Team selection test for the 54th IMO

Let $Y\in BC$ be such that $AY\parallel X{A}_1$. Thus $BY=a-b$ and since $\frac{a}{c}=\frac{c}{a-b}$ we have that $△ABC\sim △YBA$. Therefore $\angle AX{A}_1=\angle BAY=\gamma$ which implies that quadrilateral $XA{A}_{1}C$ is cyclic. It follows from $\angle XAC=\angle X{A}_{1}C=90^{\circ }-\frac{\gamma }{2}$ that $\alpha =90^{\circ }+\frac{\gamma }{2}$. Moreover $\angle A{A}_{1}C=90^{\circ }$, i.e. $A{A}_1$ is simultaneously altitude and angular bisector. Hence $\gamma =\beta$ and $90^{\circ }+\frac{\gamma }{2}+2\gamma =180^{\circ }$, i.e. $\gamma =36^{\circ }$. The angles of $△ABC$ are $36^{\circ },36^{\circ }$ and $108^{\circ }$.