# Cut The Knot!

An interactive column using Java applets
by Alex Bogomolny

# Four Construction Problems

July 2000

My morale received a little boost recently (we are getting through!) in the form of an email enquiry:

 dude, I really hate math, but that tesseract thing kicks ass! I really wonder though, is it possible to construct a physical model even though its 4-d, or can it only exist as a calculation?

(I apologize for the colloquialisms in the message. A fellow may be less guarded on the Web than in a face-to-face contact. Might this example of the perceived impersonality of email communication be an indication of how technology could affect our culture? The reference is probably to the The Tesseract page.)

Thus encouraged, let's forge on.

In the Introduction to [Yaglom] we find three construction problems:

1. Construct a triangle, given the three points in the plane that are the outer vertices of equilateral triangles constructed outward on the sides of the desired triangle.

2. Construct a triangle, given the three points in the plane that are the centers of squares constructed outward on the sides of the desired triangle.

3. Construct a heptagon (polygon of 7 sides), given the seven points that are the midpoints of its sides.

To get a better appreciation of what follows, try solving the problems. (Or consider buying Yaglom's classics at the MAA Bookstore at a discount price comparable to the cost of shipping and handling.) In the book, solutions immediately follow the formulations. Towards the end of the Introduction, Yaglom brings the three problems under a single umbrella (my paraphrase):

 Given n points M1, M2, ..., Mn (n > 2) and angles a1, a2, ..., an, construct a polygon A1, A2, ..., An, An+1 = A1 such that triangles AiMiAi+1 are isosceles (AiMi = Ai+1Mi) with the apex angle AiMiAi+1 = ai.

The first problem is obtained with n = 3, a1 = a2 = a3 = 60o, in the second problem n = 3, a1 = a2 = a3 = 90o, and in the third n = 7, ai = 180o, i = 1, 2, 3, ..., 7.

The applet has three modes. In the "Place points" mode, you define (by clicking) and move (by dragging) a sequence of points M. The order in which the points are created determines the order of traversal (the orientation) of the sequence. The set of points may have two different orientations. The angles, on the other hand, are always measured in the positive direction of the coordinate system - left handed in the applet, which means that the angles are measured clockwise.

In the "Change angles" mode, angles are displayed next to the corresponding point and can be modified by clicking (slow) or dragging the cursor (fast) a little off their central line.

In the "Drag cursor" mode, the cursor position is rotated in order through the given angles around the given points. What is shown is a broken line whose starting and ending points are denoted by the same letter. There may be several such lines.

Here's an outline of the construction. Assume A1A2...An is the desired polygon. Pick up a point A. Rotate the segment AA1 in M1 through the angle a1. Since the rotation is a motion of the plane that preserves shapes and distances, the image of the segment AA1 is equal in length to the segment itself. Rotate that image segment successively in A2, A3, and lastly in An. All intermediate segments have equal lengths. The last one again ends at A1. Let its other end be A'. We have AA1 = A'A1. This means that A1 lies on the locus of points equidistant from A and A' - the perpendicular bisector of the segment AA'. Pick up another point B and similarly construct B'. The perpendicular bisectors of AA' and BB' are bound to intersect at A1. (If the angles ai add up to a multiple of 360o, the construction breaks down. Why?)

In the general case, and especially because the vertices of the triangles in the applet are defined dynamically, the designation of outwardly constructed triangles becomes awkward. The notion of orientation saves the day. Vertices with positive angles lie to the right of the desired polygon, vertices with negative angles lie to its left. If the orientation of the vertices is counterclockwise, the former appear to be constructed outwardly, the latter inwardly on the sides of the polygon. For the clockwise orientation outside becomes inside and vice versa.

It's therefore not important which orientation we choose, but rather that the chosen orientation and the signs of the angles be in the desired relation.

A theorem by Joseph Neuberg [Honsberger, p. 273] is relevant in this context:

 On the sides of ABC first construct outwardly (inwardly) three squares with centers X, Y, and Z. On the sides of XYZ construct next three inward (outward) squares with centers P, Q, and R. Points P, Q, and R then coincide with the midpoints of ABC.

Neuberg's theorem is easily proven with Linear Polygonal Transformations. One of the constructions is defined by the circulant matrix

the other with the circulant

where c = (1 + i)/2, i being one of the square roots of -1. The bar above c denotes its conjugate, (1 - i)/2. For the product of the transformations we get

which proves the theorem.

### References

Alex Bogomolny has started and still maintains a popular Web site Interactive Mathematics Miscellany and Puzzles to which he brought more than 10 years of college instruction and, at least as much, programming experience. He holds M.S. degree in Mathematics from the Moscow State University and Ph.D. in Applied Mathematics from the Hebrew University of Jerusalem. He can be reached at alexb@cut-the-knot.com