Circular motion can be converted into a four-sided square using the Reuleaux triangle. The process relies on the property of Reuleaux triangle’s diameter being consistent across all points, and when rolled and rotated simultaneously, tracing a square area with slightly rounded corners. By slightly altering the shape of the triangle to mimic a drill bit (for cutting and extracting material), a drill with a specialized rotating chuck can create perfect, four-sided holes. These drills were pioneered by Harry Watts, and can still be found today, albeit not at your typical home improvement store.

The following is a bit of literature that helps explain and illustrate the concepts behind the Reuleaux triangle. Please check out more of Prof. Smith’s work here.

To construct a Reuleaux triangle, start with an equilateral triangle of side s (Figure 1). With a radius equal to s and the center at one of the vertices, draw an arc connecting the other two vertices. Similarly, draw arcs connecting the endpoints of the other two sides. The three arcs form the Reuleaux triangle. One of its properties is that of constant width, meaning the figure could be rotated completely around between two parallel lines separated by distance s.

It was with this property of constant width that the Reuleaux triangle was introduced in a sidebar of our geometry text (Moise and Downs, Teachers’ Edition, p. 555). “This figure has constant width,” I lectured, “just like a circle.” Without thinking, I volunteered, “Imagine it as wheels on a cart.” “What sort of cart?” “Why, a math cart, to carry my board compass and protractor,” I replied, digging myself in deeper. This was the first of several impulsive misstatements I made about the Reuleaux triangle, only to admit after a little reflection that it wasn’t so. Not in twenty years of teaching had my intuition failed me so completely.

The constant width property can be used to transport loads, but not by using Reuleaux triangles as wheels. If several poles had congruent Reuleaux triangles as cross sections, bulky items could ride atop them (Figure 2). Movement would occur as poles were transferred from back to front, providing a moveable base of constant height.

But the Reuleaux triangle cannot be a wheel. The only conceivable point for the axle, at the triangle’s centroid, is not the same distance from the Reuleaux triangle’s “sides” (Figure 3). If the sides of the equilateral triangle are s, then

2 s sqrt(3) (1) AP = - - sqrt(3) = ------- s » 0.577s, 3 2 3

while

sqrt(3) sqrt(3) PB = s - ------- s = s(1 - -------) » 0.423s. 3 3

Even if four Reuleaux triangle wheels were synchronized, the load would rise and fall continuously — you’d need Dramamine to ride this cart!

“And since it has constant width, it would just fit inside a square whose sides are that width,” I continued, trying to regain their attention. I carefully drew a square circumscribing the Reuleaux triangle (Figure 4). The triangle is normally tangent to two sides of the square with two vertices touching the square directly opposite the points of tangency (why?), as in Figure 4a. The exception is Figure 4b, where the Reuleaux triangle has one point of tangency and all three vertices on the square (one directly opposite that point of tangency).

“If the Reuleaux triangle just fits inside the square, no matter what position it’s in, couldn’t it rotate around the inside of the square?” They needed convincing — a model would have to be built. “But if it did rotate around the inside, doesn’t that mean that a sharp Reuleaux triangle could carve out a square as it rotated?” I had them. “Drill a square hole?”, one countered. “No way!”

That night I cut a four inch Reuleaux triangle from a manila folder to take to class the next day. With a lot of effort, I was able to show the triangle rotate around the inside of a four inch square. “And if this was metal at the end of a rotating shaft, it would cut out a square”, I continued, racking up two more falsehoods. Firstly, it was implied that the center of the Reuleaux triangle would coincide with the center of a drill’s shaft; it cannot. And secondly, the corners of the holes are not right angles, but slightly rounded.

Trying to show the triangle should be centered at the end of a rotating shaft, I stuck a pen through the triangle’s center which, while a student manually rotated the triangle within the square, traced the center’s path on paper beneath (Figure 5). “It’s definitely not a single point,” I had to admit, holding up the traced curve, “but it sure looks like a circle!” Falsehood #4.

Just what is the path of the centroid of a Reuleaux triangle boring a square hole? Assume the square and the equilateral triangle have sides of length 1. Center the square about the origin and position the Reuleaux triangle so vertex A is at (-1/2,0), as in Figure 6a. Using (1), the triangle’s centroid will be P (-1/2+sqrt(3)/3,0). Now imagine rotating the triangle clockwise through the position in Figure 6b, and ending up in Figure 6c, where the centroid is P”(0,-1/2+sqrt(3)/3). The path from P to P” lies in quadrant I. Let a be angle MA’B’, ß the angle formed by A’P’ and a horizontal line through A’, and c the y-coordinate of point A’. We are interested in the coordinates of P’. Note that cosa=1/2+c and that ß=270°+a+30°=300°+a. Also note that during this rotation, a goes from 60° to 30°. Because A’P’=sqrt(3)/3, if we measure from the coordinates of A'(-1/2,c), the x and y coordinates of P’ can be found.

-1 sqrt(3) -3 + sqrt(3) cos(a) + 3 sin(a) (2) x = - + ------- cos(300°+a) = ------------------------------, 2 3 6

and

sqrt(3) (3) y = c + ------- sin(300°+a) 3

sqrt(3)

= (cosa – 1/2) + ——- sin(300°+a)

3

-3 + 3 cos(a) + sqrt(3) sin(a)

= ——————————

6

as a goes from 60° to 30°. Finding the path of the triangle’s center in the other three quadrants is similar in procedure and produces equations symmetric to the origin and both axes.

3 - sqrt(3) cos(a) - 3 sin(a) Quadrant II: x = ----------------------------- 6

-3 + 3 cos(a) + sqrt(3) sin(a)

y = ——————————

6

3 – sqrt(3) cos(a) – 3 sin(a)

Quadrant III: x = —————————–

6

3 – 3 cos(a) – sqrt(3) sin(a)

y = —————————–

6

-3 + sqrt(3) cos(a) + 3 sin(a)

Quadrant IV: x = ——————————

6

3 – 3 cos(a) – sqrt(3) sin(a)

y = —————————–

6

But these equations do not describe a circle. In equations (2) and (3), when a=30°, P is on the x-axis at approximately (0.07735,0). But when a=45°,

-6 + sqrt(6) + 3sqrt(2) x = y = ------------------------, 6

which makes the distance from P’ to the origin about 0.08168. This non-circularity is also shown by graphing the four parametric equations above with a circle whose radius is slightly smaller or larger. In Figure 7, the circle is the outer curve. Notice that the centroid’s path is farther from the circle at the axes than mid-quadrant.

So the Reuleaux triangle’s centroid does not follow a circular path. How then is the Reuleaux bit contained within the square outline it’s to cut? Harry Watts designed a drill in 1914 with a patented “full floating chuck” to accommodate his irregular bits. Bits for square, pentagonal, hexagonal and octagonal holes are still sold by the Watts Brothers Tool Works in Wilmerding PA. The actual drill bit for the square is a Reuleaux triangle made concave in three spots to allow for unobstructed corner-cutting and the discharge of shavings (Figure 8).

Even the modified bit leaves slightly rounded corners. How rounded? Assume the starting position in Figure 9a, in which the Reuleaux triangle is just tangent at point C. As the triangle rotates counterclockwise, C leaves that edge of the square temporarily (labeled C’ in Figure 9b) only to rejoin another at position C” in Figure 9c. In Figure 9b, let a be angle MA’B’, ß be the angle formed by A’C’ and the horizontal line through A’, and c the y-coordinate of A’. Then ß = a+60°-90° = a-30° and cosa = 1/2+c. To generate the corner by C, a starts at 30° in Figure 9a and ends up at 60° in Figure 9c. Using A’C’ = 1 and measuring from the coordinates of A’, the coordinates of C’ are described by

-1 -1 + sqrt(3) cos(a) + sin(a) x = - + 1 cos(a-30°) = ----------------------------, 2 2

and

1 y = c + 1 sin(a-30°) = (cosa - -) + sin(a-30°) 2

-1 + cos(a) + sqrt(3) sin(a)

= —————————-.

2

The equations for the other three corners are similar, and when graphed with the rest of the square yield Figure 10.

Not only does the Reuleaux triangle have practical and interesting applications, and is easy to describe geometrically, but it generates a lot of discussion due to its nonintuitive properties. With this background, you can avoid the blunders I made. Further explorations into the topic might include other figures of constant width (see Gardner and Rademacher/Toeplitz); further identifying the curve of the Reuleaux triangle’s center as it cuts a square; and the shapes of bits for pentagonal, hexagonal and octagonal holes.

- Gardner, Martin. “Mathematical Games”. Scientific American 208:2 (February 1963): 148-156.
- How to Drill Square, Hexagon, Octagon, Pentagon Holes. Wilmerding, Penn: Watts Brothers Tool Works, 1966.
- Moise, Edwin and Floyd Downs Jr. Geometry. Reading, Mass: Addison-Wesley, 1982.
- Rademacher, Hans and Otto Toeplitz. The Enjoyment of Mathematics. New York: Dover Publications, 1990.
- Smart, James R. “Problem Solving in Geometry – a Sequence of Reuleaux Triangles.” Mathematics Teacher 79 (January 1986): 11-14.