Let $a, b, c, d$ be real numbers defined by

$$
a=\sqrt{4-\sqrt{5-a}}, b=\sqrt{4+\sqrt{5-b}}, c=\sqrt{4-\sqrt{5+c}}, d=\sqrt{4+\sqrt{5+d}}
$$

Calculate their product.

## 2 Solutions
It would be wonderful if the four reals $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ were roots of the same polynomial, whose constant term would be their product. Fortunately, this is almost true. Indeed, by squaring each equality twice (taking care to isolate the root each time), we find that $a, b, -c, -d$ are roots of the polynomial $P(X) = X^4 - 8X^2 - X + 11$. And it is clear that $a, b > 0$ and $-c, -d < 0$ (the reals $a, b, c, d$ are non-zero, since $P(0) \neq 0$). Furthermore, if $a = b$, $\sqrt{4 - \sqrt{5 - a}} = \sqrt{4 + \sqrt{5 - a}}$ so $5 - a = 0$, hence $5 = \sqrt{4 - 0} = 2$, which is absurd. Similarly, we verify that $c \neq d$.

Thus $P$ has four distinct roots $a, b, -c, -d$, so $abcd = 11$ by Vieta's formulas.

## 3 Group C

## 1 Statements