The floor is made of regular triangular stone slabs, and we throw a circular metal plate onto it. How large should the diameter of the metal plate be so that the probability of the metal plate landing entirely within one triangle equals the probability of the metal plate partially covering two triangle sides but not a vertex? (Take the side length of the triangle as a unit).
Since it is certain that the center of the metal disk will fall within or on the boundary of some triangle, it is sufficient to examine a triangle $ABC$, whose side we consider to be a unit length.

Let the sought diameter of the metal disk be $2x$. We can immediately deduce that $x$ must be smaller than the radius $\varrho$ of the inscribed circle, because otherwise the metal disk could not fall entirely within the triangle. Thus,

$$
x < \varrho = \frac{\sqrt{3}}{6} \approx 0.289
$$

The favorable area for the center of the disk, in the case where the disk is entirely within the triangle, is the regular triangle $A'B'C'$ inside the triangle $ABC$, whose sides are parallel to those of $ABC$ and are at a distance $x$ from them (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_7a7b5e16f83c430d6294g-1.jpg?height=379&width=404&top_left_y=583&top_left_x=836)

Figure 1

The side of $A'B'C'$ is $A'B' = A_1B_1 - 2A_1A' = 1 - 2x \tan 60^\circ = 1 - 2x\sqrt{3}$, and thus its area is $(1 - 2x\sqrt{3})^2 \frac{\sqrt{3}}{4}$.

In the case where the metal disk partially covers two sides but not a vertex, the favorable area for the center of the disk is the rhombus formed by the sides of $\triangle ABC$ and the extensions of the sides of $\triangle A'B'C'$, minus the $60^\circ$ sector of a circle with radius $x$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_7a7b5e16f83c430d6294g-1.jpg?height=412&width=572&top_left_y=1274&top_left_x=755)

Figure 2

The area of such a five-sided, mixed-line figure is thus $\frac{x}{\sin 60^\circ} \cdot x - \frac{x^2 \pi}{6} = \frac{2x^2 \sqrt{3}}{3} - \frac{x^2 \pi}{6}$. Since there are three such figures, the favorable area is $2x^2 \sqrt{3} - \frac{x^2 \pi}{2}$.

According to the problem, the two favorable areas are equal, so

$$
(1 - 2x\sqrt{3})^2 \frac{\sqrt{3}}{4} = 2x^2 \sqrt{3} - \frac{x^2 \pi}{2}
$$

This is a quadratic equation in $x$, which after rearrangement takes the form

$$
(4\sqrt{3} + 2\pi)x^2 - 12x + \sqrt{3} = 0
$$

from which

$$
x_{1,2} = \frac{12 \pm \sqrt{96 - 8\sqrt{3}\pi}}{8\sqrt{3} - 4\pi} \approx \frac{12 \pm 7.245}{26.43}
$$

and thus

$$
x_1 \approx \frac{4.755}{26.43} \approx 0.180, \quad [x_2 \approx \frac{19.245}{24.43} \approx 0.729]
$$

Since $x < \varrho \approx 0.289$, $x_2$ cannot be considered, so the sought diameter is $2x = 0.360$.

Remark: If the center of the metal disk falls within the area of $MNPQR$, then the metal disk partially covers three sides of the triangle (the third being a side of one of the adjacent triangles). The phrase "at least two sides of the triangle" would have been more precise.