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34 ANTHONY TODD<br />
BISECTING A TRIANGLE 35<br />
6<br />
4<br />
s ~<br />
4<br />
3<br />
3<br />
2<br />
2<br />
0 2 3 4 5 6<br />
s=6<br />
0 2 3 4<br />
FIG. 5.<br />
FIG. 7.<br />
4<br />
0.8<br />
3<br />
0.4<br />
2<br />
0.2<br />
0 0.2 0.4 0.6<br />
0 2 3 4<br />
FIG. 6.<br />
FIG. 8.<br />
by the splitters (the lines which extend from the vertex of a triangle and bisect the<br />
perimeter of a triangle) of the triangle as seen in Figure 8. In order to complete the<br />
picture we shall show the envelopes for a 3-4-5 right triangle which has only one B-line<br />
solution. In Figure 9 is plotted the incenter (denoted as a small circle) and the one<br />
B-line solution.<br />
3. <strong>Pi</strong>cture Time. We know the maximum number of B-line solutions that any<br />
triangle could have is three. It would appear that the number of two B-line solution<br />
triangles are small in comparison to the one and three B-line solution triangles since<br />
they occur only when I is concurrent with the curve which defines one of our two<br />
envelopes. What we do not know is what the triangles look like that produce one,<br />
two, or three B-line solutions.<br />
In Figure 10, is the plot of the number of B-line solutions given angle measure-<br />
ments a and 1 ( in degrees). We see from Figure 10, as stated earlier, that there are<br />
a maximum of three solutions with the majority of triangles having only one solution.<br />
The long extensions of the surface that represents three B-line solutions follows the<br />
lines that define isosceles triangles ( a = 1, a = 1r- 21, and 1 = 1r - 2a). <strong>No</strong>te<br />
that Figure 10 also shows that the solutions for the two B-line triangles is in fact the<br />
boundary of the three B-line solution surface.<br />
With the equations to the boundary of the three B-line solutions, we then are<br />
able to show what the triangles look like that have one, two, or three B-line solutions.<br />
Since we have the solutions to the angle measures, given a side AC of length 1, we<br />
can solve for the position of B to yield one, two, or three B-line solutions. To do<br />
this we need only to use the case for two solutions since we know that they form the<br />
boundary between one and three B-line solutions.<br />
Let AC be of length 1 with A at (0, 0) and C at (1, 0). The vertex B then lies
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