solved by enhanced as problem 32 here

Okay here is moving points solutions that enables us to work with complex numbers. First move $P$ on the circle $(ABC)$. Then $$P\mapsto Q\mapsto M$$is obviously projective so $M$ moves with deg $1$. Next $R$ moves with deg $4$. Now the claim is equivalent with showing that $ \infty_{ \perp AK}$ $K$ and $R$ are collinear. So we need to check problem statement in $1+1+4+1=7$ cases. Let $D,E$ and $F$ be midpoints of arcs $A$, $B$ and $C$ (that are opposite to $A$, $B$ and $C$). Claim is well known for $P=E,F$, and is trivial for $P=D$. Now something harder, let $P=A$. Then $Q=AA \cap EF$ and $R=\infty_{EF}$ so we need to show $IQ||BC$. Here we employ complex numbers. $a=x^2,b=y^2,c=z^2$ Calculating $Q$ we get $q=x^2\frac{x(y+z)+2yz}{yz-x^2}$. Its left to show $\frac{q-I}{\bar q - \bar I}=-bc$. $$q+I=x^2\frac{x(y+z)+2yz}{yz-x^2}+\sum xy$$$$yz\frac{(x+y)(x+z)}{yz-x^2}$$And now it is easily seen that we are done in this case. Next one is interesting: take $P$ on $(ABC)$ s.t. $<API=90$. Again bash it but little bit smarter this time. First calculate $t=ef\cap uw$ where $U$ and $W$ are intersections of $KL$ with $(ABC)$. We easily see $u+w=\frac{1}{2}(a+b+c+\frac{bc}{a})$ and $uw=bc$. Now we can use intersection formula to get $$t=\frac{x^2\sum xy+y^2z^2+2xyz(y+z)}{2(x^2-yz)}$$Notice that $\frac{p-a}{\bar p - \bar a}=-\frac{p-(-a)}{\bar t - \bar (-a)}$ We have avoided calculating $P$ (its not that bad either, but this is nice trick). Next it is sufficint to show $$\frac{t-a}{\bar t - \bar a}=\frac{t-a}{\bar t - \bar a}=-\frac{p-(-a)}{\bar t - (-\bar a)}$$Which is equivalent with $\frac{t-a}{I-(-a))} \in R$ But $$t-a=\frac{(x+y)^2(x+z)^2}{2(x^2-yz}$$And again as before its easily seen that we are done. And now the hardest case $P=C$ (we symmetricaly get $B$). Then calculating we get $r=\frac{x(z^2+xy)+xz(y+z)}{x-z}$ Via angle chasing we can see $IM||AB$ then its not hard to see $$m=\frac{2xyz \sum x+2z^2 \sum xy+ (x^2+y^2)(x^2+z^2) }{x^2-z^2}$$Next just calculate $\frac{m-a}{m-r} \in iR$ $$m-a=\frac{(x+z)(x+y)(\sum xy-x^2+2z^2)}{x^2-z^2}$$$$m-r=\frac{(y-x)(x+z)(\sum xy-x^2+2z^2)}{x^2-z^2}$$Now its trivial to check that we are done. Just noticed sythetic for $P=C$. Note that if $KL \cap RC=S$ then $AS\perp RC$ and claim follows easily. Note: My favorite is case $5$ (where $<API=90$).

Nice problem! Let $D$ is the intersection of $BC$ and $AM$, we prove that $\triangle AIP \sim \triangle IDP$. Let $t$ be a line passing through $D$ tangent to incircle, by second Fontene's theorem, $I$ lies on the Newton's line of it, so $IM$ is the Newton's line of quadrangle $\mathcal{Q}(BC, AC, AB, t)$. It's well-known that Miquel point and infinite of Newton's line are isogonal conjugate, hence $P$ is the Miquel point of $\mathcal{Q}(BC, AC, AB, t)$,so $\triangle AIP \sim \triangle IDP$. $J$ is the point symmetric of $I$ WRT $P$, then $A, I, D, J$ are harmonic quadrilateral, so $R$ is the circumcenter of $\odot(AIDJ)$, $RM$ is perpendicular to $AM$, we're done.

Let $F=\overline{AI}\cap\overline{BC}$, $L=\overline{AI}\cap(ABC)$, and let $X$ and $N$ be the midpoints of segments $AF$ and $AI$. Let $M'$ and $X'$ be the reflection of $M$ and $X$ over $l$, and let $O$ be the circumcenter of $\triangle ABC$. Notice that $M'$ lies on $\overline{AQ}$. Since $\angle AMR = \angle IM'R$, it suffices to show that $M'$ lies on $(INRP)$. Let $k$ be the reflection of $A$-midline over $l$; this passes through $M'$ and $X$. Claim: $k$ is the image of $(ABC)$ under inversion at $A$ with radius $\sqrt{AN \cdot AI}$. Proof. Since $\overline{AI}$ and $l$ are perpendicular to each other, $k$ is parallel to the line obtained by reflecting the $A$-midline over $\overline{AI}$. i.e. $k$ is anti-parallel to $\overline{BC}$ in $\angle A$, and hence it is perpendicular to $\overline{AO}$. Therefore, to show the claim, we just need to show that $X'$ is the inverse of $L$ under inversion, or \[AX' \cdot AL = AN \cdot AI \Longleftrightarrow \frac{AX'}{AN} = \frac{AI}{AL}\]Just notice that \[\frac{AN}{NX'} = \frac{AN}{NX} = \frac{AI}{IF} = \frac{AC}{CF} = \frac{AL}{LB} = \frac{AL}{LI}\]so $\{A, X', N\} \sim \{A, I, L\}$, and the claim follows. $\square$ As $AM' \cdot AP = AX \cdot AL = AN \cdot AI$, it follows that $M'$ lies on $(INP)$ so we are done. $\blacksquare$ [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(120); pair B = dir(225); pair C = dir(315); pair I = incenter(A,B,C); pair F = extension(A,I,B,C); pair O = circumcenter(A,B,C); pair L = 2*foot(O,A,I)-A; pair X = midpoint(A--F); pair N = midpoint(A--I); pair X1 = 2*N-X; pair S = midpoint(A--C); pair T = midpoint(A--B); pair P = dir(238); pair D = 2*foot(O,B,I)-B; pair E = 2*foot(O,C,I)-C; pair Q = extension(A,P,D,E); pair M = extension(Q,I,S,T); pair R = extension(P,rotate(90,P)*I,D,E); pair M1 = 2*foot(M,D,E)-M; real s = 0.9; pair K1 = s*unit(M1-X1) + extension(M1,X1,A,O); pair K2 = s*unit(X1-M1) + extension(M1,X1,A,O); draw(K1--K2, royalblue+1); draw(A--B--C--cycle, black+1); draw(B--D, mydash); draw(C--E, mydash); draw(R--D); draw(A--L); draw(A--P); draw(P--R); draw(T--S); draw(Q--I); draw(R--M); draw(circumcircle(A,B,C)); draw(circumcircle(P,I,N)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(315)); dot("$F$", F, dir(225)); dot("$L$", L, dir(270)); dot("$X$", X, dir(45)); dot("$N$", N, dir(60)); dot("$X'$", X1, dir(60)); dot(S); dot(T); dot("$P$", P, dir(270)); dot(D); dot(E); dot("$Q$", Q, dir(55)); dot("$M$", M, dir(225)); dot("$R$", R, dir(180)); dot("$M'$", M1, 2*dir(180)); label("$k$", K2, dir(60)); [/asy][/asy]

Here is an elementary solution. Lemma 1: Let $P, A, I, K$ be four different points on the plane such that no three of them are collinear and $\triangle PAI\sim \triangle PIK$. Then, the $I-$symmedian in the triangle $AIK$ passes through $P$. Moreover, if $O$ is the center of $(AIK)$, then $\angle OPI=90^\circ$.

Lemma 2: Let $ABC$ be a triangle with incenter $I$ and let $P$ be a point on $(ABC)$. The tangent at $I$ of $(AIP)$ intersects $BC$ at $K$. Then, $\triangle PAI\sim\triangle PIK$.

Now let's get back to the problem. Assume that the tangent at $I$ of $(AIP)$ intersects $BC$ at $K$ and $AK\cap QI=M'$. By Lemma 2, we know that $\triangle PAI\sim \triangle PIK$ and by Lemma 1, $IP$ is a symmedian of $(AIK)$. Also, $\angle PIK=\angle PAI=\angle QAI=\angle QIA=\angle M'IA$. Since $IP$ is symmedian, we find that $M'$ is the midpoint of $[AK]$. Since $K$ lies on $BC$, we know that $M'$ lies on the midsegment of $ABC$ that is parallel to $BC$. Also, $M'\in QI$. Hence, $M'=M$. Let $R'$ be the center of $(AIK)$. By Lemma 1, $\angle R'PI=90^\circ$. Also, clearly, $R'$ lies on $\ell$. Hence, $R'=R$. Then, $|RA|=|RK|$. Since $M$ is the midpoint of $[AK]$, we have $\angle RMA=90^\circ$, as desired.

Very hard problem. Let $N=AM \cap BC$ so it's enough to prove that $R$ is the circumcenter of $\triangle ANI$ because $AM=MN$. Let $P'$ the reflexion of $I$ respect $P$ then we have that $R$ is the center of $\odot (AIP')$ so it's enough to prove that $AINP'$. Now a little motivation: If $AINP'$ is cyclic, we have that $\angle IAP = \angle AIM$ and $\angle AP'I= \angle INA$ then see that $\triangle NIA$ is almost similar to $\triangle P'AI$ then let $P''=IN \cap \odot(IAM)$ then we can guess that $\triangle NP''A \sim \triangle P'AI$ but with angles: $$\angle NIM= \angle P''AM = \angle P'IA$$But $M$ is the midpoint of $AN$ so $AINP'$ is harmonic so $\triangle API \sim \triangle IPN$ (well-known) and this is sufficient to prove that $AINP'$ is cyclic. That give us the idea to use phantom points, let $N'$ a point such that $\triangle API \sim \triangle IPN'$ , we don't know if this point it's on $NC$. Then the follows steps are natural: Claim: $N' \in BC$ Proof: Let $I_A,I_B,I_C$ the excenters relative to $A,B,C$, respectively. The inversion $\Psi$ in center of $I$ that switch $A \mapsto I_A, B \mapsto I_B$ and $C \mapsto I_C$ then $P \mapsto \Psi(P)$ and $N' \mapsto \Psi(N')$. Then by angles: $$\angle I\Psi(P)I_A =\angle IAP = \angle PIN' =\angle \Psi(N')I\Psi(P)$$$$\angle\Psi(P')\Psi(N')I=\angle IPN'=\angle IPA =\angle II_A\Psi(P)$$This implies that $\Psi(N')II_A\Psi(P)$ is a parallelogram and we need to prove that $\Psi(N'),I,I_B,I_C$ are concyclic. By homothety with center $I_A$ with ratio $1/2$ we need to prove that the midpoints of $IP$,$I_AI_B$,$I_AI_C$,$I_AI$ are concyclic but by homothety with center $I$ and ratio $2$ the image of the midpoint of $PI$ it's on the circumcircle of $(I_AI_BI_C)$ and it's well-known that the others points also are on this circle. $\square$ Let $M' =AN'\cap IQ$ then $\angle PIN'=\angle PAI=\angle M'IA$ later see that the conjugate isogonal of $P$ respect $\triangle AIN'$ it's the $I$-Dumpty point of $\triangle AIN'$ so we have that $IM'$ it's a median this implies that $N'=N$ and $M=M'$.Then let $R= IP \cap \odot(AIN)$ so $ARNI$ is harmonic and $\angle API=\angle IPN$ then $P$ is the $R$-Humpty point of $\triangle ARN$ this implies that $IP=PR$ (well-known) so $P'=R$ and $AINP'$ is cyclic, as desired. $\blacksquare$

Let the reflection of $M$ over $\ell$ be $N$. Let $L=AI\cap \ell$. Then it suffices to prove $INPL$ cyclic as this would prove $INPLR$ cyclic giving $\angle AMR=\angle INR=\angle IPR=90$. Let the $A$-midline intersect $l$ at $X$. Then $XN, XM$ are reflections about $\ell$. Step 1: Preparations. $XN$ is parallel to the tangent to $(ABC)$ at $A$ because the tangent and $BC$ are anti-parallel in $\angle BAC$. Let $T=AI\cap XN$ and $D$ be the foot from $A$ onto $BC$. Then $X$ is the centre of $(AID)$ because it lies on the perpendicular bisectors of $AI$ and $AD$. Step 2: Let $XN$ intersect $AB$ and $AC$ at $E, F$ respectively. Then $EBIL, FCIL$ cyclic. Proof: Dilate at $A$ with factor $2$, sending $X\mapsto Y$ and $L\mapsto I$. But then $XI\perp AI$ so if $E, F\mapsto U, V$ then the lines $TUV$ and $BC$ are reflections about $TI$, so $TUV$ is tangent to the incircle of $\triangle ABC$. But then $$\angle AIU=180-\angle AUI-\angle IAB = 180-\angle AUV-\angle IAB - \angle IUV = 180-\angle ACB-\frac{1}{2}\angle BAC - \frac12\angle BUV$$$$ = 180-\frac12\angle ACB - \frac12 \angle BAC - \frac12 (180- \angle ACB) = 90-\frac12\angle ACB - \frac12\angle BAC = \frac12 \angle ABC = \angle ABI$$so $\triangle AIU \sim\triangle ABI$ giving $AU \times AB = AI^2$. Thus dilating back, $AE \times AB = AI \times AL$ as needed, so $EPIL$ and similarly $FCIL$ are cyclic. Step 3: Finishing: This gives $AN \times AP = AE \times AB = AK \times AL$ where the first equality comes from shooting lemma, so by power of a point, $INPL$ cyclic as needed.

Let $J$ be the midpoint of $AI$. We prove the following important claim: Claim. If $AP$ meets $(PJI)$ again at $N$, then the reflection $N'$ of $N$ across $\ell$ lies on the $A$-midline. Proof. It is easy to see by inversion at $A$ that $N'$ lies on a fixed line parallel to $BC$. Also, when $P$ is the $A$-Sharkydevil point, $\angle AJN = \angle API = 90^\circ$ implies $N'=N$, so by a homothety at $A$ with ratio $2$ it suffices to show that $W = AP \cap BC$ satisfies $\angle AIW = 90^\circ$, which is well-known. $\square$ Back to the problem, note that $R$ lies on $(PJIN)$ and $\angle NQJ = \angle JQI$; combined with the claim, this gives $M = N'$, so $\angle AMR = \angle INR = 90^\circ$.

(i didnt invert anything!)

. Let $A'$ be the reflection of $A$ over $M$, which lies on $BC$. We rely on a crucial claim within this problem: Claim: $PI$ bisects $\angle APA'$. Suppose that $NP$ intersects $BC$ at $P'$. Instead, reverse reconstruct $A'$ such that $IPP'A$ is cyclic, and $A'$ lies on $BC$. Note that this point $A'$ is unique. Now, by shooting lemma and Fact 5, $NI^2=NB^2=NP'\cdot NP$, thus $(IPP'A)$ is tangent to $AI$. Further, if $D$ is the foot of the angle bisector at $A$, then $ND\cdot NA=NP\cdot NP'$ as well, thus $ADP'P$ is cyclic by Power of a Point. This implies that $\angle PAI=\angle NP'B = \angle PIA'$, thus $(AIP)$ is tangent to $IA'$. Yet $(IPP'A)$ is tangent to $IA'$. This is sufficient to characterize $P$ as the $I$-Dumpty point of $\triangle AIA'$. Consequently, it is well-known that $IP$ is the $I$-symmedian in $\triangle AIA'$. Now we claim that $A'$ indeed satisfies $PI$ bisecting $\angle APA'$. Yet check that $\angle API=\angle NIA'=\angle IPA'$ by alternate segment theorem, which provides the result. Now, by reverse reconstruction again, if we do redefine the $M$ as the midpoint of $AA'$, and further let $IM$ intersect $AP$ at $Q$, it suffices to show that $Q$ lies on the perpendicular bisector of $AI$, or equivalently $\angle QAI=\angle AIQ$. Yet $\angle IAP = \angle A'IP=\angle QIA$ by alternate segment theorem, and symmedian property, which suffices. Thus our claim is proved. Now, consider the point $P$ again. Reconstruct $R$ as the center of $(AIA')$. It is well-known that $P$, being the $I$-Dumpty point of $\triangle AIA'$ satisfies $\angle IPR=90^\circ$. Yet we note that $R$ also lies on the perpendicular bisector of $AI$ by circumcenter definition. Yet this uniquely characterizes $R$. Further, it is well-known that $ AA'PR $ is cyclic. Now, if $PI$ intersects $(AA'PR)$ again at $X$, by Fact 5, due to our claim, $X$ is the midpoint of minor arc $AA'$. Further, $(PI, PR; PA, PA')=-1$ by right angle and bisector. Projecting this onto $(AA'PR)$, we have $RAXA'$ is harmonic, and as $X$ is midpoint of minor arc $AA'$, it is a kite, so $R,X$ are antipodes of each other. Consequently, $R$ is the midpoint of major arc $AA'$, thus as $M$ is the midpoint of $AA'$, we have $\angle AMR=90^\circ$, as desired.