# A cube is placed inside another cube

I known following problem with two square

$$Area(1)+Area(3)=Area(2)+Area(4).$$

My question. Is this problem true for two cubes?

We place a cube $XYZT.X'Y'Z'T$ into another cube $ABCD.A'B'C'D',$ is it true that

\begin{align*} & Vol(ADD'A'.XTT'X')+Vol(BCC'B'.YZY'Z')\\ =&Vol(ABCD.XYZT)+Vol(A'B'C'D'.X'Y'Z'T')\\ =&Vol(CDD'C'.ZTT'Z')+Vol(ABB'A'XYY'X'). \end{align*}

My question. If this is true for two cubes then is this true for two similar rectangular prism?

My question. And are these problems true for two hypercubes and two similar hyperrectangles in $n$-dimesion Euclidean space?

• Why the close votes? The squares result is a neat bit of elementary geometry that I'm guessing is not (yet?) well-known (at any rate it was new to me); it's natural to ask whether it generalizes to higher dimensions, and at worst there's an easy proof of counterexample, in which case the question should be settled by giving this answer rather than closing it without comment. – Noam D. Elkies Jul 5 '18 at 2:13
• Dear Noam D. Elkies, I don't close votes this question, did any body close votes it? – Tran Quang Hung Jul 5 '18 at 2:27
• What precisely do you mean by e.g. $ADD'A'.XTT'X'$? Note that in general e.g. $A,X,X',A'$ are not coplanar. – Robert Israel Jul 5 '18 at 5:58
• Please see my figure, I mean this is volume of convex polytope $ADD'A'.XTT'X'.$ – Tran Quang Hung Jul 5 '18 at 6:00
• If it is true for hypercubes in some dimension, then it is true for similar hyperrectangles. This is just a scaling of the coordinate axes, which scales all volumes by the same factor. – jarauh Jul 5 '18 at 8:34

I believe that the answer is negative. The tidiest way to calculate is as follows. We can identify $\mathbb{R}^3$ with the purely imaginary quaternions $\mathbb{H}_0$, and let $U$ be the cube with vertices $\pm i\pm j\pm k$. Then any other cube $V$ has the form $a+zU\overline{z}$ for some $a\in\mathbb{H}_0$ and $z\in\mathbb{H}\setminus\{0\}$. For any face $F$ of $U$, let $\alpha(F)$ be the volume between $F$ and $a+zF\overline{z}$, and put $\beta(F)=\alpha(F)+\alpha(-F)$. The question is whether $\beta(F)$ is independent of $F$. To calculate $\alpha(F)$, note that we have a bijection $p$ from $[0,1]\times F$ to the relevant region given by $p(t,x)=(1-t)x+t(a+zx\overline{z})$, so we just need to integrate the Jacobian determinant of $p$ over $[0,1]\times F$. To express this determinant in terms of quaternionic algebra, note that there is an isomorphism $\Lambda^3(\mathbb{H}_0)\to\mathbb{R}$ given by $x\wedge y\wedge z\mapsto\text{Re}(xyz)$.

The derivatives of $p$ with respect to $x$ depend only on $t$. Thus, we can bring the integral over $F$ inside the determinant, which has the effect of replacing $\partial p/\partial t$ by $4$ times its value at the center $c\in F$, which is $4(a+zc\overline{z}-c)$. The derivatives of $p$ with respect to components of $x$ are of the form $m(z)+c(z)t$, so the determinant is just a quadratic in $t$. If I have everything straight, we just end up with $$\beta(F) = \frac{8}{3}(1-|z|^2)((1+|z|^2)^2 - \text{Re}(z)^2 - 3\langle z,c\rangle^2)$$ (If we call the above $\phi(z,c)$, it is not hard to check that $$\phi(z,i)+\phi(z,j)+\phi(z,k)=8(1-|z|^6) = \text{vol}(U) - \text{vol}(V),$$ as it should be.)

If $z$ is real then this is independent of $c$, so the volume formula is correct when the inner cube is parallel to the outer one. The same holds if $\langle z,c\rangle^2$ is independent of $c\in\{\pm i,\pm j,\pm k\}$; this is the case when the inner cube is obtained by rotating the outer one around one of its long axes, then shrinking and translating it. But in general the formula does not hold.

• Thank you very much Prof. Neil Strickland for your answer. I have another quesion. I hope you help me. Is it true that, sum of Volume of two tetrahedrons $$Vol(BCYZ)+Vol(A'D'X'T')=const$$ when the small cube moves inside the large. In the cases $YZ\parallel BC$ and $X'T'\parallel A'D'$ then, is it true that? $$Area(BCYZ)+Area(A'D'X'T')=const.$$ – Tran Quang Hung Jul 5 '18 at 10:46
• I don't think that the tetrahedron relation is true, and unlike the original question, I don't think that there is an interesting story behind it; the formulae are just different in an unstructured way. – Neil Strickland Jul 5 '18 at 12:54

Its also easy to give a concrete counter-example: Take the cube with vertices $$\left(\frac{273}{340},\,\frac{79}{68},\,\frac{13}{20}\right) , \left(\frac{407}{340},\,\frac{57}{68},\,\frac{27}{20}\right) , \left(\frac{239}{340},\,\frac{789}{884},\,\frac{337}{260}\right) , \left(\frac{249}{340},\,\frac{621}{884},\,\frac{217}{260}\right) , \left(\frac{263}{340},\,\frac{1195}{884},\,\frac{289}{260}\right) , \left(\frac{417}{340},\,\frac{573}{884},\,\frac{231}{260}\right) , \left(\frac{431}{340},\,\frac{1147}{884},\,\frac{303}{260}\right) , \left(\frac{441}{340},\,\frac{979}{884},\,\frac{183}{260}\right)$$ inside a $0-2$-cube. Then the three volumes you ask for are $\frac{24421}{8840}\neq \frac{24563}{8840}\neq \frac{24491}{8840}$. Here is a picture of the configuration:
The outer cube has volume 8 and the inner cube has volume $1/8$. As a sanity check, lets see if everything adds up: $$\frac{24421}{8840} + \frac{24563}{8840} + \frac{24491}{8840} = \frac{14695}{1768}$$ and $$\frac{14695}{1768} + \frac{1}{8} -8 =\frac{193}{442}.$$
Since the edges of the outer cube are not coplanar with the edges, we have now double counted all the 12 tetrahedra, that arise from the fact that these pairs of edges are non-coplanar. Their volumes are $$\frac{471}{8840} , \frac{11}{170} , \frac{59}{2210} , \frac{29}{680} , \frac{49}{1768} , \frac{29}{8840} , \frac{29}{680} , \frac{29}{8840} , \frac{59}{2210} , \frac{49}{1768} , \frac{11}{170} , \frac{471}{8840}$$ and sure enough the sum of these numbers is $\frac{193}{442}$. Here is a picture of all those tetrahedra: