Two circles, the radii of which are in the ratio $9-4 \sqrt{3}$, touch each other internally. Two chords of the larger circle, equal in length and tangent to the smaller circle, are drawn. One of these chords is perpendicular to the segment connecting the centers of the circles, while the other is not. Find the angle between these chords.
Let circles with radii $R>r$ and centers $O_1$ and $O_2$ respectively touch internally at point $K$. Chord $AB$ of the larger circle is perpendicular to $O_1 O_2$ and touches the smaller circle at point $P$, and an equal chord $CD$ of the larger circle touches the smaller circle at point $Q$ and intersects chord $AB$ at point $F$. Drop a perpendicular $O_1 M$ to $CD$ and consider the right trapezoid $O_1 O_2 Q M$.

Since equal chords of a circle are equidistant from its center, we have

$$
O_1 M = O_1 P = O_1 K - P K = R - 2r
$$

Drop a perpendicular $O_1 H$ to $O_2 Q$. Then

$$
O_2 H = O_2 Q - H Q = O_2 Q - O_1 M = r - (R - 2r) = 3r - R
$$

Thus,

$$
\begin{aligned}
& \cos \angle O_1 O_2 H = \frac{O_2 H}{O_1 O_2} = \frac{3r - R}{R - r} = \frac{3 - \frac{R}{r}}{\frac{R}{r} - 1} = \\
& = \frac{4 \sqrt{3} - 6}{8 - 4 \sqrt{3}} = \frac{2 \sqrt{3} - 3}{4 - 2 \sqrt{3}} = \frac{1}{2} \cdot \frac{\sqrt{3}(2 - \sqrt{3})}{2 - \sqrt{3}} = \frac{\sqrt{3}}{2}
\end{aligned}
$$

Therefore,

$$
\angle Q F B = \angle O_1 O_2 H = 30^\circ
$$

Note that

$$
\sqrt{3} < 2 \Leftrightarrow 3 < 2 \sqrt{3} \Leftrightarrow 6 < 4 \sqrt{3} \Leftrightarrow 9 - 4 \sqrt{3} < 3
$$

Thus, $\frac{R}{r} = 9 - 4 \sqrt{3} < 3$. This means that point $H$ indeed lies on the segment $O_2 Q$.

## Answer

$30^\circ$

Two circles intersect at points $A$ and $B$. A line through point $A$ intersects the first circle again at point $C$ and the second circle again at point $D$. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ not containing point $A$, and let $K$ be the midpoint of segment $CD$. Prove that $\angle M K N = 90^\circ$. (We can assume that points $C$ and $D$ lie on opposite sides of point $A$.)

## Hint

Apply central symmetry.

## Solution

Under symmetry about point $K$, point $C$ maps to point $D$, and point $M$ maps to some point $M_1$, such that segments $DM_1$ and $CM$ are equal and parallel. Therefore, $BM = CM = DM_1$, $BN = DN$. Let $\angle CDM_1 = \alpha$, $\angle CDN = \beta$. Then $\angle ACM = \alpha$, $\angle ABM = 180^\circ - \alpha$, $\angle ABN = 180^\circ - \beta$,

$\angle MBN = 360^\circ - (180^\circ - \alpha) - (180^\circ - \beta) = \alpha + \beta = \angle NDM_1$.

Thus, triangles $MBN$ and $M_1DN$ are congruent by two sides and the included angle, so $NM = NM_1$. In the isosceles triangle $MNM_1$, the median $NK$ is also the altitude, hence $\angle MKN = 90^\circ$.