# About a Line and a Triangle

Given ΔABC, extend the side AB beyond the vertices. Now, rotate line AB around 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.

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)?

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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 b - x. Therefore, there exists one point on the line that does not move even after a single rotation. This is the fixed point of the transformation. The fixed point solves the equation x = b - x. The rotation of the line around the triangle is simply equivalent to the rotation of the line around that point through 180°.

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 b - (b - x) = x.

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

1. D. Gale, Tracking the Automatic Ant, Springer-Verlag, 1998 ### 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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

Inscribe a circle into ΔABC. There are three points of tangency where the sides of the triangle touch the inscribed circle. It helps to think of the configuration (a triangle + a circle inside) as a circle with a circumscribed triangle. Then the diagram presents a circle with three points from each of which twin tangents to the circle are drawn. Rotations around the vertices map subsequently one of the tangents in a pair into its twin. Therefore the points of tangency are mapped onto each other such that after rotations around all three vertices all three points return to their original locations. A nice way to look at the inscribed circle!

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

x + y = a
y + z = b
z + x = c

Adding up the three equations gives

x + y + z = (a + b + c)/2

Subtract from the latter x + y = a to obtain z = (b + c - a)/2. Similarly, x = (a - b + c)/2 and y = (a + b - c)/2. These expressions are of the same kind that are obtained (rather simply) when trying to locate the fixed point of the line transformation.

It is worth noting that in a different guise the problem has been posted in 1943 by H. Eves as a mathematical quickie. It is related to a construction of shapes of constant width and Conway's circles.

### Reference

1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #181 (Am Math Monthly, 50 (June, 1943), 391) 