25. Two circles of radius 1 are such that the centre of each circle lies on the other circle. A square is inscribed in the space between the circles.
What is the area of the square?
A $2-\frac{\sqrt{7}}{2}$
B $2+\frac{\sqrt{7}}{2}$
C $4-\sqrt{5}$
D 1
$$
\mathrm{E} \frac{\sqrt{5}}{5}
$$
SOlUtion
A

We let $O$ and $P$ be the centres of the circles, $Q, R, S$ and $T$ be the vertices of the square, and $U$ and $V$ be the points where the line $O P$ meets the edges of the square, as shown in the diagram.

We let $x$ be the side length of the square.
The point $V$ is the midpoint of $S T$ and therefore $V T=\frac{1}{2} x$
Because $O T$ and $O P$ are radii of the circle with centre $O$, we have $O T=O P=1$.
Because $O P=1$ and $U V=Q T=x$, we have $O U+V P=1-x$. Since $O U=V P$, it follows that $O U=\frac{1}{2}(1-x)$. Therefore $O V=O U+U V=\frac{1}{2}(1-x)+x=\frac{1}{2}(1+x)$.
Therefore, applying Pythagoras' Theorem to the right-angled triangle $O V T$, we obtain $O V^{2}+$ $V T^{2}=O T^{2}$. That is, $\left(\frac{1}{2}(1+x)\right)^{2}+\left(\frac{1}{2} x\right)^{2}=1^{2}$. This equation expands to give $\frac{1}{4}+\frac{1}{2} x+\frac{1}{4} x^{2}+\frac{1}{4} x^{2}=1$. The last equation may be rearranged to give $2 x^{2}+2 x-3=0$.
Hence, by the formula for the roots of a quadratic equation, $x=\frac{-2 \pm \sqrt{4+24}}{4}=\frac{1}{2}(-1 \pm \sqrt{7})$. Because $x>0$, we deduce that $x=\frac{1}{2}(-1+\sqrt{7})$.
Therefore the area of the square is given by
$$
x^{2}=\left(\frac{1}{2}(-1+\sqrt{7})\right)^{2}=\frac{1}{4}(1-2 \sqrt{7}+7)=\frac{1}{4}(8-2 \sqrt{7})=2-\frac{\sqrt{7}}{2} .
$$