, 2011

Place the square in the coordinate plane with $A(-1, 1)$, $B(1, 1)$, $C(1, -1)$ and $D(-1, -1)$. The equation of $\Gamma$ is $x^2 + y^2 = 2$, the equation of the line $AC$ is $y = -x$ and the equation of the line $BC$ is $y = x$. Let $M$ have coordinates $(m, n)$, where $m^2 + n^2 = 2$. The lines $AM$ and $BM$ have the following equations:
$$AM : y = \frac{n-1}{m+1}(x+1) + 1$$
$$BM : y = \frac{n-1}{m-1}(x-1) + 1.$$
The coordinates of the four given intersection points are then
$$P\left(\frac{m+n}{2+m-n}, \frac{m+n}{2+m-n}\right), \quad Q\left(\frac{n-m}{m+n-2}, \frac{m-n}{m+n-2}\right), \\ R\left(\frac{1+2m+n}{1-n}, -1\right), \quad S\left(\frac{2m-n-1}{1-n}, -1\right).$$
We now calculate the product of the gradients of the two lines PS and QR:
$$\begin{align*} m_{PS}m_{QR} &= \frac{2(m+1)(1-n)}{(m+n)(1-n) - (2m-n-1)(2+m-n)} \\ &\quad \times \frac{2(1-n)(m-1)}{(1-n)(n-m) - (1+2m+n)(m+n-2)} \\ &= \frac{4(m+1)(m-1)(n-1)^2}{4(-n^2-m^2+mn-m+n+1)(-n^2-m^2-mn+m+n+1)} \\ &= \frac{(m+1)(m-1)(n-1)^2}{(m+1)(n-1)(m-1)(1-n)} \quad \text{(since } m^2 + n^2 = 2) \\ &= -1, \end{align*}$$
which shows that the two lines are perpendicular.