Trisections of the Areas of Triangles
Given a triangle ABC, find a point D such that line segments AD, BD, and CD trisect the area of the triangle into three regions with equal areas.
Before we look at when it is in the interior of the triangle, letÕs first investigate what happens when point D is on the triangle. When D is on the triangle, only two lines will be formed to trisect the triangle. LetÕs look at the following scenarios:
Base Case: First, letÕs look at what happens when point D is on the vertex of the triangle. If D is on a vertex, letÕs say D=A, then to trisect the area, we must trisect the opposite side (segment BC here). By connecting A to the trisection points of BC, then we will have three triangles that are a third of the area of the original triangle ABC. We can see this because the heights of all these triangles are the same, and we just constructed their base lengths to be the same. See the picture below or you can see the following GSP sketch.
Case One: Now, letÕs look at what happens when point D is on the side of the triangle (letÕs say on BC) at a trisection point. If this happens, then we can just connect D to the opposite vertex (here A). We know this is a third of the area of the original triangle because the height is the same and the base is a third of the original. Now, we have two-thirds of the area of the original triangle remaining. If we just bisect this triangle on any of its sides from any of its vertices (A, B, or D), then the areas of these two triangles will also be equal to one third of the original triangle. The following triangles are the three different possibilities when point D falls on the trisection point on BC closest to C.
Case Two: Finally, letÕs look at what happens when point D is on any point on the side of the triangle other than the trisection point or vertex (here weÕll say BC). First, weÕll need to construct a triangle, which is one-third the area of triangle ABC. Then, weÕll need to bisect the remaining two-thirds portion of the triangle. If the point is to the left or right of the altitude, then the construction will look somewhat like the picture on the left, although if it were on the right side, then it would mirror this image. If the point is equal to the intersection point of the altitude of A with side BC, then the picture will look like the right picture. Please see the following GSP links to see the construction of the area trisection for the left, center, and right cases.
Case Three: Now that we have seen how to trisect the area of a triangle when point D is on the triangle, now letÕs look at how to trisect the area of a triangle when point D is on the interior of the triangle, like the initial question implies. My claim is that when D equals the centroid of triangle ABC, this is the point at which such line segments AD, BD, and CD trisect the area of a triangle into three regions with equal area. To prove this, letÕs construct triangle ABC and its medians such that D is the centroid of the triangle.
Notice that the area of the area of , and weÕll say the area is , because they have the same base length and height. Likewise, the area of the area of (which weÕll say the area is ), and the area of the area of (which weÕll say is ).
Also, we know that the medians of a triangle cut the area of that triangle in half, so . Thus, . Therefore, we know that each of these triangles is equal to 1/6 of the area of the original triangle. Therefore, the area of the area of the area of the area of . Thus, when point D is the centroid of any triangle ABC, then the segments DA, DB, and DC trisect the area of the triangle into three regions with equal area.
Given a triangle ABC. Construct two line segments parallel to the base BC to divide the triangle into three regions with equal areas.
Well, we know that will be 2/3 the area of . Since is 1/3 the area of , then is ½ the area of . From Essay 1, we know how to bisect the area of a triangle with a line parallel to the base, so if we can just construct a segment DE such that it is 2/3 the area of , then we will be able to trisect the area of the triangle with two segments parallel to the base. LetÕs compare the areas of and .
Notice that the heights of the triangle are proportional to the sides of the triangle, so we can substitute into the equation.
Now, we can see that the sides of DE is times the length of BC. We can just construct and then trisect it. To construct , we must create a spiral coming off of BC. Each of the triangles coming off are right triangles and the outer leg is always the same length as BC.
Now, we just have to trisect , and then we can construct a circle centered at C with a radius of . The distance between C and the point of intersection between the circle and BC is . We just need to translate the distance up so that we have with area 2/3 of . To do this, we can just create a parallelogram.
Now, all we have to do is bisect using the methods discussed in Essay 1.
Click this link to use a GSP tool.