CEMC Euclid, 2024

Suppose that $AP = w$, $PD = x$, $AS = y$, and $SB = z$.

We use the notation $|APXS|$ to
represent the area of $APXS$, and so
on.

Thus, $|APXS| = wy$, $|PDRX| = xy$, $|SXBQ| = wz$, and $|XRCQ| = xz$.

Then, $$|APXS| \cdot |XRCQ| = wy\cdot xz = xy \cdot wz = |PDRX| \cdot |SXQB|$$ If $|APXS| = 2$, $|PDRX| = 3$, and $|SWQB| = 6$, then $a = |XRQC| = \dfrac{2 \cdot 6}{3} = 4$.

If $|APXS| = 2$, $|PDRX| = 6$, and $|SWQB| = 3$, then $a = |XRQC| = \dfrac{2 \cdot 3}{6} = 1$.

If $|APXS| = 6$, $|PDRX| = 2$, and $|SWQB| = 3$, then $a = |XRQC| = \dfrac{6 \cdot 3}{2} = 9$.

Since we are told that there are three possible values for $a$, then these are 1, 4 and 9.

(Can you explain why there are exactly three such values?)
The $x$-intercepts of the
parabola with equation $y = x^2 - 4tx + 5t^2 - 6t$ are $$x = \dfrac{4t \pm \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2}$$ The distance, $d$, between these intercepts is their
difference, which is $$d = \dfrac{4t + \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} - \dfrac{4t - \sqrt{(-4t)^2 - 4(5t^2 - 6t)}}{2} = \sqrt{(-4t)^2 - 4(5t^2 - 6t)}$$ From this we see that
$d$ is as large as possible exactly
when the discriminant is as large as possible. Here, the discriminant,
$\Delta$, is $$\Delta = (-4t)^2 - 4(5t^2 - 6t) = 16t^2 - 20t^2 + 24t = -4t^2 + 24t$$ Completing the square, $$\Delta = -4(t^2 - 6t) = -4(t^2 - 6t + 9 - 9) = -4(t^2 - 6t + 9) + 36 = -4(t-3)^2 + 36$$ Since $(t-3)^2 \geq 0$, then $\Delta \leq 36$ and $\Delta = 36$ exactly when $(t-3)^2 = 0$ or $t = 3$.

Therefore, the discriminant is maximized when $t = 3$, which means that the distance
between the $x$-intercepts is as
large as possible when $t = 3$.