Circumcenter and incenter I need a hint on this problem:
Given a triangle ABC. CH is the altitude. CM and CN are the bisectors of $\angle ACH$ and $\angle BCH $. The circumcircle of $\triangle MNC$ and the incircle of $\triangle ABC$ have common center. Find the $\angle ACB$.
 A: Let $O$ be center of circles. Denote $\angle{CAB}=\alpha$, $\angle{CBA}=\beta$, then
$$
\angle{ACH}=\frac{\pi}{2}-\angle{CAB}=\frac{\pi}{2}-\alpha,\quad\angle{BCH}=\frac{\pi}{2}-\angle{CBA}=\frac{\pi}{2}-\beta
$$
$$
\angle{MCH}=\frac{1}{2}\angle{ACH}=\frac{\pi}{4}-\frac{\alpha}{2},\quad\angle{NCH}=\frac{1}{2}\angle{NCH}=\frac{\pi}{4}-\frac{\beta}{2}
$$
$$
\angle{CMH}=\frac{\pi}{2}-\angle{MCH}=\frac{\pi}{4}+\frac{\alpha}{2},\quad\angle{CNH}=\frac{\pi}{2}-\angle{NCH}=\frac{\pi}{4}+\frac{\beta}{2}
$$
$$
\angle{MCN}=\angle{MCH}+\angle{NCH}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2}
$$
$$
\angle{ACB}=\pi-\angle{CAB}-\angle{CBA}=\pi-\alpha-\beta.
$$

Since $O$ is the center of incircle of $\triangle{ABC}$, then
$$
\angle{ACO}=\frac{1}{2}\angle{ACB}=\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2},\quad
\angle{HCO}=\angle{ACO}-\angle{ACH}=\frac{\alpha}{2}-\frac{\beta}{2}.
$$
Denote $CH=x$, then
$$
AC=\frac{CH}{\sin\angle{CAB}}=\frac{x}{\sin\alpha},\quad 
BC=\frac{CH}{\sin\angle{CBA}}=\frac{x}{\sin\beta}
$$
$$
AH=CH\cot\angle{CAB}=x\cot\alpha,\quad BH=CH\cot\angle{CBA}=x\cot\beta 
$$
$$
MH=CH\cot\angle{CMH}=x\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right),\quad NH=CH\cot\angle{CNH}=x\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)
$$
$$
AB=AH+BH=x(\cot\alpha+\cot\beta)
$$
$$
MN=MH+NH=x\left(\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right).
$$
Since $CO$ is the radius of circumcircle of $\triangle{CMN}$, then
$$
CO=\frac{MN}{2\sin\angle{MCN}}=\frac{x\left(\cot\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)+\cot\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right)}{2\sin\left(\frac{\pi}{2}-\frac{\alpha}{2}-\frac{\beta}{2}\right)}=\frac{x}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}.
$$
In order to compute radius of incircle of $\triangle{ABC}$ we need to know its area and perimeter
$$
S_{\triangle{ABC}}=\frac{1}{2}AB\; CH=\frac{1}{2}x^2(\cot\alpha+\cot\beta)
$$
$$
P_{\triangle{ABC}}=AB+BC+CA=x\left(\cot\alpha+\cot\beta+\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}\right).
$$
Hence
$$
r=\frac{2S_{\triangle{ABC}}}{P_{\triangle{ABC}}}=\frac{x^2(\cot\alpha+\cot\beta)}{x\left(\cot\alpha+\cot\beta+\frac{1}{\sin\alpha}+\frac{1}{\sin\beta}\right)}=
\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
Consider altitudes $OP$ and $OQ$ on sides $CH$ and $AB$, then 
$$
CP=CO\cos\angle{HCO}=\frac{x\sin\left(\frac{\alpha-\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
$$
PH=OQ=r=\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}.
$$
Since $CH=CP+PH$ we get the following equation
$$
x=\frac{x\sin\left(\frac{\alpha-\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}+\frac{x\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)+\cos\left(\frac{\alpha-\beta}{2}\right)}
$$
After some laborious algebra we get
$$
\sin\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\alpha+\beta}{2}\right)
$$
$$
\frac{\alpha+\beta}{2}=\frac{\pi}{2}-\frac{\alpha+\beta}{2}
$$
Hence 
$$
\alpha+\beta=\frac{\pi}{2}
$$
$$\angle{ACB}=\pi-\alpha-\beta=\frac{\pi}{2}.
$$
