I actually now don't mind this problem. Forward: $\angle BAC = 90 \implies MA = MI \implies M = CI \cap \odot ABC \implies PX \perp CI \implies 2(\angle (AB, PX) + \angle (AC, QY)) = 2(90-b-c/2+90-c-b/2) = A = 90$ as desired. Backward: Invert about incircle. Note then $IXP \equiv IYQ$, so as $IX=r=IY$, the moving over the circle $\odot IXY$ to the left with angle equivalent to a chord with length $r$ take $IXP \mapsto IYQ \implies PQ=r$. But, $NP = NQ = r/2$ where $N$ is the ninepoint centre of $XYZ$, so $NPQ$ degenerates, so $\angle XIY=90$ as desired.

the main part is to prove that when $P,I,Q$ are collinear then $ \angle A=90 $ and my idea was same with @IDmasterz invert on $I$ with ratio $r^2$ then the incircle will go to the incircle now assume that $ M \rightarrow M' , N \rightarrow N' $ now we now that $M',N'$ are on the $ \bigcirc IXY$ and on the nine point circle of the $XYZ$ now it's known that $M',N'$ are both inside or both outside the triangle (it is serbia 2013 national p3 that they are reflected from the angle A bisector ) now assume that they are both inside we have that $ \angle MIN= \angle M'IN'=180-A/2 \Longrightarrow A/2= \angle M'XN' \leq \angle ZXK=A/2 $ where $K$ is the foot of $X$ on $ZY$ so we have that $M'=L$ and $N'=K$ so now $ \angle XKY= \angle XIY =90 $ and we are done

Here's a solution without inversion. I will only work on the backward direction. $P,Q$ are the reflections of $X,Y$ over $MI,IN$, hence $P,I,Q$ are collinear $\Rightarrow \angle MIP+\angle NIQ+\angle MIN=180$ $\Rightarrow \angle MIX+\angle NIY+\angle MIN=$ $\angle A-(\angle XMI+\angle YNI)+\angle MIN$ $=\angle A-180-2\angle MIN=180$ $\Rightarrow 180-\angle MIN=\frac {\angle A}{2}=\angle YXI,\angle XYI$ $\Rightarrow \triangle IXM\sim \triangle NYI\sim \triangle NIM$. Let $IX,IY\cap (IMN)=F,E$ resp., since $FX*IX=MX*XN=AX*XB$, therefore $F,E\in (ABI),(ACI)$ resp. Since $MIF=\angle MNI$ by the similarity, therefore $MF=MI$. Let $K$ denote the center of $(ABI)$, which is the midpoint of arc$AB$ that doesn't contain $C$, therefore $MK\perp FI\Rightarrow MK\parallel AB\Rightarrow M\equiv K\Rightarrow MA=MI$, similarly $AN=NI$, hence line $MXYN$ perpendicularly bisects $AI \Rightarrow AXIY$ is a square$\Rightarrow \angle A=90$.

Here is a weird solution... Observe the weird $P,Q$ condition is equivalent to $\angle MIN=180-(\angle A/2)$ by anglechasing. Notice that this is equivalent to $\angle MIX = \angle MNI$ which is equivalent to $MI^2=MX*MN$ but $MI^2-r^2=MX*MY$ (where $r$ is inradius) by power of a point, so it is equivalent to $\boxed{r^2=MX*YN}$. First I prove that when $\angle A=90$ it works. To see this, observe $XY$ is the perpendicular bisector of $AI$ and so $MA=MI$ but it is well known this implies $M$ is the midpoint of $AB$ so $MIC$ are collinear, also $NIB$. And so $\angle MIN= \angle BIC =135$ so we are done in this case Now assume we have $\angle A < 90$ that satisfies the solution (this is enough since $ABC$ is acute). Let $D = XY \cap IA$ and choose $D'$ on segment $AD$ such that $AD'=D'I$, which is possible because $\angle A <90$. Construct $M',N'$ on the circle $\omega$ such that $M'N' \perp AI$ at $D'$. Construct $\gamma'$ centered at $I$ with radius $r'=\sqrt{IA*ID'}>\sqrt{IA*ID}=r$ and clearly this circle contains $\gamma$. Finally, construct $B',C' \in \omega$ such that $AB',AC'$ are tangent to $\gamma$. Clearly $B'$ and $C'$ lie on the minor arcs $AB, AC$ respectively, because $\gamma'$ contains $\gamma$. Let $X',Y'$ be the tangency of $AB',AC'$ with $\gamma '$, and $X'',Y'' = AB \cap M'N', AC \cap M'N'$ respectively. Notice, as I proved above, that $AB'C'$ is a right triangle so it satisfies $r'^2=M'X'*N'Y'$. It is easy to see $r^2=MX*NY > M'X'' * N'Y'' > M'X'*N'Y' = r'^2 > r^2$, a contradiction. So we are done. Alternatively, consider the circumcircle of $IMN$ and let it cut $IY$ at $Z$ (forget about the old $Z$). Then $YZ*YI=YM*YN=YA*YC$ by power of a point so $AIZC$ is cyclic, centered at the midpoint of arc $AC$, $N_0$. Then $\angle IZN=\angle NMI = \angle ZIN$ and so $NI=NZ$ and $N_0I=N_0Z$ so $NN_0$ is perpendicular to $IZ$, which is perpendicular to $AC$, which is tangent to the tangent to $\omega$ at $N_0$, so $N_0=N$ and we finish easily.

After some basic angle chasing, we find that $P,I,Q$ are collinear iff $\triangle NYI \sim \triangle IXM$. This is equivalent to $\frac{NY}{YI} = \frac{IX}{XM} \iff xy=r^2$, where $x=MX, y=NY$. Thus we must show that $xy=r^2$ iff $\angle BAC=90$. Firstly we assume $xy=r^2$. By Power of a Point, $x(XY+y)=(s-a)(s-b)\implies x=\frac{(s-a)(s-b)-r^2}{XY} = \frac{(s-a)(s-b)-\frac{(s-a)(s-b)(s-c)}{s}}{XY} = \frac{(s-a)(s-b)c}{sXY}$. We have $XY=2(s-a)\sin\alpha = 2(s-a)\sqrt{\frac{(s-b)(s-c)}{bc}}$, so $x=\frac{c}{2s}\sqrt{\frac{bc(s-b)}{s-c}}$, and similarly, $y=\frac{b}{2s}\sqrt{\frac{bc(s-c)}{s-b}}$. Thus $r^2=xy=\frac{b^2c^2}{4s^2}\implies \frac{bc}{2s}=r=\frac{abc}{4Rs}=\frac{bc}{2s} \frac{a}{2R} \implies \frac{a}{2R}=1 \implies \sin\angle A=1 \implies \angle BAC=90$. The other direction is basically the same but I'll write it out separately to be rigorous. Assume $\angle BAC=90$. We note that the pair $(x,y)$ obtained above satisfies $xy=r^2$, and thus $x(XY+y)=(s-a)(s-b)$ and $y(XY+x)=(s-a)(s-c)$. Hence if we let $M'X, N'Y$ be the lengths $x,y$ respectively, $BM'AN'$ and $CN'AM'$ are cyclic, where $M',F,E,N'$ are collinear in that order. Thus $BM'AN'C$ is cyclic and $M'=M, N'=N$. So $MX\cdot NY=r^2$, as desired.

XmL wrote: Here's a solution without inversion. I will only work on the backward direction. $P,Q$ are the reflections of $X,Y$ over $MI,IN$, hence $P,I,Q$ are collinear $\Rightarrow \angle MIP+\angle NIQ+\angle MIN=180$ $\Rightarrow \angle MIX+\angle NIY+\angle MIN=$ $\angle A-(\angle XMI+\angle YNI)+\angle MIN$ $=\angle A-180-2\angle MIN=180$ $\Rightarrow 180-\angle MIN=\frac {\angle A}{2}=\angle YXI,\angle XYI$ $\Rightarrow \triangle IXM\sim \triangle NYI\sim \triangle NIM$. Let $IX,IY\cap (IMN)=F,E$ resp., since $FX*IX=MX*XN=AX*XB$, therefore $F,E\in (ABI),(ACI)$ resp. Since $MIF=\angle MNI$ by the similarity, therefore $MF=MI$. Let $K$ denote the center of $(ABI)$, which is the midpoint of arc$AB$ that doesn't contain $C$, therefore $MK\perp FI\Rightarrow MK\parallel AB\Rightarrow M\equiv K\Rightarrow MA=MI$, similarly $AN=NI$, hence line $MXYN$ perpendicularly bisects $AI \Rightarrow AXIY$ is a square$\Rightarrow \angle A=90$. perfect

Notice that if $C,I,M$ where collinear, similarly $B,I,N$ have to be collinear leading to that both $M,N$ lie on the perpendicular bisector of segment $AI$, Thus $X,Y$ also do!! and that establishes the fact $\angle{BAC} = 90$. for the other direction here's two solutions: the first solution So we have to prove that $C,I,M$ are collinear. First of all, let $M'=(CI) \cap (ABC)$ and $S=(MIN) \cap (AIB)$. Direct application of radical axis on $(MIN),(AIBS),(ABC)$ gives $S \in IX$. Simple angle chasing gives that: $$ \angle{MNS} = \angle{MIS} = \angle{MIX} = \angle{MIP} = \frac{1}{2} \angle{XIP} = \frac{1}{2} \angle{YNQ} = \angle{INQ} = \angle{MNI} $$Hence $MI=MS$, thus $M$ lies on the perpendicular bisector of segment $IS$ namely let it be $(\ell)$, but $M'$ is the center of $(ASBI)$ so $M'$ also does. Therefore $M,M' = (ABC) \cap (\ell)$, and since $(\ell),AB$ are parallel this forces $M'=M$, as required. $\blacksquare$ the second solution claim(1):$NQIX$ is cyclic proof: $\angle IQN=\angle IYN =180 -\angle BIA=180 -\angle IXY$ $\blacksquare$ now do an inversion wrt the incircle $N^*$ is on the ninepoint circle of $\triangle ABC$ and $X,N^*,Q$ are collinear and $N^*IXY$ are concyclic claim(2): $N*$ is the foot of the altitude from $X$ to $ZY$ proof: $\angle MXP=\angle MPX =\angle YPX=\angle XZY =90-\angle ZXQ \implies XQ \perp ZY \implies N$ is foor of the altitude $XQ$ $\blacksquare$ then$\angle XAY = \angle XIY =\angle XN^*Y=90$ and we win