Trisection of a segment
by
Bradley Johnson
I. A script tool is provided for this contruction.
The first contructions for trisecting a line segment is as follows:
Given segment AB we will place an arbitrary point H on the segment (see figure 1).
figure 1
We will now construct a circle with center A that passes through point H (see figure 2).
figure 2
Selecting an arbitrary point ( C ) on the circle, but not on the segment AB, we construct a ray through point A and point C. We use point C as the center of another circle with the same radius as the first, and label the point of intersection of this circle with the ray. We repeat this same process again (so that we now have three circles (see figure 3).
figure 3
Label the point of intersection of the ray and this third circle ( call it point E) we join E and B with a segment. We now construct parallel lines to segment EB through the other two points of intersections previously created ( intersections of the circles and ray) See figure 4.
figure 4
The points of intersectin with these paralells are the points of trisection. We know this to be true since we constructed the points along the ray to be equidistant apart, and by contruction of the parallel lines we have similar triangles.
II. A script tool for this construction is provided here.
We first construct the midpoint (m) of the given segment AB. Our next step will be to construct a circle with center B and radius BM. This gives us a new segment AC which has length 3/2(AB).
We will then use segment AC as the radius of a circle centered at point C, resulting in a segment AD which is 2(AC) = 3(AB).
If we now construct a circle centered at A and through point B, we see that this intersects with the circle from the step above. We label this point of intersection E and construct a perpendicular line to AD through point E. The intersection of this perpendicular and AD we will label as point P.
We now join points A and E with a line segment as well as points E and D resulting in the following picture in which we see two similar triangles (EAP) and (DAE)
We contructed our proof in such a way that AD = 3AB and with AE = AB it follows that AD = 3AE. Thus EAP is a dilation of scale factor 1/3 of triangle DAE and we know that AP corresponds to side AE and therefore AP is 1/3 of AB. So as desired we have constructed a segment 1/3 the length of our original segment from which we are able to trisect the segment AB.
III. A tool is provided here for this construction
This method of trisection is similar in nature to the first method presented, but slightly different.
We begin in the same manner of constructing a circle centered at point A of the given segment AB. See the picture below.
We diverge from the first method at this point in our proof. We will now construct a parallel ray through point B extending in the opposite direction as the existing ray. Along our new ray we will construct the same two circles as in the previous step.
Now we will join points D and E with a segment, along with points C and F.
By similar triangles we see that triangle GFB is a dilation scale factor 2 of triangle HEB and triangle HEB is congruent to triangle GCA. Thus we can say that the length of AG plus the length of GB is equal to the unit. Thus we have the expression AG + 2AG =AB from which we yield
AG = 1/3AB.
IV. The tool that uses the following method is provided here.
In this method we will prove that constructing a triangles centroid is a method for trisecting a line segment. In the picture below we begin with the red segment. We begin by constructing two circles with radius the length of the segment and centers the segments endpoints. From one intersection point of the two circles (point A in the diagram) we extend a ray through one of the endpoints of the original segment (creating segment AB). We will label the point of intersection of this ray and one of the circles point B. We then create a segment from point B to the not already used endpoint of the original segment. Thus we have effectively constructed a triangle. We intend to show that this triangle (ABC) has a centroid that lies 1/3 of the way across the orginal segment, and thus trisecting the segment. By the construction point D is the midpoint of segment AB and therefore our original segment is a median of triangle ABC. We know that the centroid of triangle ABC will divide each median into a 1: 2 ratio, so by constructing a second median, we find the centroid of ABC as well as the point that lies 1/3 of the way along our original segment (DC). Thus we have effectively created a trisection of the segment DC.
