4.4: Bisecting a Triangle

Pictorial solutions are especially likely for a geometric problem:

Question

What is the shortest path that bisects an equilateral triangle into two regions of equal area?

The possible bisecting paths form an uncountably infinite set. To manage the complexity, try easy cases (Chapter 2) draw a few equilateral triangles and bisect them with easy paths. Patterns, ideas, or even a solution might emerge.

Question

What are a few easy paths?

The simplest bisecting path is a vertical segment that splits the triangle into two right triangles each with base 1/2. This path is the triangle’s altitude, and it has length

\[l = \sqrt{1^{2} - (1/2)^{2} = \frac{\sqrt{3}}{2} ≈ 0.866. \label{4.42} \]

An alternative straight path splits the triangle into a trapezoid and a small triangle.

Question

What is the shape of the smaller triangle, and how long is the path?

The triangle is similar to the original triangle, so it too is equilateral. Furthermore, it has one-half of the area of the original triangle, so its three sides, one of which is the bisecting path, are a factor of \(\sqrt{2}\) smaller than the sides of the original triangle. Thus this path has length \(1/\sqrt{2} ≈ 0.707\) a substantial improvement on the vertical path with length \(\sqrt{3}/2\).

Problem 4.25 All one-segment paths

An equilateral triangle has infinitely many one-segment bisecting paths. A few of them are shown in the figure. Which one-segment path is the shortest?

Now let’s investigate easy two-segment paths. One possible path encloses a diamond and excludes two small triangles. The two small triangles occupy one-half of the entire area. Each small triangle therefore occupies one-fourth of the entire area and has side length 1/2. Because the bisecting path contains two of these sides, it has length 1. This path is, unfortunately, longer than our two one-segment candidates, whose lengths are \(1/\sqrt{2}\) and \(\sqrt{3}/2\).

Therefore, a reasonable conjecture is that the shortest path has the fewest segments. This conjecture deserves to be tested (Problem 4.26).

Multiple problems

Problem 4.26 All two-segment paths

Draw a figure showing the variety of two-segment paths. Find the shortest path, showing that it has length

\[l = 2 \times 3^{1/4} x \sin 15^{◦} ≈ 0.681. \label{4.43} \]

Problem 4.27 Bisecting with closed paths

The bisecting path need not begin or end at an edge of the triangle. Two examples are illustrated here:

Do you expect closed bisecting paths to be longer or shorter than the shortest one-segment path? Give a geometric reason for your conjecture, and check the conjecture by finding the lengths of the two illustrative closed paths.

Question

Does using fewer segments produce shorter paths?

The shortest one-segment path has an approximate length of 0.707; but the shortest two-segment path has an approximate length of 0.681. The length decrease suggests trying extreme paths: paths with an infinite number of segments. In other words, try curved paths. The easiest curved path is probably a circle or a piece of a circle.

Question

What is a likely candidate for the shortest circle or piece of a circle that bisects the triangle?

Whether the path is a circle or piece of a circle, it needs a center. However, putting the center inside the triangle and using a full circle produces a long bisecting path (Problem 4.27). The only other plausible center is a vertex of the triangle, so imagine a bisecting arc centered on one vertex.

Question

How long is this arc?

The arc subtends one-sixth \((60^{◦})\) of the full circle, so its length is \(l = \pi r/3\), where r is radius of the full circle. To find the radius, use the requirement that the arc must bisect the triangle. Therefore, the arc encloses one-half of the triangle’s area. The condition on r is that \(\pi r^{2} = 3\sqrt{3}/4\):

\[\frac{1}{6} \times \underbrace{\text{area of the full circle}}_{\pi r^{2}} = \frac{1}{2} \times \underbrace{\text{area of the triangle.}}_{\sqrt{3}/4} \label{4.44} \]

The radius is therefore \((3\sqrt{3}/4\pi)^{1/2}\); the length of the arc is \(\pi r/3\), which is approximately 0.673. This curved path is shorter than the shortest two-segment path. It might be the shortest possible path.

To test this conjecture, we use symmetry. Because an equilateral triangle is one-sixth of a hexagon, build a hexagon by replicating the bisected equilateral triangle. Here is the hexagon built from the triangle bisected by a horizontal line:

The six bisecting paths form an internal hexagon whose area is one-half of the area of the large hexagon.

Question

What happens when replicating the triangle bisected by the circular arc?

When that triangle is replicated, its six copies make a circle with area equal to one-half of the area of the hexagon. For a fixed area, a circle has the shortest perimeter (the isoperimetric theorem [30] and Problem 4.11); therefore, one-sixth of the circle is the shortest bisecting path.

Multiple problems

Problem 4.28 Replicating the vertical bisection

The triangle bisected by a vertical line, if replicated and only rotated, produces a fragmented enclosed region rather than a convex polygon. How can the triangle be replicated so that the six bisecting paths form a regular polygon?

Problem 4.29 Bisecting the cube

Of all surfaces that bisect a cube into two equal volumes, which surface has the smallest area?