Cut The Knot!An interactive column using Java applets
by Alex Bogomolny
The Theorem of Barbier
At the end of the August's column we arrived at Barbier's theorem as a surprising consequence of Count Buffon's experiment. This time around I wish to present another demonstration [Yaglom] that stems from a surprising and elegant application of algebraic concepts to geometry.
Barbier's theorem states that all shapes of constant width D have the same perimeter pD. The width of a convex figure in a certain direction is the distance between two supporting lines perpendicular to that direction. (A straight line is called supporting a convex figure if they have at least one common point and the figure lies on one side from the line. In any direction there are two supporting lines.) Shapes of constant width are convex figures that have the same width in any direction. Circle has this property. The Reuleaux triangle is the next simplest shape. Others may be constructed starting with equilateral (but not necessarily equiangular) stars as demonstrated by the applet below.
The algebraic concept fundamental to the proof is Minkowski's addition of convex sets. With a fixed origin O, the sum of two shapes is the collection of all endpoints of vectors
Properties of Minkowski's Addition
The shape of the sum does not depend on the location of the origin.
When the origin is shifted, the sum moves in the opposite direction by the same distance.
The sum of convex figures is convex.
Due to the idiosyncrasies of implementation, in the applet above, the sum is always convex even when the addend are not. In fact the result shown is the convex hull of the sum. This is quite sufficient for my purposes as the shapes of constant width are, by definition, convex.
The width of the sum in any direction equals the sum of widths of the addends in that direction.
This is quite clear when the two figures are polygons. Indeed, the sum is a polygon whose vertices are obtained as the sums of vertices of the addends. We may claim more.
Call the points common to a convex figure and its supporting line extreme in the direction perpendicular to the supporting line. In any direction their is a pair of (opposing) extreme points. The applet offers a convincing demonstration that the extreme points of the sum in some direction are formed as the sums of the extreme points of the addends in the same direction.
For example, if vertices 23 and 11 are extreme points of the sum in a certain direction, then we may be sure that vertices 2 and 1 are extreme for the "left" polygon in that direction, and the same is true of the vertices 3 and 1 of the "right" polygon. (Note that the vertex labels of the sum are formed by concatenation of vertex labels of the two addends.)
Assume, as in the original display, the "left" polygon is a triangle 012, while the right polygon is a quadrilateral 0123. Consider the polygons 03-13-23, which is just a translation of Δ012, and the quadrilateral 10-11-12-13, which is just a translation of 0123. The two share vertex 13 that serves as an extreme point for both polygons (together with 23 for the "left" one and together with 11 for the "right" one) in the direction at hand.
The perimeter of the sum equals the sum of perimeters of the addends.
This is clear for the polygons. In the most general case, the sum of an n-gon and m-gon is an (n+m)-gon. Furthermore, each side of the sum is equal and parallel to one and only one of the n+m sides of the addends. If one of the sides of the n-gon is parallel to a side of the m-gon, the sum will have a side which is parallel to either and whose length is the sum of the two sides' lengths.
We extrapolate this result to more general convex sets in the hope that their perimeter may be adequately defined as the limit of the perimeters of circumscribed polygons. It can be shown that if the circumscribed polygons of two convex figures have parallel sides and the same orientation, their sum happens to be a circumscribed polygon of the sum of the two given shapes [Yaglom, Problem 44].
The sum of a figure and its centrally symmetrical image has central symmetry.
To complete the toolbox for a proof of Barbier's theorem we only need one additional fact about shapes of constant width:
Circle is the only centrally symmetric shape of constant width.
Indeed, let a shape of constant width has a diameter (the line connecting two extreme points in a given direction) AB. AB must pass through the center of symmetry. Otherwise, its central image A'B' is another diameter in the same direction. In the parallelogram ABA'B' one of the angles A or B is not less than 90°. Assuming it's A, from ΔBAB',
BB' > AB.But this leads to a contradiction, because in a shape of constant width no two points may be at the distance exceeding its diameter (common width in any direction.)
Therefore, all diameters of a centrally symmetric shape of constant width pass through the center of symmetry. Because of the symmetry, each of the diameters is divided in half by that point. Thus not only the shape has diameters, it also has radii. It's a circle.
We are now in a position to prove Barbier theorem. Let K be a shape of constant width D. Let L be obtained from K by central symmetry. L is also a shape of constant width with diameter D. The following can be claimed about the sum K+L:
First of all, K+L is a shape of constant width 2D (by 3).
K+L is a circle (by 5 and 6).
The sum of perimeters of K and L is p·2D (by 4).
If p(K) is the perimeter of K and P(L) is that of L, then
- M.Gardner, The Unexpected Hanging and Other Mathematical Diversions, The University of Chicago Press, 1991
- R.Honsberger, Ingenuity in Mathematics, MAA, New Math Library, 1970
- H.Rademacher and O.Toeplitz, The Enjoyment of Mathematics, Dover Publications, 1990.
- I.M.Yaglom and V.G.Boltyansky, Convex Figures, Holt, Rinehart and Winston, 1961
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