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About a Line and a TriangleGiven It is virtually obvious that although the line now occupies exactly the same position as before, something has changed. After three rotations, the line turned around 180o. So, for example, the point A will now lie on a different side from B than before. We say that turning the line around the triangle changed its orientation. It appears that the line occupies the same position but not quite: points on the line did not preserve their locations. However, since there are just two possible orientations of the line, we come up with an interesting question: what happens to the line after it turns around the triangle twice? Will it occupy its original position exactly (point-for-point)? The answer is easily obtained from the following observation. After the first rotation the line occupies the same position but with a different orientation. Let's turn the line into coordinate axis. In other words, let's choose the origin - point O, the unit of measurements, and the positive direction. If, after the rotation, the point O maps into the point b (positive or negative) units away from its original position, then the point originally at the distance x from O will be now located at the position Since the second rotation is exactly the same as the first one, after two such rotation the line will occupy exactly the same position, point-for-point. This is also seen algebraically. The point with the coordinate x moves first into the point with the coordinate (b - x) which then moves into What about the fixed point? In the applet, besides vertices of the triangle, I also indicate the mid-point of the side AB. Sometimes but not in general, the mid-point is fixed. So, how can be the fixed point described in general? In his book (see the reference below), David Gale makes the following remark: this really is a quickie. Well, I must confess that it took me a while to figure this one out. (I just calculated the location of the fixed point.) However, after the fact, I now see what he meant and can't help smiling. This really is a quickie if you look at the situation from the right angle. Reference
Copyright © 1996-2008 Alexander Bogomolny
HintInscribe a circle into I did not see that solution until after locating the fixed point algebraically. Following the line as it rotates around three vertices (The easiest way is to follow the point C as it only rotates once!), measuring some distances and eventually solving a linear equation brought up an expression that reminded me of the following. Let the sides of the triangle be a, b, and c. Let x, y, and z denote the length of tangents from the vertices of the triangle to the inscribed circle. Then we obviously have something like
Adding up the three equations gives
Subtract from the latter x + y = a to obtain It is worth noting that in a different guise the problem has been posted in 1943 by H. Eves as a mathematical quicky. It is related to a construction of shapes of constant width and Conway's circles. Reference
Copyright © 1996-2008 Alexander Bogomolny |
ABC, extend the side AB beyond the vertices. Now, rotate the line AB around the vertex A until it falls on the side AC. Next rotate it (from its new position) around C until it falls on the side BC. Lastly, rotate it around B till it takes up its erstwhile position.
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