An interactive column using Java applets
Cut The Knot!
by Alex Bogomolny
Four Construction Problems
My morale received a boost recently (Are we getting through!) in the form of an email inquiry:
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 Tesseract page.)
Thus encouraged, let's forge on.
In the Introduction to [Yaglom] we find 3 construction problems:
Construct a triangle, given three points in the plane that are the outer vertices of equilateral triangles constructed outward on the sides of the desired triangle.
Construct a triangle, given three points in the plane that are the centers of squares constructed outward on the sides of the desired triangle.
Construct a heptagon (polygon of 7 sides), given seven points that are the midpoints of its sides.
To get a better appreciation of what follows, try to solve the problems. (Or consider buying Yaglom's classic 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):
Given n points M1, M2, ..., Mn (n > 2) and angles a1, a2, ..., an, construct a polygon A1, A2, ..., An,
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 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.
On the sides of ΔABC first construct outwardly (inwardly) three squares with centers X, Y, and Z. On the sides of ΔXYZ construct next three inward (outward) squares with centers P, Q, and R. Points P, Q, and R then coincide with the midpoints of ΔABC.
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.
Copyright © 1996-2017 Alexander Bogomolny