Task 4 - 250914 Three circles with the given radius $r$ may lie in a plane such that each touches the other two. A common tangent is drawn to each pair of these three circles, which does not have any point in common with the third circle. With these three tangents, an equilateral triangle is constructed.

Calculate the area of this triangle in dependence on $r$!

Hint: For numerical values that occur in the area specification, the use of approximate values is permitted (but not required); if so, however, with an indication - and justification - of how many decimal places the approximate value is accurate to.
Let the circles be $h, k, l$ with their centers $P, Q, R$. The common tangents to $k, l$ and to $l, h$ and to $h, k$ are $e$, $f$, and $g$ respectively. The intersection points of $f$ with $g$, $g$ with $e$, and $e$ with $f$ are $A$, $B$, and $C$ respectively. The triangle whose area is to be calculated is then $\triangle A B C$.

The perpendicular from $P$ and $Q$ to $g$ have their feet at $S$ and $T$, which are the points of tangency of $g$ with $h$ and $k$ respectively. The perpendicular from $P$ to $f$ has its foot at $U$, which is the point of tangency of $f$ with $h$ (see figure).

Therefore, $S T Q P$ is a rectangle with $P S = Q T = r$, $P Q = S T = 2r$, and the triangles $A P S$ and $A P U$ are congruent right triangles by the SSS (side-side-side) congruence theorem (where the right angles $\angle A S P$ and $\angle A U P$ are opposite the longest side), which implies $\angle P A S = \angle P A U$.

Since the sum of these two angles is the interior angle $\angle B A C = 60^\circ$ of the equilateral triangle $A B C$, it follows that $\angle P A S = \angle P A U = 30^\circ$ and thus by the angle sum theorem $\angle A P S = 60^\circ$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1064.jpg?height=577&width=646&top_left_y=243&top_left_x=705)

Therefore, for the point $P'$, which is the image of $P$ under reflection across $g$, the triangle $A P P'$ is equilateral. The altitude $A S$ to the side $P P'$ is also a median; hence, it follows that $A P = P P' = 2 P S = 2r$ and thus (by the Pythagorean theorem or the formula for the altitude in an equilateral triangle) $A S = r \sqrt{3}$. Similarly, we obtain $B T = r \sqrt{3}$.

Therefore, the side length of the triangle $A B C$ is

$$
a = r \sqrt{3} + 2r + r \sqrt{3} = 2r(1 + \sqrt{3})
$$

and thus, by the area formula for equilateral triangles, the area is

$$
F = \frac{1}{4} a^2 \sqrt{3} = \frac{1}{4} \cdot 4r^2 (1 + \sqrt{3})^2 = r^2 \cdot (4 \sqrt{3} + 6)
$$

Example using approximate values:

It holds that $\sqrt{3} \approx 1.732$ to three decimal places. Therefore, $1.7315 < \sqrt{3} < 1.7325$ and thus $12.296 < 4 \sqrt{3} + 6 < 12.93$. Hence, $F \approx 12.93 r^2$, and the numerical value 12.93 is accurate to two decimal places.

Solutions of the First Round 1985 adopted from [5]

### 6.27.2 Second Round 1985, Class 9