac{(\alpha + \beta)}{2} \cdot \sin \frac{(\beta - \alpha)}{2} = 4\cos \frac{(\alpha + \beta)}{2} \cdot \sin \frac{\beta}{2} \cdot \cos \frac{\alpha}{2}$ $\Leftrightarrow$ $2\cos \frac{(\alpha + \beta)}{2} \cdot (2\sin \frac{\beta}{2} \cdot \cos \frac{\alpha}{2} - \sin \frac{(\beta - \alpha)}{2}) = \sin (\alpha + \beta)$ $\Leftrightarrow$ $2\cos \frac{(\alpha + \beta)}{2} \cdot (2\sin \frac{\beta}{2} \cdot \cos \frac{\alpha}{2} - \sin \frac{(\beta - \alpha)}{2}) = 2\sin \frac{(\alpha + \beta)}{2} \cdot \cos \frac{(\alpha + \beta)}{2}$ $\Leftrightarrow$ $2\sin \frac{\beta}{2} \cdot \cos \frac{\alpha}{2} - \sin \frac{(\beta - \alpha)}{2} = \sin \frac{(\alpha + \beta)}{2}$ $\Leftrightarrow$ $2 \cdot \frac{1}{2} \cdot (\sin \frac{(\alpha + \beta)}{2} + \sin \frac{(\beta - \alpha)}{2}) - \sin \frac{(\beta - \alpha)}{2} = \sin \frac{(\alpha + \beta)}{2}$, which is obviously correct. Using this we get that $MI_2 = \frac{MD + MC - DC}{2\cos \beta}$, $MI_1 = \frac{MA + MD - AD}{2\cos \alpha}$ and $MI_3 = \frac{MB + MC - BC}{2\cos \gamma}$. Now $MI_2 \cdot \sin \angle I_1MI_3 = \frac{MD + MC - DC}{2\cos \beta} \cdot \sin (\alpha + 2\beta + \gamma) = \frac{MD + MC - DC}{2\cos \beta} \cdot \cos \beta = \frac{MD + MC - DC}{2}$, similarly $MI_1 \cdot \sin \angle I_2MI_3 = \frac{MA + MD - AD}{2}$ and $MI_3 \cdot \sin \angle I_1MI_2 = \frac{MB + MC - BC}{2}$. Since we wanted to show that $MI_1 \cdot \sin \angle I_2MI_3 + MI_3 \cdot \sin \angle I_1MI_2 = MI_2 \cdot \sin \angle I_1MI_3$ it is left to show that $\frac{MA + MD - AD}{2} + \frac{MB + MC - BC}{2} = \frac{MD + MC - DC}{2}$ $\Leftrightarrow$ MA + MD + MB + MC - AD - BC = MD + MC - DC $\Leftrightarrow$ MA + MD + MB + MC + DC = MD + MC + AD + BC $\Leftrightarrow$ MA + MB + DC = AD + BC $\Leftrightarrow$ MA + MB + DC = AB + DC $\Leftrightarrow$ MA + MB = AB, which is correct $\Rightarrow$ we are ready and $MI_1I_2I_3$ is cyclic.