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Cut The Knot!An interactive column using Java appletsby Alex Bogomolny |
Four Construction Problems
July 2000
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My morale received a boost recently (Are we getting through!) in the form of an email inquiry:
(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 Tesseract page.) Thus encouraged, let's forge on. In the Introduction to [Yaglom] we find 3 construction problems:
To get better appreciation of what follows, try solving the problems. (Or consider buying Yaglom's classics at the MAA Bookstore at the discount price comparable to the cost of shipping and handling.) In the book, solutions follow immediately the formulations. Towards the end of the Introduction, Yaglom brings the three problems under a single umbrella (my paraphrase):
The first problem is obtained with The applet below is to assist in solving the general problem: 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 (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 display 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 M2, M3, and lastly in Mn. All intermediate segments have equal lengths. The last one again ends at A1. Let its other end be A'. We have 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 not therefore important which orientation we choose, but that the chosen orientation and the signs of the angles be in the desired relation, and the sheer fact that there are exactly two possible orientations. A theorem by Joseph Neuberg [Honsberger, p. 273] is relevant in this context:
Neuberg's theorem is easily proven with Linear Polygonal Transformations. (There's also a synthetic proof.) 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
Copyright © 1996-2008 Alexander Bogomolny |
AiMiAi+1 = ai.
ABC first construct outwardly (inwardly) three squares with centers X, Y, and Z. On the sides of 
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