THE Tenth ROMANIAN MASTER OF MATHEMATICS

Let the tangents to ${\gamma }_i$ at $A$ and $B$ meet at $C_i$. Since $f_1(x)$ is convex and $f_2(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $A{C}_{1}B{C}_2$.



Lemma. If $CA$ and $CB$ are the tangents drawn from a point $C$ to the graph $\gamma$ of a quadratic trinomial $f(x)=p{x}^2+qx+r$, $A,B\in \gamma$, $A\ne B$, then the abscissa of $C$ is the arithmetic mean of the abscissae of $A$ and $B$.

Proof. Assume, without loss of generality, that $C$ is at the origin, so the equations of the two tangents have the form $y=k_{a}x$ and $y=k_{b}x$. Next, the abscissae $x_A$ and $x_B$ of the tangency points $A$ and $B$, respectively, are multiple roots of the polynomials $f(x)-k_{a}x$ and $f(x)-k_{b}x$, respectively. By the Vieta theorem, $x_A^2=r/p=x_B^2$, so $x_A=-x_B$, since the case $x_A=x_B$ is ruled out by $A\ne B$.

The Lemma shows that the line $C_1{C}_2$ is parallel to the $y$-axis and the points $A$ and $B$ are equidistant from this line.



Suppose, if possible, that the incentre $O$ of the quadrangle $A{C}_{1}B{C}_2$ does not lie on the line $C_1{C}_2$. Assume, without loss of generality, that $O$ lies inside the triangle $A{C}_1{C}_2$ and let $A^{\prime }$ be the reflection of $A$ in the line $C_1{C}_2$. Then the ray $C_{i}B$ emanating from $C_i$ lies inside the angle $A{C}_i{A}^{\prime }$, so $B$ lies inside the quadrangle $A{C}_1{A}^{\prime }{C}_2$, whence $A$ and $B$ are not equidistant from $C_1{C}_2$ – a contradiction.

Thus $O$ lies on $C_1{C}_2$, so the lines $A{C}_i$ and $B{C}_i$ are reflections of one another in the line $C_1{C}_2$, and $B=A^{\prime }$. Hence $y_A=y_B$, and since $$f_i(x)=y_A+p_i(x-x_A)(x-x_B),i=1,2,$$ the line $C_1{C}_2$ is the axis of symmetry of both parabolas, as required.



---

Alternative solution.

Use the standard equation of a tangent to a smooth curve in the plane, to deduce that the tangents at two distinct points $A$ and $B$ on the parabola of equation $y=p{x}^2+qx+r$, $p\ne 0$, meet at some point $C$ whose coordinates are $$x_C=\frac{1}{2}(x_A+x_B)\text{and}{y}_C=p{x}_A{x}_B+q\cdot \frac{1}{2}(x_A+x_B)+r.$$ Usage of the standard formula for Euclidean distance yields $$CA=\frac{1}{2}|x_B-x_A|\sqrt{1+(2p{x}_A+q{)}^2}\text{and}CB=\frac{1}{2}|x_B-x_A|\sqrt{1+(2p{x}_B+q{)}^2},$$ so, after obvious manipulations, $$CB-CA=\frac{2p(x_B-x_A)|x_B-x_A|(p(x_A+x_B)+q)}{\sqrt{1+(2p{x}_A+q{)}^2}+\sqrt{1+(2p{x}_B+q{)}^2}}$$ Now, let the tangents to ${\gamma }_i$ at $A$ and $B$ meet at $C_i$, write the condition in the statement in the form $C_{1}B-C_{1}A=C_{2}B-C_{2}A$, apply the above formula and clear common factors to get $$\frac{p_1(p_1(x_A+x_B)+q_1)}{\sqrt{1+(2p_1{x}_A+q_1{)}^2}+\sqrt{1+(2p_1{x}_B+q_1{)}^2}}=\frac{p_2(p_2(x_A+x_B)+q_2)}{\sqrt{1+(2p_2{x}_A+q_2{)}^2}+\sqrt{1+(2p_2{x}_B+q_2{)}^2}}$$ Next, use the fact that $x_A$ and $x_B$ are the solutions of the quadratic equation $(p_1-p_2)x^2+(q_1-q_2)x+r_1-r_2=0$, so $x_A+x_B=-(q_1-q_2)/(p_1-p_2)$, to obtain $$\frac{p_1(p_1{q}_2-p_2{q}_1)}{\sqrt{1+(2p_1{x}_A+q_1{)}^2}+\sqrt{1+(2p_1{x}_B+q_1{)}^2}}=\frac{p_2(p_1{q}_2-p_2{q}_1)}{\sqrt{1+(2p_2{x}_A+q_2{)}^2}+\sqrt{1+(2p_2{x}_B+q_2{)}^2}}$$ Finally, since $p_1{p}_2<0$ and the denominators above are both positive, the last equality forces $p_1{q}_2-p_2{q}_1=0$; that is, $q_1/p_1=q_2/p_2$, so the two parabolas have the same axis.