Reuleaux triangle

By Daina Taimina and David W. Henderson

What is this triangle?

If an enormously heavy object has to be moved from one spot to another, it may not be practical to move it on wheels. Instead the object is placed on a flat platform that in turn rests on cylindrical rollers. As the platform is pushed forward, the rollers left behind are picked up and put down in front. An object moved this way over flat horizontal surface does not bob up and down as it rolls along. The reason is that cylindrical rollers have a circular cross section, and a circle is closed curve "with constant width". What does it mean? If a closed convex curve is placed between two parallel lines and the lines are moved together until they touch the curve, the distance between the parallel lines is the curve's "width" in one direction. Because a circle has the same width in all directions, it can be rotated between two parallel lines without altering the distance between the lines.

Figure 1

Is the circle the only curve with constant width? Actually there are infinitely many such curves. The simplest noncircular such curve is named the Reuleaux triangle. Mathematicians knew it earlier (some references goes back to Leonard Euler in 18th century) but Reuleaux was the first to demonstrate its constant width properties.

 

How to construct Reuleaux triangle

A Reuleaux triangle can be constructed starting with an equilateral triangle of side s and then replacing each side by a circular arc with the other two original sides as radii.

Figure 2

The corners of a Reuleaux triangle are the sharpest possible on a curve with constant width. Extending each side of an equilateral triangle a uniform distance at each end can round these corners. The resulting curve has a width, in all directions, that is the sum of the same two radii.

Figure 3

Other symmetrical curves of constant width result if you start with a regular pentagon (or any regular polygon with an odd number of sides) and follow similar procedures. This has practical application in some British coins.

Figure 4

This result can be generalized to the case when the polygon is not regular, but each of its vertices is the endpoint of two diagonals of the same length h (and the lengths of other diagonals are less than h). Of course, polygon has to have odd number of vertices.

But here is one really surprising method of constructing curve with a constant width:

Draw as many straight lines as you please all mutually intersecting. Each arc is drawn with the compass point at the intersection of the two lines that bound the arc. Start with any arc, then proceed around the curve, connecting each arc to the preceding one. If you do it carefully, the curve will close and will have a constant width. (You can try to prove it! It is not difficult at all.) The curves drawn in this way may have arcs of as many different circles as you wish. Here is one example with three first steps in drawing such curve but you will really enjoy making your own! After you have done that, you can make couple more copies of it and check that your wheels really roll!

Figure 5

Another interesting result about curves of constant width is that inscribed and circumscribed circles of an arbitrary figure of constant width h are concentric and the sum of their radii is equal to h.

Some more links about Reuleaux triangle:

Rolling with Reuleaux triangle http://www.nas.com/~kunkel/reuleaux/reuleaux.htm

http://www.maa.org/mathland/mathland_10_21.html

http://www.cut-the-knot.org/do_you_know/cwidth.shtml

http://www.wundersamessammelsurium.de/Mathematisches/Reuleaux/

Drilling square holes http://upper.us.edu/faculty/smith/reuleaux.htm

Here we collected some

Problems about Reuleaux triangle and other curves of constant width

  1. For the Reuleaux triangle, find its area. Does it exceed the area of the circle of the same width (diameter)?
  2. Show that among curves with constant width circle bounds the region of greatest area and the Reuleaux triangle bounds the region of least area.
  3. For the Reuleaux triangle, derive a formula linking the length of the perimeter to its width. This theorem is due to Joseph Emile Barbier (1839-1889): all shapes of constant width D have the same perimeter, L = o D. (Proof of this theorem in Lyusternik L.A. Convex figures and polyhedra. New York, Dover, 1963, p. 31-35)
  4. The angle between two intersecting curves is defined as the angle between their tangents at the point of intersection. Find internal angles of the Reuleaux triangle.
  5. Assume that the equilateral triangle's side is 50. Each side was extended 10 units in each direction and then the shape was rounded to construct a curve with constant width. What is the width of the resulting shape?
  6. Show that for every point on the boundary of a figure of constant shape there exists another boundary point with the distance between the two equals to the width of the shape.
  7. Show that the distance between any two points inside the shape of constant width never exceeds its width.
  8. Show that the length of the boundary of shapes of constant width depends only on their width.
  9. Show that a curve of constant breadth has just one point in common with each of its supporting lines.
  10. The distance between any two points on a curve of constant breadth b is at most equal to b.
  11. If a line joins the two points of contact of two parallel supporting lines of a curve of constant breadth, then it is perpendicular to the supporting line.
  12. There is at least one supporting line through every point of a curve of constant breadth.
  13. Through every point P of a curve of constant breadth, a circle of radius b can be drawn that encloses the curve and that is tangent, at P, to the supporting lines of the curve, or to a predetermined supporting line if there are more than one.
  14. If a circle has three (or more) points in common with a curve of constant breadth b, then the length of the radius of the circle is at most b.
  15. The circle is the only curve of constant width with a center of symmetry. (Before proving this theorem it is useful first to prove lemma: Any two diameters of a curve of constant width must intersect on the curve or in the interior of the curve. If they intersect on the curve, then the point of intersection is a corner point of the curve. )
  16. A sequence of triangles and associated Reuleaux triangles can be drawn by using midpoints of the sides of the original triangle as the vertices for the next triangle. The pattern can be repeated indefinitely, with each new triangle smaller than the preceding one. How does the total area of all smaller Reuleaux triangles compare with the area of the original Reuleaux triangle? The key observation that suggests a solution to the problem is realizing that, in addition to the sequence of Reuleaux triangles, we have a related sequence of equilateral triangles.

The three dimensional analogue of a curve with constant width is the solid of constant width. A sphere is not the only such solid that will rotate within a cube, at all times touching all six sides of the cube; all solids of constant width share this property. Rotating the Reuleaux triangle around one of its axes of symmetry generates the simplest example of a nonspherical solid of this type. There are an infinite number of others. The solids of constant width that have the smallest volumes are derived from the regular tetrahedron in somewhat the same way the Reuleaux triangle is derived from the equilateral triangle. Spherical caps are first placed on each face of the tetrahedron, then it is necessary to alter three of the edges slightly. These altered edges may either form a triangle or radiate from one corner. Since all curves of the same constant width have the same perimeter, it might be supposed that all solids of the same constant width have the same surface area. It was proved by Hermann Minkowski that all the shadows of solids of constant width (when the projecting rays are parallel and the shadow falls on a plane perpendicular to the rays) are curves of the same constant width. All such shadows have equal perimeters. Michael Goldberg has introduced the term "rotor" for any convex figure that can be rotated inside a polygon or polyhedron while at all times touching every side or face. The Reuleaux triangle is the rotor of least area in a square. The least area rotor for the equilateral triangle is a biangle – lens shaped figure formed with two 60-degree arcs of a circle having a radius equal to the triangle’s altitude.. As it rotates its corners trace the entire boundary of the triangle, with no rounding of corners.

Closely related to the theory of rotors is a famous problem named the Kakeya needle problem. Japanese mathematician Soichi Kakeya first posed it in 1917: what is the plane figure of least area, in which a line segment of length 1 can be rotated 360 degrees?

The rotation obviously can be made inside a circle of unit diameter, but that is far from the smallest area. Ten years after Kakeya problem was posed Russian mathematician Abram Besicovitch proved that the problem had no answer. He showed that there is no minimum area as an answer to Kakeya’s needle problem.

Reuleaux Tetrahedron:

http://mathworld.wolfram.com/ReuleauxTetrahedron.html

http://www.fastgeometry.com/Reuleaux/ConstantBreadth3DShapes.htm

Kakeya needle problem:

http://mathworld.wolfram.com/KakeyaNeedleProblem.html

http://www.math.hmc.edu/funfacts/ffiles/20003.3-2.shtml

http://members.wri.com/wellin/Kakeya.html

http://www.math.ubc.ca/~ilaba/kakeya

http://www.ams.org/notices/200103/fea-tao.pdf

Where I can read more about Reuleaux triangle?

  1. Blaschke Wilhelm, Kreis und Kugel. Leipzig, 1916; Berlin: W. de Gruyter, 1956.
  2. Boltyanski Vladimir, Soifer Alexander, Geometric Etudes in Combinatorial Mathematics, Center for Excellence in Mathematical Education, Colorado Springs, 1991.
  3. Cadwell J. H., Topics in Recreational Mathematics. Cambridge, England : Cambridge University Press, 1966, Ch.15.
  4. Dossey John A., "What? A Roller With Corners? -Closed curves of constant width", Mathematics Teacher, No.65, 1972, pp. 720 - 724.
  5. Gardner Martin, The Unexpected Hanging and Other Mathematical Diversions, Simon&Schuster, New York, 1969.
  6. Goldberg Michael, "N-gon Rotors Making n+1 Contacts with Fixed Simple Curves", American Mathematical Monthly, Vol.69, June-July 1962, pp. 486-91.
  7. Goldberg Michael, "Rotors in Polygons and Polyhedra," Mathematical Tables and Other Aids to Computation, Vol. 14, July 1960, pp.229-39.
  8. Goldberg Michael, "Trammel Rotors in Regular Polygons," American Mathematical Monthly, Vol. 64, February 1957, pp. 71-78.
  9. Rademacher Hans, Toeplitz Otto, The Enjoyment of Mathematics, Princeton, N. J.: Princeton University Press, 1957, pp 163-77, 203.
  10. Reuleaux Franz, The Kinematics of Machinery. New York: Macmillan, 1876; Dover Publications, 1964, pp.129-46.
  11. Smart James R., " Problem Solving in Geometry--a Sequence of Reuleaux Triangles: Investigation of area relations for a sequence of Reuleaux triangles associated with an equilateral triangle and a sequence of medial triangles", Mathematics Teacher, No.79, 1986, pp.11 - 14.
  12. Smith Stanley A., "Rolling Curves - Activities involving curves of constant width", Mathematics Teacher, No.67, 1974, pp. 239 - 242.
  13. Smith, Scott G., "Drilling square holes: Using a Reuleaux triangle", Mathematics Teacher, No.86, 1993, pp. 579 - 583.
  14. Yaglom, I., M., Boltyanskii V. G., Convex Figures, New York: Holt, Rinehart& Winston, 1961,Ch. 7 and 8.

Where can I learn more about Kakeya's needle problem?

  1. Besicovitch, A. S., "The Kakeya Problem", American Mathematical Monthly, Vol. 70, August-September 1963, pp. 697-706.
  2. Blank A. A., "A Remark on the Kakeya Problem," American Mathematical Monthly, Vol. 70, August- September 1963, pp. 706-11.
  3. Cadwell J. H., Topics in Recreational Mathematics, Cambridge, England: Cambridge University Press, 1966. Pp. 96-99.
  4. Yaglom I. M., Boltyanskii, V. G., Convex Figures, New York: Holt, Rinehart& Winston, 1961. Pp. 61-62, 226-27.