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, $$a^2-c^2=2a+1.(\ast )$$ Moreover, $AH=a-c$ and $BH=c+a$. By Pythagorean theorem, from the right-angled triangle $HCA$ and $HCB$ we obtain $$C{A}^2=(a^2-c^2{)}^2+(a-c{)}^2\text{and}C{B}^2=(a^2-c^2{)}^2+(c+a{)}^2.$$ By the law of cosines, from $△ABC$ we obtain $$A{B}^2=C{A}^2+C{B}^2-2CA\cdot CB\cdot \cos \angle ACB.$$

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