16
$\begingroup$

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.

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

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $n\geq 3$. $\endgroup$ – BS. Sep 3 at 6:40
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Robin Houston Sep 2 at 16:28
  • 2
    $\begingroup$ No problem! That just makes it more interesting. $\endgroup$ – Daniil Rudenko Sep 2 at 16:33
  • 3
    $\begingroup$ 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. $\endgroup$ – Daniil Rudenko Sep 2 at 16:53
  • $\begingroup$ 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 ? $\endgroup$ – BS. Sep 2 at 17:30
  • $\begingroup$ I should add that $\tan(\theta)=\sqrt{x^2-2}$, and should have said "this equation takes place". $\endgroup$ – BS. Sep 2 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.