Reuleaux Triangles and Constant Width, math that surprises you lesson

Lay a board on round logs and the board stays at the same height however the logs turn. It seems that only a circle could do this, but a triangle with curved sides, the Reuleaux triangle, does it too. Can a shape that is not round really keep two parallel surfaces the same distance apart in every position? This page builds the Reuleaux triangle from three arcs, says exactly what width means, shows why its width never changes, and explains why it still does not make an ordinary wheel on a fixed axle.

Key terms

Reuleaux triangle

Definition Start with an equilateral triangle of side $w$ and draw a circle of radius $w$ around each of its corners. Each circle passes through the other two corners, and the region that lies inside all three circles is the Reuleaux triangle of width $w$ . Its boundary consists of three circular arcs of radius $w$ . Each arc joins two corners and is centered at the third corner, and it spans an angle of $60^\circ$ at that center. For every direction, the two parallel lines that touch the shape on opposite sides are exactly $w$ apart, so the Reuleaux triangle has constant width $w$ . It can turn between a floor and a board at height $w$ while touching both, so rollers with this cross-section keep a board level at height $w$ . It has no center that is equally far from all of its boundary. The center of the original triangle is $w/\sqrt3 \approx 0.577w$ from each corner but only $(1 - 1/\sqrt3)w \approx 0.423w$ from the middle of each arc. So while the board stays level, the marked center of a roller rises and falls between those two heights above the floor, and an axle fixed there would bob by $(2/\sqrt3 - 1)w \approx 0.155w$ . The arcs must be exactly these circular arcs. A shape that only looks similar, such as an ordinary triangle with rounded corners, usually has a width that changes with direction. The shape is named after the engineer Franz Reuleaux.

In plain words It is a triangle whose sides bulge outward as arcs of circles, drawn so that each corner is exactly one width away from every point of the opposite side. However it is turned, it measures the same between the parallel jaws of a pair of calipers.

Example Take $w = 1$ and put the center of the triangle at the origin, with corners

\begin{aligned} &\left(0, \tfrac{1}{\sqrt3}\right), \\[0.4em] &\left(\pm\tfrac12, -\tfrac{1}{2\sqrt3}\right). \end{aligned}

Stand it corner up. The horizontal line through the top corner, at height $1/\sqrt3$ , touches the shape there. The bottom arc is centered at that top corner with radius $1$ , so its lowest point is at height $1/\sqrt3 - 1$ , and the horizontal line through it touches the shape from below. The two lines are $1$ apart. On a floor, the center is then $1 - 1/\sqrt3 \approx 0.423$ above the floor. Turned corner down, the center is $1/\sqrt3 \approx 0.577$ above the floor, and in both positions the top of the shape is at height $1$ .

Supporting line

Definition A shape in the plane is convex if, for any two of its points, the whole segment between them belongs to the shape. Let $K$ be a convex region, a convex shape that is bounded, contains its boundary and has some area. A supporting line of $K$ is a line that has at least one point in common with $K$ while all of $K$ lies on one side of it, in one of the two closed half-planes the line bounds. For every direction there are exactly two supporting lines parallel to it, one on each side of $K$ . Where the boundary is smooth, the supporting line through a boundary point is the tangent line there. At a corner, many supporting lines pass through the same point.

In plain words Slide a ruler toward the shape from far away, keeping its direction fixed, and stop the moment it touches. The ruler now lies along a supporting line. It touches the shape but never cuts into it.

Example For the Reuleaux triangle of width $1$ standing corner up, the horizontal line through the top corner and the horizontal line touching the bottom arc at its middle are the two horizontal supporting lines. The two arcs meet at the top corner at an angle of $120^\circ$ , so a line through that corner remains a supporting line while it tilts up to $30^\circ$ either way from horizontal. A line through the center is never a supporting line, because the shape lies on both sides of it.

Width of a convex shape

Definition Let $K$ be a convex region and choose a direction. The width of $K$ in that direction is the distance between the two supporting lines of $K$ that are perpendicular to it. It equals the length of the shadow of $K$ on a line in that direction, when every point of $K$ is projected onto the line at right angles. Every segment inside $K$ parallel to the chosen direction is at most as long as the width in that direction, and it can be shorter. The largest width over all directions is the diameter of $K$ , the greatest distance between two of its points.

In plain words Width is what calipers measure when their two parallel jaws close on the shape from opposite sides. A ruler laid across the inside of the shape measures something else, and it may give a smaller number.

Example Here is why the Reuleaux triangle of width $1$ has width $1$ in every direction. Give each supporting line its outward direction, perpendicular to the line and pointing away from the shape. The outward directions of the supporting lines through a corner $C$ fill an angle of $60^\circ$ , and the outward directions of the tangent lines along the arc centered at $C$ fill the opposite $60^\circ$ . Three corners and three arcs account for all $360^\circ$ , so for every direction one supporting line passes through some corner $C$ while the parallel line on the other side touches the arc centered at $C$ , at a point $P$ . At the change from one corner to the next, $P$ is an end of the arc, which is also a corner. The radius $CP$ has length $1$ and is perpendicular to the tangent at $P$ , which is the second supporting line, so it is perpendicular to the first one as well. The distance between the two lines is therefore the length of $CP$ , which is $1$ . Checking only the symmetric positions, corner up or corner down, would not be enough, and this argument covers every direction at once.

Curve of constant width

Definition A convex region has constant width $w$ if its width is $w$ in every direction, and its boundary is then called a curve of constant width $w$ . Equivalently, every two parallel supporting lines of the region are $w$ apart. Every such curve has diameter $w$ , and by Barbier’s theorem, published by Joseph-Émile Barbier in 1860, its length is $\pi w$ , the same as for a circle of diameter $w$ . The circle is the simplest example. Others are the Reuleaux triangle and the Reuleaux polygons built in the same way from regular polygons with an odd number of sides. Among all curves of constant width $w$ , the Reuleaux triangle encloses the least area, by the Blaschke-Lebesgue theorem, and the circle encloses the most. Constant width does not mean that some center is the same distance from every point of the boundary. Only a circle has such a center.

In plain words These are the shapes that always fit exactly between calipers set to $w$ , however they are turned. A circle is one of them, but far from the only one.

Example A curve of constant width $w$ can turn inside a square of side $w$ while touching all four sides, because in every position its width across and its width up and down are both $w$ . Unless it is a circle, it also has to shift a little as it turns. For the Reuleaux triangle of width $1$ , the three arcs each span $60^\circ$ , a third of $\pi$ radians, at radius $1$ , so its boundary has length

3 \cdot \frac{\pi}{3} = \pi,

just as Barbier’s theorem says. Its area is one equilateral triangle plus three circular segments, each a $60^\circ$ sector minus a triangle,

\begin{aligned} A &= \frac{\sqrt3}{4} + 3\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) \\[0.4em] &= \frac{\pi - \sqrt3}{2} \approx 0.705, \end{aligned}

less than the area $\pi/4 \approx 0.785$ of the circle of diameter $1$ .

Check yourself

Change the size. Build the Reuleaux triangle from an equilateral triangle of side $2$ . Find its width, the length of its boundary and its area, compare them with a circle of diameter $2$ , and find how far its center rises and falls as it rolls between a floor and a board.

Show the answer

The arcs have radius $2$ , so the width is $2$ and the board rests at height $2$ . Each arc spans $60^\circ$ , a third of $\pi$ radians, so the boundary has length $3 \cdot 2 \cdot \pi/3 = 2\pi \approx 6.283$ , the same as the circle of diameter $2$ . Every length has doubled, so the area is $4$ times as large,

\begin{aligned} 4 \cdot \frac{\pi - \sqrt3}{2} &= 2\left(\pi - \sqrt3\right) \\[0.4em] &\approx 2.819, \end{aligned}

less than the circle’s area $\pi \approx 3.142$ . The center is $2/\sqrt3 \approx 1.155$ from each corner and $2 - 2/\sqrt3 \approx 0.845$ from the middle of each arc. It is therefore at height $0.845$ when a corner points up and $1.155$ when a corner points down, and it rises and falls by $4/\sqrt3 - 2 \approx 0.309$ while the board stays at height $2$ .

Spot the mistake. Someone stands the Reuleaux triangle of width $1$ corner up, lays a ruler horizontally through its center, and measures less than $1$ from one side to the other. Does that disprove constant width?

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No. The ruler measures a chord, not a width. With the corners as in the first example above, the right end of that chord lies on the arc centered at the lower left corner $\left(-\tfrac12, -\tfrac{1}{2\sqrt3}\right)$ , so its $x$ satisfies

\left(x + \tfrac12\right)^2 + \tfrac{1}{12} = 1.

That gives $x = \sqrt{11/12} - \tfrac12 \approx 0.457$ , and by symmetry the chord has length

\sqrt{11/3} - 1 \approx 0.915.

The width in the horizontal direction is measured between the two vertical supporting lines. They pass through the two lower corners, at $x = -\tfrac12$ and $x = \tfrac12$ , and are exactly $1$ apart. A chord can be shorter than the width in its direction, but never longer.

Extend the pattern. Start from a regular pentagon of side $1$ . From each corner, draw the arc centered there that joins the two corners farthest away. What is the width of the resulting Reuleaux pentagon, how long is its boundary, and why does the same recipe fail for a square?

Show the answer

The two corners farthest from a given corner are the ends of the opposite side, both a diagonal away. In a regular pentagon of side $1$ the diagonal is the golden ratio,

d = \frac{1 + \sqrt5}{2} \approx 1.618,

so every arc has radius $d$ . The pentagon’s corner angle of $108^\circ$ is split by the two diagonals into three angles of $36^\circ$ , so each arc spans $36^\circ$ , a fifth of $\pi$ radians. The same argument as for the triangle pairs every corner with its opposite arc, so the width is $d \approx 1.618$ in every direction, and the boundary has length $5 \cdot d \cdot \pi/5 = \pi d \approx 5.083$ , as Barbier’s theorem requires. In a square, each corner has only one farthest corner, the one across the diagonal, so there is no opposite side for an arc to span. The recipe needs an odd number of corners.