## PROBLEM 12. OPAQUE SQUARE

A set of straight line segments inside or on the boundary of a square with side length 1 is called "opaque" if any straight line that intersects the square also intersects at least one of these segments. For example, the two diagonals form an opaque set
![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-22.jpg?height=430&width=918&top_left_y=1472&top_left_x=567)

Fig. 14

(Fig. 14, a). Another opaque set is shown in Fig. 14, b. The total length of the diagonals is $2 \sqrt{2} \approx 2.82$. It is fairly easy to show using elementary calculations that the minimum sum of the lengths of the segments of the opaque symmetric set in Fig. 14, b is $1+\gamma \overline{3} \approx 2.73$. Find an opaque set with the sum of the lengths of the segments less than $1+\sqrt{3}$.
Solution. An opaque set consisting of two adjacent sides of a square and a remote half-diagonal between them (Fig. 15) has a length of

$$
2+\frac{\sqrt{2}}{2}=2+\frac{1.42}{2}=2.71<1+\sqrt{3}
$$

Sides $AB$ and $BC$ have a common point with any line intersecting triangle $ABC$. If the entire triangle $ABC$ is considered, then the half-diagonal $OD$ is sufficient to make the second half of the square opaque.

Now let's consider a more efficient way to make $\triangle ABC$ opaque. The point $P$ inside triangle $ABC$, from which each side is visible at an angle of $120^{\circ}$, is called the Fermat point of this triangle and is the point for which

![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-23.jpg?height=531&width=557&top_left_y=591&top_left_x=378)

Fig. 15

![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-23.jpg?height=522&width=551&top_left_y=590&top_left_x=1141)

Fig. 16

the sum of its distances to the vertices of the triangle is minimal (Fig. 16):

$X A+X B+X C$

is minimal for $X \equiv P$.

The minimum sum is the length of the segment $B B^{\prime}$, where $B^{\prime}$ is the vertex of the equilateral triangle constructed on the exterior side $AC$ (Fig. 17). Clearly, in the considered case, $B B^{\prime}$ is the perpendicular bisector of $AC$ and $\triangle ABE$ is isosceles. Thus,

$$
B E=A E=\frac{1}{2} A C=\frac{\sqrt{2}}{2}
$$

and

$$
B B^{\prime}=B E+E B^{\prime}=\frac{\sqrt{2}}{2}+3\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}(1+\sqrt{3})
$$

Including the half-diagonal $OD=\sqrt{2} / 2$, we get an opaque set with a total length of

$$
\frac{\sqrt{2}}{2}(2+\sqrt{3}) \approx 2.64
$$

This elegant solution was proposed by Maurice Pouret, a school teacher from the city of Orléans (in the province of Ontario, Canada).