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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 $\angle ACB$.

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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. $

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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}. $

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    @AdamAndersson, I've written a new solution.2012-06-08