# Which right square pyramids are scissors congruent to a cube?

Consider a right square pyramid whose base has side length $$2r$$ and whose height is $$h$$. Let the dihedral angle between the base and each triangular side be $$\theta$$, and the dihedral angle between adjacent triangular sides be $$\phi$$. We have $$\cos(\theta)=\frac{r}{\sqrt{r^2 + h^2}}$$ and $$\cos(\phi)=\frac{-r^2}{r^2+h^2}$$ so in particular $$\cos^2(\theta)=-\cos(\phi).$$

This square pyramid is scissors congruent to a cube iff its Dehn invariant is zero, for which it is necessary that $$\pi$$ be expressible as a rational linear combination of $$\theta$$ and $$\phi$$.

If $$r=h$$ then this condition is satisfied, and indeed such a square pyramid is scissors congruent to a cube. There is no other nontrivial way to satisfy this condition with both $$\theta$$ and $$\phi$$ being rational multiples of $$\pi$$, as shown by this answer on MSE.

It seems unlikely that there are any other solutions at all, but I have no idea how to even begin to prove such a thing. To ask a precise

Question: Are there any pairs $$\theta$$, $$\phi$$ such that $$\cos^2(\theta)=\cos(\phi)$$ and $$\pi$$ is expressible as a rational linear combination of $$\theta$$ and $$\phi$$, except where $$\cos(\phi)\in\{0,\frac12,1\}$$?

Equivalently, are there any pairs $$\theta$$, $$\phi$$ such that $$\cos^2(\theta)=\cos(\phi)$$, and $$\pi$$ is expressible as a rational linear combination of $$\theta$$ and $$\phi$$, but $$\theta$$ and $$\phi$$ are not rational multiples of $$\pi$$.

Added later: I asked the question above in the expectation that the answer would be no, which would then imply the nonexistence of other right square pyramids that are scissors congruent to a cube. Unfortunately it is quite easy to see that the answer is actually yes, as Daniil Rudenko points out below.

This shows that I asked the wrong precise question. The right question is more complicated, and is described – and answered – by user145307.

• Nitpicking : I think that for interior angles, $\phi>\pi/2$, and accordingly $\cos(\phi)=-r^2/(r^2+h^2)=-\cos^2(\theta)$. – BS. Sep 2 '19 at 9:53
• @BS Fair point! I’ve edited the question accordingly. – Robin Houston Sep 2 '19 at 10:02
• A remark: it is easy to see that a right square pyramid is scissor congruence to a Shlafli orthoscheme. – Daniil Rudenko Sep 3 '19 at 16:07
• I asked a similar question to yours about arbitrary tetrahedron. – Daniil Rudenko Sep 3 '19 at 16:08
• So... you say that "The right question is ... answered by user145307" and then you don't accept the answer. Keeping it classy, MO! – user145307 Sep 4 '19 at 12:49

This is not the question you want to ask. (The actual question asked is easy by a continuity argument.) If the sides of the pyramid have length $$2r$$ and the height is $$h$$, then the other side lengths of the pyramid have length $$\sqrt{2r^2 + h^2}$$. You want to ask whether the element

$$\xi = (2r \otimes \theta) + (\sqrt{h^2 + 2 r^2} \otimes \phi)$$

is trivial in the Dehn group. For this to be true, either:

1. $$\theta$$ and $$\phi$$ are both rational multiples of $$\pi$$ (case already done).
2. $$r$$ is a rational multiple $$0 < v < 1/\sqrt{2}$$ of $$\sqrt{h^2 + 2 r^2}$$, and so

$$\xi = \sqrt{h^2 + 2 r^2} \otimes (2 v \theta + \phi);$$

then one moreover demands $$2 v \theta + \phi$$ is a rational multiple of $$\pi$$. This is a much more stringent requirement. In fact, it never happens, by the following elementary but tedious argument.

By scaling, we can assume that $$h = 1$$. Hence it follows that $$r^2$$ is a rational multiple of $$1 + 2 r^2$$, which certainly implies that $$r^2$$ is rational. So let $$r^2 = t$$. Thus we require that

$$\frac{r}{\sqrt{1 + 2 r^2}} = \sqrt{\frac{t}{1 + 2 t}} = v$$

is rational. It follows that we have $$r^2 = t = \frac{v^2}{1 - 2 v^2}$$ for some rational $$v$$. Thus we can rephrase the problem as follows:

Find all rational $$0 < v < 1/\sqrt{2}$$ with $$\cos(\theta) = \frac{r}{\sqrt{r^2 + 1}} = \frac{v}{\sqrt{1 - v^2}},$$ $$\cos(\phi) = \frac{-r^2}{r^2 + 1} = \frac{-v^2}{1 - v^2},$$ and such that $$(2 v \theta + \phi) \in {\mathbf{Q}} \pi.$$

Let us do an example to explain how one can eliminate any specific $$v$$. Take the case of $$v = 1/2$$. We find that $$\cos(\theta) = 1/\sqrt{3}, \quad \cos(\phi) = -1/3,$$ from which we deduce (for example) that $$\cos(2 v \theta + \phi) = \cos(\theta + \phi) = - \frac{5}{3 \sqrt{3}}.$$ If $$\alpha = 2v \theta + \phi$$ is a multiple of $$\pi$$, then $$\cos(\alpha) = \frac{\zeta + \zeta^{-1}}{2},$$ where $$\zeta = e^{i \alpha}$$ is a root of unity. But knowing $$\cos(\alpha)$$ one can solve for $$\zeta$$ and then determine if it is a root of unity or not from its minimal polynomial. In the example above, we win immediately because $$2 \cos(\alpha)$$ should be an algebraic integer and it is not. The cases $$v = 1/3$$ and $$v = 2/5$$ can be handled in a very similar way: if $$v = 1/3$$, then $$\cos(\theta) = 1/(2 \sqrt{2})$$ and $$\cos(\phi) = -1/8$$, and, with $$\alpha = 3 (2v \theta + \phi)$$, $$\cos(\alpha) = \cos(2 \theta + 3 \phi) = \frac{87}{256},$$ and if $$v = 3/5$$, then $$\cos(\theta) = 3/5$$ and $$\cos(\phi) =9/25$$, and with $$\alpha = 5 (2v \theta + \phi)$$, $$\cos(\alpha) = \cos(6 \theta + 5 \phi) = \frac{3617721}{4194304}.$$ In both cases, the corresponding $$\zeta$$ is manifestly not a root of unity because $$2 \cos(\alpha)$$ is not an algebraic integer.

Returning to the original problem, the first thing we will do is prove that $${\mathbf{Q}}(e^{2 v i \theta})$$ is an abelian extension. We may write $$e^{2 v i \theta} = e^{(2 v \theta + \phi) i} \cdot e^{- \phi i}.$$ Since $$2 v \theta + \phi \in {\mathbf{Q}} \pi$$, the first factor is a root of unity and so lies in an abelian extension. On the other hand, the second term is simply $$\cos(\phi) - i \cdot \sin(\phi).$$ Since $$\cos(\phi) \in {\mathbf{Q}}$$, it follows that $$i \cdot \sin(\phi) = \sqrt{\cos^2(\phi) - 1}$$ lives in an imaginary quadratic extension of $${\mathbf{Q}}$$. In particular, $$e^{- i \phi}$$ clearly lies inside an abelian extension. Taken together, we deduce:

The extension $${\mathbf{Q}}(e^{2 v i \theta})$$ is abelian.

We also have the explicit formulae $$e^{i \theta} = \cos(\theta) + i \sin(\theta) = \frac{v + i \sqrt{1 - 2 v^2}}{\sqrt{1 - v^2}},$$ and then squaring: $$e^{2 i \theta} = \frac{ 3 v^2 - 1 + 2 v \sqrt{2 v^2 - 1}}{1 - v^2}.$$

Let $$v = a/b$$ with $$(a,b) = 1$$, this becomes $$e^{2 i \theta} = \frac{ 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 }}{(b^2 - a^2)}.$$ Let $$E = {\mathbf{Q}}(\sqrt{2 a^2 - b^2 })$$, which is an imaginary quadratic extension of $${\mathbf{Q}}$$. (The condition that $$a/b = v < 1/\sqrt{2}$$ implies that $$b^2 > 2 a^2$$.) Let us write $$x = 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 } \in \mathcal{O}_E.$$ Note that $$N(x) = (b^2 - a^2)^2$$. Secondly, note that $$e^{2 v i \theta} = (e^{2 i a\theta})^{1/b}$$

By Galois theory, $$E(\alpha^{1/b})$$ can only be an abelian extension of $${\mathbf{Q}}$$ (or even of $$E$$) under the following conditions:

1. $$\alpha$$ is a perfect $$b$$th power in $$E$$.
2. $$b$$ is even and $$\alpha$$ is a perfect $$b/2$$th power in $$E$$.

(Added explanation: You can work prime by prime on the divisors $$p$$ of $$b$$. Suppose that $$\alpha$$ is not a perfect $$p$$th power for a prime $$p > 3$$. Then the Galois closure of $$E(\alpha^{1/p})$$ is $$E(\alpha^{1/p},\zeta_p)$$ and contains the automorphism $$\tau: \alpha^{1/p} \rightarrow \zeta_p \alpha^{1/p}$$. But then Galois group of $$E(\zeta_p)$$ over $$E$$ contains an element $$\sigma$$ which fixes $$\alpha$$ and sends $$\zeta_p$$ to $$\zeta^i_p \ne \zeta_p$$. Then $$\tau$$ and $$\sigma$$ do not commute. The same argument works if $$p = 3$$ as long as $$E(\zeta_p) \ne E$$, which can only happen if $$E = \mathbf{Q}(\sqrt{-3})$$. But $$2 a^2 - b^2 \ne - 3 c^2$$ by $$3$$-adic considerations. The same argument works for "$$p = 4$$" as well, as long as $$E(\zeta_4) \ne E$$, which can only happen if $$E = \mathbf{Q}(\sqrt{-1})$$. But $$2 a^2 - b^2 = - c^2$$ can't happen when $$b$$ is even (which is the only relevant case) because then $$a$$ is odd and $$2 a^2 - b^2$$ is $$2 \mod 4$$.)

Moreover, since$$(a,b) = 1$$, if $$e^{2 i a \theta}$$ is a perfect $$b$$ or $$b/2$$th power, then so is $$e^{2 i \theta}$$. In particular, $$e^{4 i \theta}$$ is a $$b$$th power in $$E$$. Note also that $$e^{4 i \theta} = \frac{x^2}{N(x)} = \frac{x^2}{x \overline{x}} = \frac{x}{\overline{x}},$$ so $$x/\overline{x}$$ is a perfect $$b$$th power in $$E$$. Our goal is now to prove that the ideal $$(x)$$ is (almost) a $$b$$th power, and deduce that $$N(x) = (b^2 - a^2)^2$$ is (almost) a $$b$$th power.

Suppose that $$\mathfrak{p}$$ is a prime ideal of $$\mathcal{O}_E$$ which divides both $$x$$ and $$\overline{x}$$. I claim that $$\mathfrak{p}$$ divides $$2$$. Note that $$N(x) = (b^2 - a^2)^2$$, so so $$a^2 \equiv b^2 \mod \mathfrak{p}$$. But then $$x + \overline{x} = 2(3 a^2 - b^2) \equiv 4 a^2 \mod \mathfrak{p}.$$ If $$a \in \mathfrak{p}$$ then also $$b \in \mathfrak{p}$$ contradicting that $$(a,b) = 1$$. Hence $$(x,\overline{x})$$ is only divisible by primes above $$2$$. Since $$x/\overline{x}$$ is a $$b$$th power, and $$(x,\overline{x})$$ is supported at primes above $$2$$, we deduce that $$(x) = I^b \cdot J$$ where $$J$$ is supported at primes above $$2$$. We deduce that

$$N(x) = (b^2 - a^2)^2 = n^b \cdot 2^k.$$

where $$n$$ is an odd integer. Let us first consider the case when $$a$$ and $$b$$ are not both odd. In this case $$k$$ is trivial, and $$(b^2 - a^2)^2 = n^b$$. Since $$b^2 - a^2 = 1$$ has no solutions in positive integers, it follows that $$b^2 - a^2 > 1$$, but then $$b^4 > (b^2 - a^2)^2 \ge 2^b,$$ from which we deduce that $$b < 16$$. Now suppose that $$a$$ and $$b$$ are both odd. Since $$(b-a,a+b) = 2$$ in this case, it must be the case that $$b-a = r^b 2^u, b+a = s^b 2^v,$$ where one of $$u$$ and $$v$$ must be equal to $$1$$.

1. Case 1: $$u = 1$$. If $$b-a=2$$, then from the inequality $$v < 1/\sqrt{2}$$ and $$a < b/\sqrt{2}$$, we deduce that $$2 = b - a > b(1 - 1/\sqrt{2}),$$ and thus $$b < 7$$. If $$b-a=2 \cdot r^b$$ with $$r > 1$$, then $$b \ge b - a \ge 2^{b+1},$$ which is impossible.

2. Case 2: $$v = 1$$. We have $$a+b = 2 \cdot s^b$$, and now $$2b > a + b \ge 2^{b+1},$$ which once again is impossible.

Putting this together, we deduce that $$b < 16$$. Checking all the cases with $$b < 16$$ (noting that $$a/b < 1/\sqrt{2}$$ and $$(a,b) = 1$$) we find that the only possible pairs are $$(a,b) = (1,2), (1,3), (3,5), \ \text{or} \ v =1/2, 1/3, 3/5.$$ But these cases have already been considered previously.

• Impressive answer, a real "proof from The Book"... Welcome to MO ! May I ask details (reference/name) for the Galois theory result that $E(\alpha^{1/b})$ abelian implies $\alpha$ (almost) perfect $b$-th power in $E$ ? This is not (by far) my area of expertise... As a tiny optimisation, note you can obtain $b<8$ instead of $16$ using $n\geq 3$. – BS. Sep 3 '19 at 6:40

I am afraid I am missing something, but let me try nonetheless. Take $$\phi=\theta+r\pi$$ for any rational $$r\in(0,\frac{1}{2}).$$ We are trying to find a solution of the equation $$f(\theta)=\cos^2(\theta)+\cos(\theta+r\pi)=0.$$ Clearly, $$f(0)>0,$$ and $$f(\frac{\pi}{2})<0,$$ so it has a solution.

• Ha! I feel rather embarrassed I didn’t seriously consider the possibility that there are solutions. Sadly I fear this shows only that I asked the wrong precise question, since this is not a sufficient condition for the Dehn invariant to be zero. – Robin Houston Sep 2 '19 at 16:28
• No problem! That just makes it more interesting. – Daniil Rudenko Sep 2 '19 at 16:33
• So, if I understand correctly, the condition "Dehn invariant is zero" does not imply that $\theta,$ $\phi$ are rationally dependent. And indeed, understanding when $$2r \otimes \theta +\sqrt{h^2+2r^2} \otimes \phi$$ vanishes seems like a hard problem. – Daniil Rudenko Sep 2 '19 at 16:53
• Indeed this amounts (if my calculations are right) to find positive integers $m,n$ such that with $x=2n/m$, $x^2>2$ and $a(x)^{2m}=b(x)^{2n}$ for $a(x)=(1-i\sqrt{x^2-2})/\sqrt{x^2-1}$ and $b(x)=(1-ix\sqrt{x^2-2})/(x^2-1)$ (complex numbers of modulus $1$). These equations take place in the imaginary quadratic field $\mathbb{Q}(i\sqrt{x^2-2})$, hence in a cyclotomic one, and one can use automorphisms of the latter to derive more... Or maybe elliptic curves with complex multiplication ? – BS. Sep 2 '19 at 17:30
• I should add that $\tan(\theta)=\sqrt{x^2-2}$, and should have said "this equation takes place". – BS. Sep 2 '19 at 17:55