Inscribing an Equilateral Triangle

I was reminded of the cylindrical wedge that casts shadows of a triangle, a square, and a circle, and it got me wondering: What if I wanted to create such a shape with an equilateral triangle as one of its shadows? The wedge shown either casts an isosceles (but not equilateral) triangular shadow, or it casts a rectangular (but not square) shadow. To cast an equilateral triangle shadow, we would need to start with a cylinder with a square shadow (i.e., one where its diameter is equal to its height) and cut an equilateral triangle out of it.

equilateral-corner

This turns out to be fairly straightforward, although more difficult to cut than the easy wedge. Start at one corner (\(V_1\)) and connect to points on the opposite sides. What are these points?

Using a unit square, the legs of the two congruent right triangles have lengths of \(1\) and \(a\). The isosceles right triangle has legs of \(b\). Since these are right triangles, and since \(V_1V_2V_3\) is equilateral, we can solve for \(a\) using the Pythagorean Theorem: \[a^2 + 1^2 = V_1V_3^2 = b^2 + b^2 \\ a^2 + 1 = 2b^2\]

Furthermore, since \(b = 1 – a\): \[a^2 + 1 = 2(1 – a)^2 = 2 – 4a + 2a^2 \\ a^2 – 4a + 1 = 0\]

Using the Quadratic Formula: \[a = \frac{4 \pm \sqrt{(-4)^2 – 4}}{2} = 2 \pm \sqrt{3} \]

Since \(a \le 1\), \(a = 2 – \sqrt{3}\) and \(b = \sqrt{3} – 1\).

This is the only solution that would retain the entire square, but I extended my thinking: Are there any other ways to make an equilateral triangle?

equilateral-side

My next thought was to make one side of the triangle parallel to one side of the square. The math here, too, is fairly straightforward. We know that \(s = 1\), so too that \(V_1V_2 = 1\). We can see that \(V_3\) is the midpoint of \(\overline{DC}\), because the other two sides of these right triangles are congruent. So \(DV_3 = \frac{1}{2}\). If \(DV_1 = a\), then \(a^2 + 0.5^2 = 1^2\) and hence \(a^2 = \frac{3}{4}\) and \(a = \frac{\sqrt{3}}{2}\).

I could see that there are an infinite number of triangle, simply by dragging the vertex \(V_1\) between the two versions. But I wanted a general formula.

After a bunch of math following a bunch of red herrings, I finally created a model in GeoGebra and explored. At this point, it occurred to me that the three vertices each have a range of possible values. Create a variable, \(k\), that is in the range (0,1). The zero state is in the first diagram; the 0.5 state is in the second; the 1 state is the vertical reflection of the 0 state.

We can see, for instance, that \(V_3\) starts at \(2 – \sqrt{3}\), goes through \(0.5\), and ends at \(\sqrt{3} – 1\). My conjecture was that, if we map this range onto (0, 1), as well as the corresponding ranges for the other two vertices, the result would always form an equilateral triangle.

I was right!

\(V_1\) is in the range \((0, 2 – \sqrt{3})\), so this is easy to map: Taking A as the origin and \(V_1 = (0, a)\), then \(a = k(2 – \sqrt{3})\).

\(V_3\) is in the same range, but is mirrored. If \(V_2 = (1, b)\) then \(b = (1-k)(2 – \sqrt{3})\).

To map \(V_2\), we need to first find the width of the range, then find the left end. To map the range, we’ll take the width of the range and add the left end. The width is \(\sqrt{3} – 1 – 2 + \sqrt{3} = 2\sqrt{3} – 3\); the left end is \(2 – \sqrt{3}\). So if \(V_3 = (c, 1)\), then \(c = k(2\sqrt{3} – 3) + 2 – \sqrt{3} = 2 – 3k + (2k – 1)\sqrt{3}\).

In summary, with k in the range (0, 1), we can make a set of equilateral triangles using the points \(\{(0, (2 – \sqrt{3})k), (1, 2 – \sqrt{3} -(2 – \sqrt{3})k), (2 – 3k + (2k – 1)\sqrt{3}, 1)\}\)

rotation-equilateralEach of these triangles has four rotations, so for fun I strung them all together in an animation.

Each of the triangles in this animation is equilateral (at least within the tolerance of GeoGebra’s calculations).

I’ll also note: While the original diagram shows how to cut an equilateral triangle such that the desired shadows would be cast, the object would not sit flat. It’s certainly possible to do this using computer graphics, but a real world object would tip over unless we used wires or magnets or some other trick.

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