## Problem 1.

The lines $\mathrm{t}$ and $\mathrm{t}'$, tangent to the parabola with the equation $y = \mathrm{x}^{2}$ at points A and B, intersect at point C.

The median of the triangle $\triangle \mathrm{ABC}$ corresponding to vertex C has length $\mathrm{m}$.

Determine the area of the triangle $\triangle \mathrm{ABC}$ in terms of $\mathrm{m}$.
## Solution:

Let $\mathrm{A}\left(\mathrm{a}, \mathrm{a}^{2}\right) ; \mathrm{B}\left(\mathrm{b}, \mathrm{b}^{2}\right)$. The equations of $\mathrm{t}$ and $\mathrm{t}^{\prime}$ are:

$$
t: y=2 a x-a^{2}, \quad t^{\prime}: y=2 b x-b^{2}
$$

and their intersection C is: $C\left(\frac{a+b}{2}, a b\right)$.

The median CM lies on the line: $x=\frac{a+b}{2}$, parallel to the

OY axis. The coordinates of M are: $\left(\frac{a+b}{2}, \frac{a^{2}+b^{2}}{2}\right)$.

We have: $m=C M=\frac{(a-b)^{2}}{2}$ and the height of the triangle $\triangle \mathrm{BMC}$ is: $h=\frac{b-a}{2}=\sqrt{\frac{m}{2}}$. Denoting the area of the triangle with vertices $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ as $\lfloor X Y Z\rfloor$, we finally get:

$$
[A B C]=2[B M C]=2 \frac{1}{2} m \sqrt{\frac{m}{2}}=\sqrt{\frac{m^{3}}{2}}
$$