Hero’s Formula and Mirror Triangles

Here’s a problem with an interesting solution. You’re given two triangles, T1 and T2. The sides of T1 are 25, 25, and 30. The sides of T2 are 25, 25, and 40. Which has the greater area?

The impulsive answer is probably to say that T2 is larger, since the third side is larger. However, the two triangles have the same area. If you haven’t thought about this before (as I hadn’t), it may seem odd that two triangles with two sides the same and the third side different would have the same area. It turns out, though, that for all but one length of the third side, every triangle can be paired with another triangle with the same two sides and a third side of a different length.

That sounds a bit confusing, so let me try rephrasing it in the more rigorous symbols of mathematics: Given two line segments of lengths \(s_1\) and \(s_2\), and given a third line segment of length \(h_1\) that forms a triangle with the first two, there is exactly one line segment of length \(h_2\) that forms a triangle of the same area with the first two, with one exception.

Let’s think about why this is, and what the exception is.

First, imagine a case where \(s_1\) and \(s_2\) are on top of each other, that is, where the angle between them is 0. \(h_1\) has to be as long as the difference between the two, that is \(|s_1 – s_2|\). There’s no actual triangle here, but we can say the area of the shape is 0.

Next, imagine a case where \(s_1\) and \(s_2\) are end to end, that is, where the angle between them is \(\pi\). \(h_1\) has to be as long as the sum of the two, that is, \(s_1 + s_2\). Once again, there’s no actual triangle, but we can say the area of the shape is 0.

Now, imagine a case where \(s_1\) and \(s_2\) form some angle between 0 and \(\pi\). This triangle will have some area. If your two segments are on a hinge moving from 0 to \(\pi\), the resulting triangles will have areas that start at 0, get larger for a while, then get smaller for a while, then settle back to 0. There will be a point where the triangle is as large as it’s going to get: That’s the exception.

Consider Hero’s Formula for calculating the area of a triangle based on its side lengths. One form of it is \(T=\frac{1}{4}\sqrt{(a^2 + b^2 + c^2)^2 – 2(a^4 + b^4 + c^4)}\). For sake of example, take the case where two sides are 1 unit long. This gives us: \[T=\frac{1}{4}\sqrt{(1 + 1 + c^2)^2 – 2(1 + 1 + c^4)}\\ = \frac{1}{4}\sqrt{(2 + c^2)^2 – 2(2 + c^4)} \\ = \frac{1}{4}\sqrt{4 + 4c^2 + c^4 – 4 – 2c^4}  \\ = \frac{1}{4}\sqrt{4c^2 – c^4}  \\ = \frac{c}{4}\sqrt{4 – c^2}\]

We can see right away that the roots are at 0 (\(\frac{0}{4} = 0 \)) and 2 (\(\sqrt{4 – 2^2} = 0\)). Graphing this, we see that the area starts at 0, rises to a peak, then falls back to 0 as we vary the length of the third side. In this case, the maximum is at \(\sqrt{2}\), meaning that the triangle with the largest size with two sides 1 unit long is a right triangle, and the area of this triangle is 0.5.

If we instead use sides of 1 and 2, Hero’s Formula gives us \(T = \frac{1}{4}\sqrt{10c^2 – c^4 – 9}\) as the area, which has a maximum at \(\sqrt{5}\)… again, a right triangle. As we would expect, the roots of \(10c^2 – c^4 – 9\) are at 1 and 3: \[10c^2 – c^4 – 9 = 0 \Rightarrow \\ 10c^2 – c^4 = 9 \\ c^2(10 – c^2) = 9 \\ c = 1 \supset 1(10 -1) = 1(9) = 9 \\ c = 3 \supset 3^2(10 – 3^2) = 9(10 -9) = 9(1) = 9\]

Indeed, it also makes sense that the triangle with greatest area with two given side lengths is a right triangle. The area of a triangle is one-half its width times its altitude. Take one of the known legs to be its constant width; that means as the other leg sweeps around, the altitude changes. The point at which the altitude is greatest is the point at which the two legs are at right angles.

So, where \(h_1 = \sqrt{s_1^2 + s_2^2}\), there will be no other triangle with the same side lengths \(s_1\) and \(s_2\) and the same area. For all other \(h_1\) that form a valid triangle with \(s_1\) and \(s_2\), there will be exactly one \(h_2\) that forms a triangle with \(s_1\) and \(s_2\) of the same area.

Note that this has an interesting ramification with infinity. \(\sqrt{s_1^2 + s_2^2} – |s_1 – s_2|\) is smaller than \(s_1 + s_2 – \sqrt{s_1^2 + s_2^2}\). However, even though the gaps are different, each point in the first interval matches to exactly one point in the second interval. Infinity is weird that way: When we match an infinite number of points in one set to an infinite number of points in another set, we can do it one-to-one even though the two sets don’t seem to be the same “size”. This nuanced understanding of infinity belongs more in the arena of calculus than to that of geometry.

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