Let $\Theta_1$ and $\Theta_2$ be circumferences with centers $O_1$ and $O_2$, exteriorly tangents. Let $A$ and $B$ be points in $\Theta_1$ and $\Theta_2$, respectively, such that $AB$ is common external tangent to $\Theta_1$ and $\Theta_2$. Let $C$ and $D$ be points on the semiplane determined by $AB$ that does not contain $O_1$ and $O_2$ such that $ABCD$ is a square. If $O$ is the center of this square, compute the possible values for the angle $\angle O_1OO_2$.
Problem
Source: Rio de Janeiro Mathematical Olympiad 2018, Level 4, #5
Tags: geometry
Letteer
17.11.2019 23:30
Let $r_1$ and $r_2$ be the radii of the two circles and $a$ be the side length of the square. We have,
$$a^2 = (r_1+r_2)^2-(r_1-r_2)^2 = 4r_1r_2$$
Apply the cosine rule to the triangles $OAO_1$ and $OBO_2$,
$$OO_1^2 = r_1^2 + \frac12a^2-2r_1\frac{a}{\sqrt2}\cos135=r_1^2+2r_1r_2+2r_1\sqrt{r_1r_2}$$$$OO_2^2 = r_2^2 + \frac12a^2-2r_2\frac{a}{\sqrt2}\cos135=r_2^2+2r_1r_2+2r_2\sqrt{r_1r_2}$$
Now, apply the cosine rule to the triangle $OO_1O_2$ to express the angle $\theta = \angle O_1OO_2$,
$$\cos\theta = \frac {OO_1^2+OO_2^2-(r_1+r_2)^2}{2\>OO_1\cdot OO_2}$$$$=\frac{1+\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_2}{r_1}}}{\sqrt{\left(1+2\frac{r_1}{r_2}+2\sqrt{\frac{r_1}{r_2}}\right)\left(1+2\frac{r_1}{r_2}+2\sqrt{\frac{r_1}{r_2}}\right)}}$$$$=\frac{1+\sqrt r+\frac{1}{\sqrt r}}{\sqrt{ \left( 1+2r+2\sqrt{r} \right) \left( 1+\frac 2r +\frac{2}{\sqrt r} \right) }}$$
where $r=\frac{r_1}{r_2}$, the ratio of the two circles. Now, examine the two limits.
1) If $r=1$, $\cos\theta = \frac35$;
2) If $r=\infty$, $\cos\theta = \frac{1}{\sqrt{2}}$.
Therefore, the possible values of the angle $\theta = \angle O_1OO_2$ are in the range
$$45^\circ < \theta \le \cos^{-1}\frac35$$