79. Construct a curve of constant width $h$, composed of five, seven, or any odd number of arcs of circles of radius $h$. What is the length of each of these curves?

Curves of constant width $h$, composed of arcs of circles of radius $h$, will play a significant role in the subsequent exposition (see pp. 98-102, the text printed in small type).
79. Let $A B C D E$ be a regular pentagon, the larger of whose diagonals is equal to $h$. From each vertex of the pentagon, we draw an arc of a circle with radius $h$, connecting two opposite vertices (Fig. 253). The resulting convex curve will be a curve of constant width, because of any two parallel supporting lines, one passes through a vertex of the pentagon, and the other is tangent to the opposite arc; therefore, the distance between them is equal to $h$. The angle $CAD$ is $36^{\circ}$, so each of the five arcs that make up the curve has a length of $\frac{2 \pi h}{10}$, and the length of the entire curve is $\pi h$.

The same construction can be carried out for any regular polygon with an odd number of sides. (The polygon must have an odd number of sides because each vertex must have two opposite vertices that are connected by an arc of a circle with radius $h$.) If we start with a regular $(2 n-1)$-gon, the resulting curve will consist of $2 n-1$ equal arcs of circles with radius $h$; the central angle of each arc will be $\frac{2 \pi}{2}(2 n-1)$, and the total length of the curve will be $(2 n-1) \cdot \frac{2 \pi h}{2(2 n-1)}=\pi h$.

Note that in our construction, there is no need to require that the polygon with an odd number of sides be regular. It is only necessary that all diagonals of the polygon, connecting each vertex with one of the two opposite vertices (the number of such diagonals is equal to the number of vertices of the polygon), have the same length $h$, and that all other diagonals and all sides of the polygon be shorter than $h$. It is not difficult to verify that many polygons that are not regular possess these properties. Draw from each vertex of such a polygon an arc of a circle with radius $h$, connecting two opposite vertices (Fig. 254). The resulting convex curve is a curve of constant width. From each pair of parallel supporting lines, one passes through a vertex of the polygon, and the other is tangent to the arc of the circle connecting two opposite vertices, and therefore the distance between them is equal to $h$.

We will show that the length of the resulting curve is also equal to $\pi h$. Let $\varphi_{k}$ be the angle $A_{n+k-1} A_{k} A_{n+k}\left(A_{n+k-1}\right.$ and $A_{n+k}$ - vertices opposite to vertex $A_{k}$; Fig. 254).

Then the arc connecting points $A_{n+k-1}$ and $A_{n+k}$ subtends the central angle $\varphi_{k}$. Thus, the curve in question consists of $2 n-1$ arcs of circles with radius $h$, which together subtend an angle $\varphi_{1}+\varphi_{2}+\ldots+\varphi_{2 n-1}$. If we prove that

$$
\varphi_{1}+\varphi_{2}+\ldots+\varphi_{2 n-1}=180^{\circ}
$$

then it will follow that the length of our curve is equal to the length of a semicircle with radius $h$, i.e., equal to $\pi h$.

Consider triangles whose vertices are an arbitrary vertex $A_{k}$ and the opposite vertices $A_{n+k-1}$ and $A_{n+k}$ (such a triangle, corresponding to $k=1$, is shaded in Fig. 254). The number of such triangles is $2 n-1$, so the sum of the angles of all these triangles is $(2 n-1) \cdot 180^{\circ}$. On the other hand, the sum of the angles of all these triangles is the sum of all the angles of the polygon $A_{1} A_{2} \ldots A_{2 n-1}$, and the angles $\varphi_{1}, \varphi_{2}, \ldots, \varphi_{2 n-1}$ appear in this sum three times (for example, the angle $\varphi_{1}$ is included in the sum of the angles of triangles $A_{1} A_{n} A_{n+1}, A_{n+1} A_{1} A_{2}$, $A_{n} A_{1} A_{2 n-1}$). Thus, the sum of the angles of all the specified triangles is the sum of the angles of the $(2 n-1)$-gon plus twice the sum $\varphi_{1}+\varphi_{2}+\ldots+\varphi_{2 n-1}$.

Thus,

$$
\begin{aligned}
& (2 n-1) \cdot 180^{\circ}= \\
& \quad=[(2 n-1)-2] \cdot 180^{\circ}+2\left(\varphi_{1}+\varphi_{2}+\ldots+\varphi_{2 n-1}\right)
\end{aligned}
$$

from which we obtain:

$$
\varphi_{1}+\varphi_{2}+\ldots+\varphi_{2 n-1}=180^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_03412fca680f64158016g-280.jpg?height=515&width=526&top_left_y=499&top_left_x=524)

Fig. 253.

![](https://cdn.mathpix.com/cropped/2024_05_21_03412fca680f64158016g-281.jpg?height=560&width=596&top_left_y=545&top_left_x=122)

Fig. 254.