Belarus — Final Round

Answer: $\angle ACB = 45^\circ$.
Let $A(a; a^2)$ and $C(c; c^2)$ (see the Fig.). Since $AB \parallel Ox$, we see that $A$ and $B$ are symmetric with respect to the axis $Oy$. Then their ordinates are equal and abscissae differ by sign.
Hence $B(-a; a^2)$. Since $H$ belongs to $AB$ and $CH \parallel Oy$, we have $H(c; a^2)$.
Then $CH = a^2 - c^2$ and (without loss of generality we assume $a > 0$) $AB = 2a$. By condition,
$$a^2 - c^2 = 2a + 1. \quad (*)$$
Moreover, $AH = a - c$ and $BH = c + a$. By Pythagorean theorem, from the right-angled triangle $HCA$ and $HCB$ we obtain
$$CA^2 = (a^2 - c^2)^2 + (a - c)^2 \quad \text{and} \quad CB^2 = (a^2 - c^2)^2 + (c + a)^2.$$
By the law of cosines, from $\triangle ABC$ we obtain
$$AB^2 = CA^2 + CB^2 - 2CA \cdot CB \cdot \cos \angle ACB.$$

Replacing the lengths of $AB$, $CA$ and $CB$ by their obtained expressions and taking into account $(*)$, we obtain
$$\begin{align*} 4a^2 &= (a^2 - c^2)^2 + (a-c)^2 + (a^2 - c^2)^2 + (c+a)^2 - \ &\quad 2\sqrt{(a^2 - c^2)^2 + (a-c)^2}\sqrt{(a^2 - c^2)^2 + (c+a)^2} \cos \angle ACB = 2(a^2 - c^2)^2 \ &\quad + 2(a^2 + c^2) - 2(a-c)\sqrt{(a+c)^2 + 1}(a+c)\sqrt{(a-c)^2 + 1} \cos \angle ACB = \ &\quad \phantom{+ 2(a^2 + c^2)} 2(a^2 - c^2)^2 + 2(a^2 + c^2) - \ &\quad -2(a^2 - c^2)\sqrt{(a^2 - c^2)^2 + (a+c)^2 + (a-c)^2 + 1} \cos \angle ACB \ &\stackrel{(*)}{=} 2(2a+1)^2 + \ &\quad +2(a^2 + a^2 - 2a - 1) - 2(2a+1)\sqrt{(2a+1)^2 + 2a^2 + 2c^2 + 1} \cos \angle ACB = \ &\stackrel{\text{not, then from (3)}}{=} 12a^2 + 4a - 2(2a+1)\sqrt{4a^2 + 4a + 1 + 2a^2 + 2(a^2-2a-1) + 1} \cos \angle ACB = \ &\stackrel{\text{not, then from (3)}}{=} 12a^2 + 4a - 2(2a+1)\sqrt{8a^2} \cos \angle ACB = 12a^2 + 4a - 4\sqrt{2a}(2a+1) \cos \angle ACB. \end{align*}$$
Thus,
$$4\sqrt{2}a(2a+1) \cos \angle ACB = 8a^2 + 4a,$$
whence $\cos \angle ACB = 1/\sqrt{2}$, and therefore, $\angle ACB = 45^\circ$.