# What's wrong in this effort to describe equilateral triangles algebraically?

Today I was working on a spreadsheet to generate equilateral triangles. I perceived the question in terms of finding values for a, b, f and g in the following diagram so that the red dots are equidistant.

Each distance is a hypotenuse and so describable with the Pythagorean Theorem. (The given values are the squares of the hypotenuse, but I think we can use them instead of the roots.)

From there it seems a system of three equations. But something funny happens when I try to simplify.

It seems to me that tautologies arise for two reasons: all numbers are solutions, or we substituted an equation into itself. Neither seems to be the case here. So what happened? Can you suggest a comparable (but correct) way to generate the vertices of equilateral triangles in Excel?

• @DougM what about $a=g=1$ and $b = f = \frac{-1+\sqrt3}{2}$, as just one example of a nontrivial solution? – Misha Lavrov Mar 23 '17 at 19:47
• @DougM Sure, but what suggests to you that the OP is interested in integer or rational solutions? – Alex Kruckman Mar 23 '17 at 19:50
• I must be misreading the OP's intent and will remove my previous comments. – Doug M Mar 23 '17 at 19:51

## 1 Answer

You did substitute an equation into itself, in a way. Although you have written down three equations, one of them is redundant: if $H_1, H_2, H_3$ are the lengths of the three hypotenuses, and we know $H_1 = H_2$ and $H_2 = H_3$, then $H_1 = H_3$ follows.

You've simplified two of the equations and substituted the results into the third, which yields a tautology.

The reason we only have two equations is that the system has two degrees of freedom.

• The first is scaling: if $(a,b,f,g)$ is a solution, then $(ax, bx, fx, gx)$ is also a solution.
• The first is rotation: for example, if we fix vertex $3$, we can choose the angle between the $x$-axis and side $13$ of the triangle arbitrarily. (At the very least, it can vary between $0^\circ$ and $30^\circ$; past that, we'd have to draw a different diagram.)

So it makes sense to solve for two of the variables in terms of the other two. The nicest choice turns out to be to solve for $a$ and $g$ in terms of $b$ and $f$ (or vice versa), giving us (by means of equation-solving software) $$a = b + f \sqrt3, \qquad g = b \sqrt3 + f.$$

This is easier to obtain by a different argument. The vector $1\to 2$, with coordinates $(-b, -f-g)$, is a $60^\circ$ counter-clockwise rotation of the vector $1\to 3$, with coordinates $(-a-b, -f)$. So we may write down the matrix equation $$\begin{bmatrix}-b \\ -f-g\end{bmatrix} = \begin{bmatrix}\cos 60^\circ &-\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ\end{bmatrix} \begin{bmatrix}-a-b \\ -f\end{bmatrix}$$ which gives us two linear equations to solve: $$\begin{cases} b = \frac12(a+b) - \frac{\sqrt3}{2}f \\ f+g = \frac{\sqrt3}{2}(a+b) + \frac12f\end{cases}$$ and solving for $a$ and $g$ yields the same solution.

• Can you suggest an efficient way to find lots of sets of values that work? – Chaim Mar 23 '17 at 21:49
• @Chaim I've edited the solution with a suggestion. – Misha Lavrov Mar 23 '17 at 22:16
• That certainly answers the question I asked; but I'm curious why the solution comes from equation-solving software. Is it possible to solve by hand? And by the way, @DougM deleted his comment before I saw it; did he explain that integers and rational solutions are impossible? – Chaim Mar 24 '17 at 12:15
• It's probably possible to solve by hand, but it requires either creativity or patience and I turned out not to have enough of either. Integer and rational solutions are impossible; here's a proof. – Misha Lavrov Mar 24 '17 at 14:28
• (And I guess this solution is also a proof, since if $b, f \in \mathbb Q$, then $b + f \sqrt 3 \notin \mathbb Q$.) – Misha Lavrov Mar 24 '17 at 14:35