Subfield: Geometry


Question: In the Cartesian plane, let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be the graphs of the quadratic functions $f_{1}(x)=p_{1} x^{2}+q_{1} x+r_{1}$ and $f_{2}(x)=p_{2} x^{2}+q_{2} x+r_{2}$, where $p_{1}>0>p_{2}$. The graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ cross at distinct points $A$ and $B$. The four tangents to $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ at $A$ and $B$ form a convex quadrilateral which has an inscribed circle. Prove that the graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ have the same axis of symmetry.

Solution:
Let $\mathcal{A}_{i}$ and $\mathcal{B}_{i}$ be the tangents to $\mathcal{G}_{i}$ at $A$ and $B$, respectively, and let $C_{i}=\mathcal{A}_{i} \cap \mathcal{B}_{i}$. 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 $C A$ and $C B$ are the tangents drawn from a point $C$ to the graph $\mathcal{G}$ of a quadratic trinomial $f(x)=p x^{2}+q x+r, A, B \in \mathcal{G}, A \neq B$, 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 $A \neq B$.

<img_3615>



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 $f_{i}(x)=y_{A}+p_{i}\left(x-x_{A}\right)\left(x-x_{B}\right)$, the line $C_{1} C_{2}$ is the axis of symmetry of both parabolas, as required.
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 $y=p x^{2}+q x+r$,







$p \neq 0$, meet at some point $C$ whose coordinates are



$$

x_{C}=\frac{1}{2}\left(x_{A}+x_{B}\right) \quad \text { and } \quad y_{C}=p x_{A} x_{B}+q \cdot \frac{1}{2}\left(x_{A}+x_{B}\right)+r

$$



Usage of the standard formula for Euclidean distance yields



$$

C A=\frac{1}{2}\left|x_{B}-x_{A}\right| \sqrt{1+\left(2 p x_{A}+q\right)^{2}} \quad \text { and } \quad C B=\frac{1}{2}\left|x_{B}-x_{A}\right| \sqrt{1+\left(2 p x_{B}+q\right)^{2}},

$$



so, after obvious manipulations,



$$

C B-C A=\frac{2 p\left(x_{B}-x_{A}\right)\left|x_{B}-x_{A}\right|\left(p\left(x_{A}+x_{B}\right)+q\right)}{\sqrt{1+\left(2 p x_{A}+q\right)^{2}}+\sqrt{1+\left(2 p x_{B}+q\right)^{2}}}

$$



Now, 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}\left(p_{1}\left(x_{A}+x_{B}\right)+q_{1}\right)}{\sqrt{1+\left(2 p_{1} x_{A}+q_{1}\right)^{2}}+\sqrt{1+\left(2 p_{1} x_{B}+q_{1}\right)^{2}}}=\frac{p_{2}\left(p_{2}\left(x_{A}+x_{B}\right)+q_{2}\right)}{\sqrt{1+\left(2 p_{2} x_{A}+q_{2}\right)^{2}}+\sqrt{1+\left(2 p_{2} x_{B}+q_{2}\right)^{2}}}

$$



Next, use the fact that $x_{A}$ and $x_{B}$ are the solutions of the quadratic equation $\left(p_{1}-p_{2}\right) x^{2}+$ $\left(q_{1}-q_{2}\right) x+r_{1}-r_{2}=0$, so $x_{A}+x_{B}=-\left(q_{1}-q_{2}\right) /\left(p_{1}-p_{2}\right)$, to obtain



$$

\frac{p_{1}\left(p_{1} q_{2}-p_{2} q_{1}\right)}{\sqrt{1+\left(2 p_{1} x_{A}+q_{1}\right)^{2}}+\sqrt{1+\left(2 p_{1} x_{B}+q_{1}\right)^{2}}}=\frac{p_{2}\left(p_{1} q_{2}-p_{2} q_{1}\right)}{\sqrt{1+\left(2 p_{2} x_{A}+q_{2}\right)^{2}}+\sqrt{1+\left(2 p_{2} x_{B}+q_{2}\right)^{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.