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 $$x_1=c\cos \alpha ,x_2=c\sin \alpha ,y_1=c\cos \beta ,\text{ and }{y}_2=c\sin \beta .$$ Therefore $$\begin{array}{rl}S& =(1-x_1)(1-y_1)+(1-x_2)(1-y_2)=2-(x_1+x_2+y_1+y_2)+x_1{y}_1+x_2{y}_2\\ & =2-c(\cos \alpha +\sin \alpha +\cos \beta +\sin \beta )+c^2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )\\ & =2-\sqrt{2}c\left(\sin \left(\alpha +\frac{\pi }{4}\right)+\sin \left(\beta +\frac{\pi }{4}\right)\right)+c^2\cos (\alpha -\beta )\\ & =(2-c^2)-2\sqrt{2}c\sin \left(\frac{\alpha +\beta }{2}+\frac{\pi }{4}\right)\cos \left(\frac{\alpha -\beta }{2}\right)+2c^2{\cos }^2\left(\frac{\alpha -\beta }{2}\right)\\ & =(2-c^2)-2\sqrt{2}cst+2c^2{t}^2\end{array}$$ where $$s=\sin \left(\frac{\alpha +\beta }{2}+\frac{\pi }{4}\right),\text{ and }t=\cos \left(\frac{\alpha -\beta }{2}\right)$$ 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 $$S=(2-c^2)-s^2+(\sqrt{2}ct-s{)}^2$$ Therefore, the minimum of $S$ is equal to $1-c^2=-2012$, and is reached when $s=\pm 1$ and $t=\frac{\sqrt{2}s}{2c}=\frac{\pm \sqrt{4026}}{4026}$. That is precisely when $$x_1=y_2=\frac{1+\sqrt{4025}}{2}\text{ and }{x}_2=y_1=\frac{1-\sqrt{4025}}{2},$$ or vice-versa.

Second solution. For the maximum, we have by Cauchy-Schwartz inequality $$\begin{array}{rl}S& =2-(x_1+x_2+y_1+y_2)+x_1{y}_1+x_2{y}_2\\ & \le 2+\sqrt{4\cdot (x_1^2+x_2^2+y_1^2+y_2^2)}+\sqrt{(x_1^2+x_2^2)(y_1^2+y_2^2)}\\ & \le 2015+2\sqrt{4026}\end{array}$$ and the equality holds when $x_1=x_2=y_1=y_2=-\frac{\sqrt{4026}}{2}$.

For the minimum, we have $$\begin{array}{rl}S& =2-(x_1+y_1)-(x_2+y_2)+\frac{(x_1+y_1{)}^2}{2}+\frac{(x_2+y_2{)}^2}{2}-c^2\\ & =(1-c^2)+\frac{(x_1+y_1-1{)}^2}{2}+\frac{(x_2+y_2-1{)}^2}{2}\\ & \ge 1-c^2=-2012\end{array}$$ 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 $$x_1=y_2=\frac{1+\sqrt{4025}}{2}\text{ and }{x}_2=y_1=\frac{1-\sqrt{4025}}{2}$$