Taiwan IMO Selection Camp

Let the side length of square $ABCD$ be $a$, let the projection length of $\overline{PA}$ onto $\overline{BC}$ be $t$, and let the projection length of $\overline{PB}$ onto $\overline{CD}$ be $v$. As shown in the figure

then $x, y, z$ can be expressed as follows:
$$x^2 = t^2 + (a - v)^2 \qquad (1)$$
$$y^2 = v^2 + (a - t)^2 \qquad (2)$$
$$z^2 = t^2 + v^2 \qquad (3)$$

From (1) and (3) we solve for
$$v = \frac{a^2 + z^2 - x^2}{2a}$$
From (2) and (3) we solve for
$$t = \frac{a^2 + z^2 - y^2}{2a}$$
Substituting back into (3) gives
$$z^2 = \left(\frac{a^2 + z^2 - x^2}{2a}\right)^2 + \left(\frac{a^2 + z^2 - y^2}{2a}\right)^2$$
After simplification we obtain the equation in $a$
$$2a^4 - 2a^2(x^2 + y^2) + (z^2 - x^2)^2 + (z^2 - y^2)^2 = 0$$
Regarding this as a quadratic equation in $a^2$, we solve to get
$$a^2 = \frac{x^2 + y^2 + \sqrt{(x^2 + y^2)^2 - 2(z^2 - x^2)^2 - 2(z^2 - y^2)^2}}{2}$$
Since $a^2$ has a solution, the discriminant is non-negative. If this equation has a repeated root, that means
$$a^2 = \frac{x^2 + y^2}{2}$$
But from (1)+(2) we obtain
$$x^2 + y^2 = 2a^2 + 2t^2 + 2v^2 - 2av - 2at = 2a^2 + 2(t(a - v) + v(v - a)) < 2a^2$$
which is a contradiction. Therefore the discriminant of this quadratic equation is greater than zero, that is
$$(x^2 + y^2)^2 - 2(z^2 - x^2)^2 - 2(z^2 - y^2)^2 > 0$$
This is equivalent to
$$((x - y)^2 - 2z^2)((x + y)^2 - 2z^2) < 0$$
Hence $(x - y)^2 < 2z^2 < (x + y)^2$