Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

A Sampling from a Geometry Collection

September 2003

To jog the understanding is a greater feat, than to jog the memory: for it takes more to make a man think, than to make him remember.

The Art of Worldly Wisdom,
Baltazar Gracian
Barnes & Noble, 1993, p. 38

Very recently I had the pleasure of reviewing a remarkable collection of articles, The Changing Shape of Geometry edited by Chris Pritchard. The book is a highly recommended resource for mathematics teachers; it may serve as a great source of enjoyment and a possible starting point for further explorations for mathematics fans. In this column I'll take a closer look at three topics from the book.

An Isoperimetric Theorem
John Hersee, pp. 40-41

The word isoperimetric combines [Schwartzman] three Greek roots: isos "equal", peri "around", and metron "a measure". The well known Isoperimetric Theorem deals with the shapes that have the same perimeter. The subject of its much simpler namesake is the configuration in which shapes with equal perimeters arise naturally.

Draw a semicircle and on its diameter a few (two or more or infinitely many) smaller semicircles that abut each other in sequence. By construction, the sum of the diameters of the small semicircles equals the diameter of the large one. What about their perimeters, or, say, their total arc lengths?

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

Answer: the arc lengths and, therefore, the perimeters are equal. In other words, a semicircle and a family of abutting semicircles constructed on its diameter are isoperimetric. The proof follows from the formula


P = πD

that links the diameter D and the circumference P of a circle. For the arc lengths involved, we have

½πD1 + ½πD2 + ... = ½π(D1 + D2 + ... ) = ½πD,

where D is the diameter of the big semicircle, while D1, D2, ... are the diameters of the small ones. We thus obtain a family of isoperimetric shapes whose boundary consists of circular arcs and straight line segments, the derivation being based on the well known, but mostly misrepresented, identity (1).

When rewritten as


P/D = π,

the identity says that P/D is an invariant of the circular shape. Because of the importance and ubiquity of the latter, this invariant in the case of the circle was designated with a special symbol π. However, the fact that P/D is a shape invariant is not peculiar to the circles. For example, for a square, P/D = 2·2, while for a rectangle with sides proportional to 3 and 4, P/D = 14/5. (Diameter of a shape is the longest distance between any two of its points, and also the segment joining any two points, for which the longest distance is attained. Circle has infinitely many diameters in the latter sense. Shapes of constant width share that property (as well as (1) and (2)) with the circle. But square has only two diameters, a regular 2n-gon has n.)

It is often misunderstood that the existence of π is an immediate consequence of the fact that all circles are similar. For example, in an otherwise excellent textbook, Jacobs introduces (1) along with the definition of the circumference as the limit of the perimeters of inscribed regular n-gons in a chapter on regular polygons, while a chapter on similarity mostly (if not exclusively) deals with similar triangles. Curiously, an exercise in that chapter is based on Euclid, VI.31 which implicitly assumes that the areas A of similar shapes stand, as was proven in Euclid, VI.19 for triangles, in the same ratio as the squares of their diameters:


A1/A2 = D12/D22,

or, in other words, that A/D2 is another shape invariant. Interestingly, there is no mentioning in Euclid's Elements that P/D is a shape invariant, whereas analogues of (3) are brought up several times. In particular, Euclid XII.2 asserts (3) for the circle. The value of A/D2, or that of P/D, is not mentioned in the Elements at all. That the two are related has been established by Archimedes in his Measurement of a Circle, Proposition 1. In Proposition 3, he also proved an estimate


223/71 < π < 22/7.

(Proposition 2 that claims A/D2 = 11/14 is inconsistent with the latter and can't be seen as authentic.)

In geometry teaching the hands-on estimation of π most universally comes at the expense of the basic understanding expressed by (2). Even working mathematicians often overlook its significance. For example, [Arndt, p. 165] writes, "In the beginning, π = 3. ... The real history of π only begins when better approximations than 3 were discovered". I disagree. P. Beckmann [Beckmann, pp. 11-12], too, speculates that historically the realization that P/D is a shape invariant must have preceded any attempts at numerical estimation of this ratio. Be as it may, common emphasis on hands-on estimation of π misses an opportunity to get students involved with real mathematics at an elementary level.

(The applet allows for experimentation with two other families of isometric curves. An independent proof in case of two sets of parallel lines must be obvious. The small hollow circles are draggable. The red freehand curve is also modifiable by dragging any of its points. The number of break points could be changed by clicking on the number at the bottom of the applet.)

Two Right Tromino Theorems
Solomon Golomb, pp. 343-345

Polyominoes have been invented by Solomon Golomb half a century ago. Beautiful in its simplicity, a right tromino theorem has been first published in American Mathematical Monthly 61, 10 (December, 1954.) Its proof by induction is now considered a prototypical classics.

A polyomino is a "rook"-connected set of equal squares. (In computer science the kind of connectedness where neighboring squares are required to share an edge is also known as the 4-connectedness. If two squares that share a vertex are also considered neighbors, we get the 8-connectedness.)

Trominoes are polyominoes that consist of three squares. There are two kinds of trominoes: a 3×1 rectangle and an L-shaped piece. The Right Tromino Theorem in the title is concerned with the latter.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.


If a single square is removed from a 2n×2n board, n any natural number, the rest can be tiled by trominoes.

The proof is by induction. The case of n = 1 is trivial. An 2n×2n board can be split into 4 boards with side length of 2n-1. The missing square belongs to one of them. The remaining three boards form a "big" L-shaped tromino. Place a tromino at the concave corner of that shape. It will cover one square from each of the three 2n-1×2n-1 component boards. Thus we are getting the 2n×2n board split into four 2n-1×2n-1 boards, each missing a square, plus a tromino, which assures the inductive step.

I have little to add here, except perhaps, that on an occasion I showed the applet and explained the theorem to seven and ten years old brothers. Both showed a grasp for the inductive step right away. In my judgement, filling a sequence of boards of doubling side lengths (or playing with the above applet) constitutes a very meaningful activity that combines a game with significant mathematics.

General Introduction
Chris Pritchard, pp. 1-9

(In the applet below, the eight blue vertices and the blue rectangle are draggable.)

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

Given two similar rectangles, a big one WXYZ and a small one ABCD, toss the smaller onto the larger. There exists one and only one point E in ABCD that falls on top of the point corresponding to it in WXYZ. Find the point.


If the corresponding sides of the two rectangles are parallel the construction is trivial. The more general case may be reduced to this one.

First, let BA produced meet XW at B'.

∠BEX = ∠BB'X = angle of rotation.

These angles are subtended by the same chord, XB. So XBEB' is cyclic quadrilateral.

Similarly, let AD produced meet WZ at A'.

∠AEW = ∠AA'W.

These angles are subtended by the same chord, AW. So WAEA' is cyclic quadrilateral.

Since E lies on the circumference of both circles, it is located at one of the points of intersection.

I have two critical remarks regarding this construction. First, a solution to a construction problem is best split into three stages: analysis, construction, and a proof of the validity of the construction. While the sentence that immediately follows the diagram could be considered as suggesting a proof of the validity of sorts, the first too stages are conspicuously intermixed: the point E appears too early in the construction and appears to play an important role right off.

Second, and perhaps, more important. The construction is asymmetric with regard to the elements involved, so that I believe Einstein might have called it ugly. Indeed, why have we produced sides AB and AD and not any other pair? Why not a diagonal? The fact that there was so much freedom of choice should have set off alarm. The problem is egregiously redundant. While the question with similar rectangles is of course legitimate, it obscures a deeper result. What is it?

Why, there are just so many ways to obtain one similar shape from another. If the shapes are of different sizes, there bound to be a spiral similarity that maps one onto the other and whose center remains fixed under the transformation. But to determine the center of a spiral similarity one only needs one segment and its image. The construction is in fact much the same. True, putting it in a more general framework may remove from the problem a degree of glamor, but there's a sensible payoff: unexpectedly there are several circles that all pass through the same point.

The element of surprise is still present, but the problem becomes just a dot in a cobweb of several related problems, which, who knows, the reader may risk to plunge into for the sake of further investigation.


  1. P. Beckmann, {A History of} π (pi), St. Martin's Griffin, 1971
  2. J. Arndt, C. Haenel, π   Unleashed, Springer, 2000
  3. W. Dunham, The Mathematical Universe, John Wiley & Sons, Inc., 1994
  4. H. R. Jacobs, Geometry, 3rd edition, W. H. Freeman and Company, 2003
  5. C. Pritchard (ed.), The Changing Shape of Geometry, Cambridge University Press, 2003
  6. S. Schwartzman, The Words of Mathematics, MAA, 1994

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