It seems that the only difficulty is computation.

The main lemma is the following. Claim: If the tangents to $G_1$ at $A$ and $B$ meet at $C$, then the $x$-coordinate of $C$ coincides with the midpoint of $A$ and $B$. (Proof) Suffice to prove for $y = kx^2$ by shifting transformation. If the two points are $(a, ka^2)$ and $(b, kb^2)$ one checks that $\left( \frac{a+b}{2}, kab \right)$ is fine. Let $APBQ$ be the quadrilateral obtained. Then $PQ$ bisects $AB$ at $M$, say. If $\angle AMQ < 90^\circ$ then $AQ < AP$ and $BP < BQ$, giving $AQ + BP < AP + BQ$ which is absurd. So we conclude ${PQ} \perp {AB}$, meaning that ${AB}$ is the perpendicular bisector of ${PQ}$.

My solution : Let $(x,y)$ coordinates of points $A,B$ be $(x_a,y_a),(x_b,y_b)$.We have that the line tangent at $A$ to $f_1$ has slope of $f_1'(x)=2p_1x+r_1$ at point $x_a$, so the equation for that line is $y_{a1}(x)=f_1'(x_a)x +y_a-f_1'(x_a)x_a$. Now the intersection of lines tangent to $f_1$ at $A,B$ have coordinates(it has intersection since $A\neq B$) $y_{a1}(x)=y_{b1}(x)$ $ f_1'(x_a)x +y_a-f_1'(x_a)x_a=f_1'(x_b)x +y_b-f_1'(x_b)x_b$ $x=\frac{y_a-f_1'(x_a)x_a-y_b+f_1'(x_b)x_b}{f_1'(x_a)-f_1'(x_b)}$ $x=\frac{y_a-y_b+f_1'(x_b)x_b-f_1'(x_a)x_a}{f_1'(x_a)-f_1'(x_b)}$ but since $y=p_1x^2+q_1x+r_1$ $x=\frac{p_1(x_a^2-x_b^2)+q_1(x_a-x_b)+2p_1(x_b^2-x_a^2)+q_1(x_b-x_a)}{2p_1(x_a-x_b}$ $x=\frac{p_1(x_b^2-x_a^2)}{2p_1(x_a-x_b)}$ which implies $x=\frac{x_a+x_b}{2}$. Let midpoint of $AB=M$ and let intersections of tangents be $X,Y$ and $\angle XMA=x$. Since $XBYA$ is tanget we have that $XA+YB=XB+YA$.From law of cosine one $\triangle XMA,\triangle XMB,\triangle AMY,\triangle BMY$ we have that $XA=\sqrt{XM^2+\frac{AB^2}{4}-\frac{AB\cdot XM\cdot \cos{x}}{2}}$ and similiar for other which gives us $\sqrt{XM^2+\frac{AB^2}{4}-\frac{AB\cdot XM\cdot \cos{x}}{2}}+\sqrt{YM^2+\frac{AB^2}{4}-\frac{AB\cdot YM\cdot \cos{x}}{2}}= \sqrt{XM^2+\frac{AB^2}{4}+\frac{AB\cdot XM\cdot \cos{x}}{2}}+\sqrt{YM^2+\frac{AB^2}{4}+\frac{AB\cdot YM\cdot \cos{x}}{2}}$ which implies $\cos{x}=0$(since $LHS>RHS$ or vice versa if $\cos{x}\neq 0$) so we have that $XY\perp AB$. Now since $X,Y,M$ have same $x$ coordinates we have that $XY\perp x-axis$ so $AB\parallel x-axis$ which implies that $XM$ is axis of symmetry of $f_1$ and $YM$ is axis of symmetry of $f_2$ but $X,Y,M$ are colinear hence the conclusion.

Let $C, D$ be the points of intersection of the tangents to $G_1, G_2$ at $A, B$ respectively. For the standard parabola, $y=x^2$ the intersection of tangents at $A(a, a^2)$ and $B(b, b^2)$ is $C(h, k)$ then by using slopes, $$\frac{k-a^2}{h-a}=2a,$$and $$\frac{k-b^2}{h-b}=2b,$$solving which gives $h=\frac{a+b}{2}$. This holds for a general parabola by appropriate affine transformations. It follows that $CD$ bisects $AB$ and it is parallel to the $y$-axis. Since $ABCD$ is inscribed, fixing $A, C, D$ we see that $AC-AD=BC-BD$ so $B$ lies on the branch of a hyperbola and a line (dilation of $CD$ about $A$ of ratio $2$), hence it is uniquely determined. It follows that $ACBD$ is a kite and so $AB$ is parallel to the $x$-axis, hence $CD$ is the common axis of symmetry for the two parabolas.

What's Romanian Masters?

Problem 4. Lemma 1. A convex quadrilateral $ACBD$ has an incircle, and $CD$ bisects $AB$. Then $AB \perp CD$. Proof of Lemma 1. Let $X=AB \cap CD$. Assume the contrary and WLOG $\angle AXC = \angle BXD > 90$. Then, by using LoC and $AX=BX$, we can easily get $AC>BC$ and $BD>AD$. Thus we have $AC+BD>BC+AD$, which is a contradiction to the fact that $ACBD$ has an incircle. Lemma 2. Let $A, B$ be two points on a quadratic curve. Let $C$ be the intersection of the two tangents to the quadratic curve at $A$ and $B$. Then, $C=\frac{A+B}{2}$ in terms of $x$ coordinates. Proof of Lemma 2. Translate the quadratic curve to $y=px^2$, and the rest is simple calculation. Now back to the problem, let the convex quadrilateral be $ACBD$. By Lemma 2, we know that $C=D=\frac{A+B}{2}$ in terms of $x$ coordinates, so $CD$ bisects $AB$. By Lemma 1, we now have $CD \perp AB$. Since $C=D$ in terms of $x$ coordinates, $CD$ is parallel to the $y$-axis. This implies that $AB$ is parallel to the $x$ axis, so $A$ and $B$ are symmetric wrt the two axis of symmetry. Therefore, we now know that the two axis of symmetry are actually the same. Done.

Ever since my nightmarish experience of struggling with coordinate bash in a contest, Cartesian axes tend to freak me out. So let's throw out all those $p_i$'s and $q_i$'s and go synthetic. The only important piece of information is that the two parabolas have their axes parallel to each other; let $X_{\infty}$ be their intersection point (yes, we've sneaked into the projective plane). Let $E$ be the intersection of the tangents to $G_1$ at $A,B$. Suppose $EX_{\infty}$ cuts $G_1$ at $P$ and $AB$ at $C$. Note that $APBX_{\infty}$ is a harmonic quadrilateral; so $AB$ and the tangents to $G_1$ at $P$ and $X_{\infty}$ are concurrent. The tangent at $X_{\infty}$ is just the line at infinity, $\ell_{\infty};$ name the tangent at $P$ as $\ell$. Then $AB\cap\ell$ lies on $\ell_\infty$ means $AB||\ell$. But since $PAX_\infty B$ is harmonic, we can project from $P$ to get that $\ell,PC,PA,PB$ makes a harmonic pencil; so $C$ is the midpoint of $AB$. This can be rephrased as $EC||\text{the axis of }G_1$, where $C$ is the midpoint of $AB$. Similarly, if you agree to call the other intersection of tangents to $G_2$ by the name $F$, we have $FC||\text{the axis of }G_2||\text{the axis of }G_1$. So $EC$ and $FC$ are actually the same line; so $EF$ bisects $AB$. The remaining part is similar to what almost everybody else did; assume WLOG $\angle ACF>90^{\circ}$, reason that $AF>FB$ and $BE>AE$, so $AF+BE=FB+AE$ can't hold; so we must have $EF\perp AB$. The rest follows. $\blacksquare$ To sum up, the lesson is simple: every time you see Cartesian coordinates, build a synthetic wall around yourself, and make Descartes pay for it.

MY SOLUTION: Lemma : the reflection of the focus across a tangent to a parabola lies on the directrix and is the foot of the perpendicular from the touchpoint of the tangent to the directrix Proof: well known, but we can use contradiction Lemma: If tangents from P to a parabola meet the parabola at A,B, and the points A',B' are the feet of perpendiculars from A and B to the directrix and F is the focus of the parabola, then P is the circumcenter of FB'C'. Proof : Obvious due to the previous lemma Lemma: The line joining the midpoint of AB and P is parallel to the axis of the parabola. Proof: The previous lemma implies that the midpoint of A'B' lies on the line through P and parallel to the axis so the same line bisects AB. Back to the problem: We have the quadrilateral as ACBD. By the lemma just proved we have CD parallel to the y axis. Now if AB and CD are not perpendicular, an obvious inequality implies that ACBD can't be inscribed. Thus AB and CD are perpendicular and thus CD is the common axis of the parabolas.

Just note that in the above solutions, there is another way to prove that $AB\perp PQ$ Let $I$ be the center of circle inscribed in quadrilateral $APBQ$ Note that midpoint of $AB$ lies on $PQ$, and so does midpoint of $PQ$. Since the center of incircle lies on Newton-Gauss line, we know that $I\in PQ$, and then the quadrilateral have symmetry axis $PQ$. So we get $PQ\perp AB$. (This way, we have some minor flaw when $AB$ also bisect $PQ$. But in that case, we get $APBQ$ must be a parallelogram (and then, rhombus). Then we easily get the claim.)

OK, here's my ridiculously over-projective solution to what was really an algebra problem: Recall that a parabola is identically a conic tangent to the line at infinity. Since these parabolas have axes of symmetry parallel to the y-axis, they are both tangent to the line at infinity on the y-axis, call this point $\infty_y$. Let the intersections of the tangents with respect to $G_1$ and $G_2$ be $C,D$ respectively, so by any of the above arguments $C,D$ have the same x-coordinate $\implies C,D,\infty_y$ collinear. I coordinate bashed this but there are synthetic arguments, and even projective ones (see Ankoganit's). In the remainder of this solution, we let $\infty_{PQ}$ be the point at infinity along line $PQ$, for any points $P,Q$. We apply brianchon's theorem with respect to $G_1$ and degenerate hexagon $CB\infty_{CB}\infty_y \infty_{AC} A$. This tells us that lines $C \infty_y, A \infty_{CB}, B \infty_{AC}$ concurrent. Similarly $D \infty_y, A \infty_{BD}, B \infty_{AD}$ collinear. Then we will take the projective dual of these results with respect to the inscribed circle. Let the circle have center $O$ and be tangent to lines $AC,CB,BD,DA$ at points $P,Q,R,S$ respectively. Then taking the dual of the first brianchon result, we get that the pole of line $C D \infty_y$ lies on the line through $V=PS \cap OQ, W=QR \cap OP$ (this follows by repeated application of LaHire and the known polars of the points $A,B,C,D,O$). However, this pole also lies on the polars of $C$ and $D$, which are lines $PQ$ and $RS$. Thus $PQ,RS,VW$ concur, call this point $T$. Now, let $Q'$ and $P'$ be the reflections of $Q$ and $P$ over $O$, respectively. Applying pascal's theorem to hexagon $SPP'Q'QR$, we get that $SP\cap Q'Q=V$, $PP' \cap QR=W,$ and $P'Q' \cap SR$ are collinear. Therefore, $T \in P'Q'$, so $T=PQ \cap P'Q'$ lies on the line at infinity. Finally, since $T=PQ \cap RS$, we know that $PQ \parallel RS$. Since $PQRS$ is cyclic, it is an isosceles trapezoid; therefore by symmetry $AB \perp CD$, so since $CD$ is vertical, $AB$ is horizontal, and we're done, with absolutely no algebra!

A fast solution with Pascal: Let $PA,PB$ be tangent to $G_2$ and $QA,QB$ be tangent to $G_1$. Let $M$ be the midpoint of $AB$ and let $C$ be the point at infinity lying on the y-axis, so that the line at infinity is tangent to $G_1,G_2$ at $C$. Let $A_1,B_1$ be points with $A_1\in AP, B_1\in BP$ and $AB_1,BA_1$ parallel to the y-axis. By Pascal's Theorem on $AABBCC$ in conic $G_2$ we deduce that $AA\cap BC=A_1, BB\cap AC = B_1, CC\cap AB = \infty_{AB}$ are collinear, implying $A_1B_1||AB$ so $ABA_1B_1$ is a parallelogram. Then it follows that $PM$ is also parallel to the y-axis, as is $QM$. Now the finish is easy; if $PQ$ is not perpendicular to $AB$ then WLOG assume $PA>PB$, so since $P,Q$ are on opposite sides of $M\infty_{\perp AB}$ we deduce that $QA<QB \implies PA+QB>QA+PB$, a contradiction, so we're done.

@OP, seems like $p_2x_2$ should be $p_2x_2$. @below: rip when your correction is corrected

MSTang wrote: @OP, seems like $p_2x_2$ should be $p_2x_2^2$. I believe it should be $p_2x^2$? But yeah.

Somewhat of a stupid problem. Let $AXBY$ be the quadrilateral in question. Claim. $\overline{XY}$ bisects $\overline{AB}$. Proof. It's well known that the $x$-coordinate of the intersection of two tangents to a parabola at $x=a$ and $x=b$ has $x$-coordinate $\frac{a+b}2$. Hence $\overline{XY}$ is a vertical line that passes through the midpoint of $\overline{AB}$. $\blacksquare$ Suppose that $\ell = \overline{XY}$ is not the perpendicular bisector of $\overline{AB}$. Then for $M$ the midpoint of $\overline{AB}$ and $\overrightarrow{AB}$ having a negative $y$-component, any point $X$ on $\ell$ will satisfy $AX > BX$, and any point $Y$ on $\ell$ above $M$ will satisfy $BY > AY$. This implies that $AX+BY>BX+AY$ if $AXBY$ is convex, which is a clear contradiction. Hence $\ell$ is the perpendicular bisector of $\overline{XY}$, so we're done.