Selection tests for the International Mathematical Olympiad, 2013

We present for this problem two solutions, one using trigonometric functions and the other using classical inequalities.

First solution. The condition $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = c^{2}$, here $c^{2} = 2013$, is equivalent to saying that there exist $\alpha, \beta \in \mathbb{R}$ such that
$$x_{1} = c \cos \alpha, \quad x_{2} = c \sin \alpha, \quad y_{1} = c \cos \beta, \quad \text{ and } \quad y_{2} = c \sin \beta.$$
Therefore
$$\begin{aligned} S & = (1 - x_{1})(1 - y_{1}) + (1 - x_{2})(1 - y_{2}) = 2 - (x_{1} + x_{2} + y_{1} + y_{2}) + x_{1} y_{1} + x_{2} y_{2} \\ & = 2 - c(\cos \alpha + \sin \alpha + \cos \beta + \sin \beta) + c^{2}(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \\ & = 2 - \sqrt{2} c \left( \sin \left( \alpha + \frac{\pi}{4} \right ) + \sin \left( \beta + \frac{\pi}{4} \right ) \right ) + c^{2} \cos (\alpha - \beta) \\ & = (2 - c^{2}) - 2 \sqrt{2} c \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right ) \cos \left( \frac{\alpha - \beta}{2} \right ) + 2 c^{2} \cos^{2} \left( \frac{\alpha - \beta}{2} \right ) \\ & = (2 - c^{2}) - 2 \sqrt{2} c s t + 2 c^{2} t^{2} \end{aligned}$$
where
$$s = \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right ), \quad \text{ and } \quad t = \cos \left( \frac{\alpha - \beta}{2} \right )$$
are two independent variables, since $\alpha + \beta$ and $\alpha - \beta$ are independent, taking all the real values between $-1$ and $1$ included.

Hence, the maximum of $S$ is equal to $2 + c^{2} + 2 \sqrt{2} c = 2015 + 2 \sqrt{4026}$, and is reached when $s = -t = \pm 1$. That is precisely when $x_{1} = x_{2} = y_{1} = y_{2} = -\frac{\sqrt{4026}}{2}$.

For the minimum, notice that
$$S = (2 - c^{2}) - s^{2} + (\sqrt{2} c t - s)^{2}$$
Therefore, the minimum of $S$ is equal to $1 - c^{2} = -2012$, and is reached when $s = \pm 1$ and $t = \frac{\sqrt{2} s}{2 c} = \frac{\pm \sqrt{4026}}{4026}$. That is precisely when
$$x_{1} = y_{2} = \frac{1 + \sqrt{4025}}{2} \quad \text{ and } \quad x_{2} = y_{1} = \frac{1 - \sqrt{4025}}{2},$$
or vice-versa.

Second solution. For the maximum, we have by Cauchy-Schwartz inequality
$$\begin{aligned} S & = 2 - (x_{1} + x_{2} + y_{1} + y_{2}) + x_{1} y_{1} + x_{2} y_{2} \\ & \leq 2 + \sqrt{4 \cdot (x_{1}^{2} + x_{2}^{2} + y_{1}^{2} + y_{2}^{2})} + \sqrt{(x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2})} \\ & \leq 2015 + 2 \sqrt{4026} \end{aligned}$$
and the equality holds when $x_{1} = x_{2} = y_{1} = y_{2} = -\frac{\sqrt{4026}}{2}$.

For the minimum, we have
$$\begin{aligned} S & = 2 - (x_{1} + y_{1}) - (x_{2} + y_{2}) + \frac{(x_{1} + y_{1})^{2}}{2} + \frac{(x_{2} + y_{2})^{2}}{2} - c^{2} \\ & = (1 - c^{2}) + \frac{(x_{1} + y_{1} - 1)^{2}}{2} + \frac{(x_{2} + y_{2} - 1)^{2}}{2} \\ & \geq 1 - c^{2} = -2012 \end{aligned}$$
where $c^{2} = 2013$.

Thus the minimum of $S$ is $-2012$ and is reached when $y_{1} = 1 - x_{1}, y_{2} = 1 - x_{2}$ and $x_{1}^{2} + x_{2}^{2} = y_{1}^{2} + y_{2}^{2} = 2013$. This is equivalent to
$$x_{1} = y_{2} = \frac{1 + \sqrt{4025}}{2} \quad \text{ and } \quad x_{2} = y_{1} = \frac{1 - \sqrt{4025}}{2}$$