32. Given an equilateral triangle with side 1. What is the smallest $d$ such that a segment of length $d$ can, by sliding its ends along the sides of the triangle, sweep over the entire triangle?
32. Answer: $d=2 / 3$.

If a segment sweeps the entire triangle, then in some position it passes through its center. We will prove that among all segments with endpoints on the sides and passing through the center $O$ of an equilateral triangle, the shortest one is the segment $A B$ parallel to its side.

Let $A^{\prime} B^{\prime}$ be any segment with endpoints on the same sides and passing through the center of the triangle, and let $O B^{\prime}B B^{\prime}$. It is clear that even the projection of this segment onto the line $A B$ has a length greater than $2 / 3$. Therefore, a segment that sweeps the entire triangle cannot be shorter than $2 / 3$.

On the other hand, a segment of length $2 / 3$ can sweep the entire triangle. To prove this, we need to show that through any point $P$ inside the triangle, a segment of length $2 / 3$ with endpoints on the sides of the triangle can be drawn. Draw through $P$ two lines parallel to the closest and the farthest sides from $P$. They intersect the triangle along segments, the length of the first of which is greater than $2 / 3$, and the second is less than $2 / 3$. By rotating the line passing through $P$ from the first position to the second, at some intermediate position, the length of the segment by which it intersects the triangle will be exactly $2 / 3$.

$\nabla$ Note that, despite the obviousness of the last statement, its rigorous proof requires considerations of continuity (P5).

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-117.jpg?height=344&width=688&top_left_y=704&top_left_x=274)

Fig. 33