## Task A-3.7.

Let $A B C D$ be a parallelogram such that $|A B|=4,|A D|=3$, and the measure of the angle at vertex $A$ is $60^{\circ}$. Circle $k_{1}$ touches sides $\overline{A B}$ and $\overline{A D}$, while circle $k_{2}$ touches sides $\overline{C B}$ and $\overline{C D}$.

Circles $k_{1}$ and $k_{2}$ are congruent and touch each other externally. Determine the length of the radius of these circles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_f93036d069fffd92d333g-26.jpg?height=569&width=1111&top_left_y=1680&top_left_x=401)

Let $O_{1}$ and $O_{2}$ be the centers of circles $k_{1}$ and $k_{2}$, respectively. Let $M$ be the point of tangency of circle $k_{1}$ with side $\overline{A B}$, $N$ be the point of tangency of circle $k_{2}$ with side $\overline{C D}$, and $T$ be the point of tangency of the two circles. Let $r$ be the radius of these circles.

Notice that $T$ is the midpoint of segment $\overline{O_{1} O_{2}}$.

Since segments $\overline{O_{1} M}$ and $\overline{O_{2} N}$ are perpendicular to the parallel lines $A B$ and $C D$ (because the circles are tangent to these lines at points $M$ and $N$), we conclude that segments $\overline{O_{1} M}$ and $\overline{O_{2} N}$ are parallel. Therefore, $\varangle M O_{1} T = \varangle T O_{2} N$.

Since $k_{1}$ is tangent to sides $\overline{A B}$ and $\overline{A D}$, the center of circle $O_{1}$ lies on the angle bisector of the angle at vertex $A$. Similarly, point $O_{2}$ lies on the angle bisector of the angle at vertex $C$, so the angles $\varangle O_{1} A M = \varangle O_{2} C N = 30^{\circ}$.

Since $|O_{1} M| = |O_{2} N| = r$, we conclude that triangles $A M O_{1}$ and $C N O_{2}$ are congruent right triangles. In particular, $|A O_{1}| = |C O_{2}|$ and $\varangle A O_{1} M = \varangle C O_{2} N$.

Consider triangles $A O_{1} T$ and $C O_{2} T$. We know that $|A O_{1}| = |C O_{2}|$, and segments $\overline{O_{1} T}$ and $\overline{O_{2} T}$ are radii of circles $k_{1}$ and $k_{2}$, so they are of equal length. Finally, we have

$$
\varangle A O_{1} T = \varangle A O_{1} M + \varangle M O_{1} T = \varangle C O_{2} N + \varangle N O_{2} T = \varangle C O_{2} T
$$

so these triangles are congruent by the S-A-S (Side-Angle-Side) congruence theorem.

1 point

In particular, $|A T| = |T C|$ and $\varangle A T O_{1} = \varangle C T O_{2}$. Since points $O_{1}, T$, and $O_{2}$ are collinear, from the equality of angles we conclude that points $A, T$, and $C$ are collinear. Therefore, $T$ is the midpoint of diagonal $\overline{A C}$ of parallelogram $A B C D$, from which we conclude that $T$ is the intersection of the diagonals of the parallelogram.

Since $T$ is the midpoint of diagonal $\overline{B D}$ and segment $\overline{O_{1} O_{2}}$, we conclude that quadrilateral $O_{1} B O_{2} D$ is a parallelogram. In this parallelogram, the parallelogram law holds:

$$
|O_{1} O_{2}|^{2} + |B D|^{2} = 2 (|O_{1} B|^{2} + |O_{1} D|^{2})
$$

The length of segment $\overline{B D}$ can be determined using the cosine rule in triangle $A B D$:

$$
|B D|^{2} = |A B|^{2} + |A D|^{2} - 2 \cdot |A B| \cdot |A D| \cos \varangle B A D = 3^{2} + 4^{2} - 2 \cdot 3 \cdot 4 \cdot \frac{1}{2} = 13.
$$

The length of segment $\overline{O_{1} B}$ is determined using the Pythagorean theorem in triangle $O_{1} B M$. Since

$$
|M B| = |A B| - |A M| = |A B| - |O_{1} M| \tan \varangle O_{1} A M = 4 - r \sqrt{3},
$$

we have

$$
|O_{1} B|^{2} = |O_{1} M|^{2} + |M B|^{2} = r^{2} + (4 - r \sqrt{3})^{2}.
$$

Similarly, we conclude that $|O_{1} D|^{2} = r^{2} + (3 - r \sqrt{3})^{2}$. Substituting these into the parallelogram law, we get an equation for $r$:

$$
\begin{aligned}
13 + 4 r^{2} & = 2 (r^{2} + (4 - r \sqrt{3})^{2} + r^{2} + (3 - r \sqrt{3})^{2}) \\
13 + 4 r^{2} & = 4 r^{2} + 50 - 28 \sqrt{3} + 12 r^{2} \\
0 & = 12 r^{2} - 28 \sqrt{3} r + 37
\end{aligned}
$$

The solutions to the quadratic equation are

$$
r_{1,2} = \frac{28 \sqrt{3} \pm 24}{24} = \frac{7 \sqrt{3}}{6} \pm 1
$$

We discard the solution $\frac{7 \sqrt{3}}{6} + 1$ because in this case the radius of the circle would be greater than 2, meaning the diameter of the circle would be greater than the length of both sides of the parallelogram, so in this case $k_{1}$ and $k_{2}$ cannot be inside the parallelogram.

Therefore, the final conclusion is that the desired radius is

$$
r = \frac{7 \sqrt{3}}{6} - 1
$$

Note: The claim that $T$ is the intersection of the diagonals without proof is worth 1 point, which corresponds to the fifth point in the above scoring scheme. The proof can also be done by introducing a central symmetry with respect to the same point $T$, but in that case, the proof of this claim must be provided for a full score.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - A variant
January 5, 2023.

## IF THE STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO SCORE AND EVALUATE THAT APPROACH APPROPRIATELY.