Favorite problem: Serbian mathematical competition 2007 - Regional level, year 2, B category

Let $O, I$ be the circumcenter and incenter of $\triangle ABC,$ respectively. By the reflection condition, it is clear that $\triangle ABC$ is isosceles, as $AI$ is an angle bisector, median, and altitude. Let $\angle CAB = \angle CBA = \theta.$ We have

$\begin{aligned} \angle AIO &= \angle AOI \quad \small{\color{cyan} \text{(reflection)}} \\ &= 2 \angle CBA \quad \small{\color{cyan} \text{(inscribed angle)}} \\ &= 2 \theta, \end{aligned}$

and $\angle IAB = \dfrac{\theta}{2}.$ Since these two angles are the angles of a right triangle, we have $2 \theta + \dfrac{\theta}{2} = 90^{\circ},$ or $\theta = 36^{\circ}.$ Therefore, $\angle C = 180^{\circ} - 2 \theta = 180^{\circ} - 2(36^{\circ}) = \boxed{108^{\circ}}.$

for simplicity define the triangle as the points ${A,C,B}= (0,0) , (x_1 , y_1), (x_2,0)$ . by the given condition reflecting the incircle( $I$ over $AB$ is just reflecting over the x axis. let the circumcenter be $O$ . since the circumcenter lies on the perpendicular bisector, it's abscissa lies midway of $AB$ , and since $O$ is just $I$ reflected around the x axis, the abscissa of $I$ lies in the middle of $AB$ . this means $\Delta ABC$ is isosceles. since this system now has 2 degree of freedom, we can work with two variables and later cancel them out in some division to get a formula for the angle that is constant. instead we can call the ordinate of $C$ 1 and make the math easier. it will be justified later as to why the solution is unique. so we have ${A,C,B}= (0,0) , (x , 1), (2x,0)\to {AC,CB,AB}= (\sqrt{x^2+1},\sqrt{x^2+1},2x)$ notice that by definition of circumradius $OC= R , OA=R$ , and by definition of inradius $ID=r$ , and due to the reflection $OD= r$ . then it is easy to see $CD+DO=CO \to 1+r=R$ . we can easily find the circumradius by using triangle $\Delta CDA$ first $\sin (\angle A)= \dfrac{CD}{CA}=\dfrac{1}{\sqrt{x^2+1}} = \dfrac{CB}{2R}= \dfrac{\sqrt{x^2+1}}{2R}\to R= \dfrac{x^2+1}{2} \to r = \dfrac{x^2-1}{2}$ the semiperimeter is $\dfrac{\sqrt{x^2+1}+\sqrt{x^2+1}+2x}{2}= \sqrt{x^2+1}+x$ . we can relate $r$ and $R$ by two different area formulas, i.e $rs= \dfrac{AB\times BC\times CA}{4R} \to \dfrac{x^2-1}{2}\left(\sqrt{x^2+1}+x\right)=\dfrac{2x\times \sqrt{x^2+1}\times \sqrt{x^2+1}}{4\times\dfrac{x^2+1}{2}} \to (x^2-1)\left(\sqrt{x^2+1}+x\right) =2x \\ \to x^2-1 = 2x\left(\sqrt{x^2+1}-x\right) \to 3x^2-1=2x \sqrt{x^2+1} \to (3x^2-1)^2=4x^2(x^2+1)$ if we define $z= x^2+1$ then $(3z-7)^2= 4(z-1)z \to 5z^2-38z+49 = 0 \to z =x^2+1= \dfrac{\pm 2 +2\sqrt{5}}{\sqrt{5}}$ notice that the smaller solution gives $R=\dfrac{z}{2} \approx .55$ but since the circumradius is suppose to be outside the triangle for this case, it can be neglected. this leaves only one choise, and using a previous formula $\sin(\angle A) = \dfrac{1}{\sqrt{z}} \to \angle A= \dfrac{\pi}{5} \to \angle C = \pi - 2\times \dfrac{\pi}{5} = \dfrac{3\pi}{5} = \boxed{108 \deg}$ and this is unique because suppose on the choice of another $CD\neq 1$ a different solution was found, then that triangle should be able to be scaled down to $CD=1$ and the new triangle will still be a solution to the given conditions. but then it must appear as a solution to the $CD=1$ system, which only has one solution. hence the constraint has only one solution for the angle.