# Tricky 3d geometry problem

1,545

## Solution 1

If we place the cube with its main diagonal from $(0,0,0)$ to $(1,1,1)$ and three edges along the axes, then we can parametrize two edges and a diagonal: \begin{align} edge_1&:s\mapsto\begin{pmatrix}s\\0\\0\end{pmatrix},\quad s\in[0,1]\\ edge_2&:s\mapsto\begin{pmatrix}1\\0\\s\end{pmatrix},\quad s\in[0,1]\\ diag&:x\mapsto\frac{1}{\sqrt 3}\begin{pmatrix}x\\x\\x\end{pmatrix},\quad x\in[0,\sqrt 3] \end{align} For a given $s\in[0,1]$ one can minimize the quadratic expression (just pick the vertex) $$|diag(x)-edge_1(s)|^2$$ with respect to $x$ to find that $s=\sqrt 3 x$ and with this the distance $f(x)$ between the point $diag(x)$ on the diagonal and the point $edge_1(s)$ on $edge_1$ is $$f(x)=\sqrt 2 x$$ Similarly, one may deduce that for $$|diag(x)-edge_2(s)|^2$$ to be minimized wrt. $x$ for a fixed $s\in[0,1]$ we must have $s=\sqrt 3 x-1$ and so the distance $g(x)$ between the diagonal and $edge_2$ is $$g(x)=\sqrt{2(x^2-\sqrt 3x+1)}$$ By symmetry, we may conclude that the curve we are rotating is $$h(x)= \begin{cases} \sqrt 2 x&\text{ for }x\leq\tfrac13\sqrt 3\\ -\sqrt 2(x-\sqrt 3)&\text{ for }x\geq \tfrac23\sqrt 3\\ \sqrt{2(x^2-\sqrt 3x+1)}&\text{ in between} \end{cases}$$ defined on the domain $x\in[0,\sqrt 3]$ which is illustrated here:

Remark: Fixing $s$ and varying $x$ fixes a point on an edge and varies a point on the diagonal until the nearest point is found. Doing it the other way around would result in a wrong construction of fixing a point on the diagonal and finding the nearest point on the given edge, which minimizes distance orthogonal to an edge instead of orthogonal to the diagonal/axis of rotation.

To demonstrate how it fits, here is an overlay in a dynamic 3D-model of it:

The red curve is the function $h(x)$ derived above corresponding to the "union" case of the solid formed by the uncountable union of all positions of a full rotation of the cube. The purple lines describe the "intersection" case, the uncountable intersection of all positions in a full rotation of the cube.

## Solution 2

Let's take the unit centered cube, with vertexes at $\pm 1$. To rotate it so that its main diagonal gets aligned with the $x$ axis (vertical axis in the figure) we can use two rotations along two axes, the first by 45 degrees, the second by $\tan^{-1}(\sqrt{1/2})=\sin^{-1}(\sqrt{1/3})$. We get then the rotation matrix:

$$Q= \begin{pmatrix} \sqrt{\frac{1}{3}} & 0 & \sqrt{\frac{2}{3}} \\ 0 & 1 & 0 \\ -\sqrt{\frac{2}{3}} & 0 & \sqrt{\frac{1}{3}} \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \\ \end{pmatrix}= \begin{pmatrix} \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} \\ 0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\ -2\sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ \end{pmatrix}$$

Indeed, we can verify that the matrix is orthogonal and $Q \, (1, 1,1)'=(\sqrt{3} ,0, 0)'$

Lets consider first the upper part. That corresponds to the points spanned by the edges starting on the upper vertex, hence they correspond to the revolution of:

$$\begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}}(\alpha+2) \\ \frac{1}{\sqrt{2}}(\alpha-1)\\ \frac{1}{\sqrt{6}}(\alpha-1) \end{pmatrix}$$

Then $\alpha=\sqrt{3} x-2$, in the range $\alpha \in[-1,1]$, or $x\in[1/\sqrt{3},\sqrt{3}]$.

The rotations along the $x$ axis will keep $r^2=y^2+z^2$ constant, and so

$$r^2= \frac{2}{3}(\alpha-1)^2=\frac{2}{3}(3-\sqrt{3}x)^2$$

Or $$r = \sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right)$$

For the next part, we consider another edge, starting from a neighbour vertex, say from $(1,1,-1)'$:

$$\begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}}\alpha \\ \frac{1}{\sqrt{2}}(\alpha+1)\\ \frac{1}{\sqrt{6}}(\alpha-3) \end{pmatrix}$$ with $\alpha=\sqrt{3} x$, in the range $\alpha \in[-1,1]$, or $x\in[-1/\sqrt{3},1/\sqrt{3}]$.

$$r^2= y^2+z^2=\frac{2}{3}\alpha^2+2=2x^2+2$$

$$r(x)=\begin{cases} \sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right) & \mbox{if } 1/\sqrt{3} \le x \le \sqrt{3}\\ \sqrt{2x^2+2} & \mbox{if } 0 \le x \le \sqrt{1/3} \end{cases}$$

(This seems to agree with String's answer.)

So, yes the middle cross section is an hyperbola.

To compute the total volume you need to integrate : $V = 2 \int_0^\sqrt{3} \pi r(x)^2 dx$ and scale the result by multiplying it by $(L/2)^3$ (because our cube has edge length $2$, instead of $L$)

## Solution 3

Hint:

This rotation-surface is called Katenoid.

It's parametrization is given by

$$x(u,v)=(a \cos (u) \cosh (v),a \sin (u) \cosh (v),a v)$$

To calculate the volume, use the function:

$$f(x)=a \cosh \left(\frac{x-A}{a}\right)+B$$.

Rotate it around the x-Axis.

Then a formulae for volume is given by:

$$V=\pi \int _a^b\ f(x)^2{dx}$$

Please, look at my comment again. Here is another rotation-surface rotation-hyperboloid.

$$x(\text{s},\text{v})\text{=}(\cos (s)-v \sin (s),v \cos (s)+\sin (s),v)$$

This one can be generated by straight-lines:

I don't know, which surface to use.

## Solution 4

The central portions are hyperboloids of 1 sheet capped by two cones at the ends. ( not catenoids). The hyperboloid of one sheet is a ruled surface formed by rotation of a skew line about the rotational axis of symmetry.(Examples are cooling towers, hypoid gears etc).

Equations

Hyperboloid of one single sheet $\dfrac{(x^2 + y^2)}{a^2} -\dfrac{z^2}{c^2}=1$ with one negative sign,

Hyperboloid of two separate sheets$\dfrac{-(x^2 + y^2)}{a^2} +\dfrac{z^2}{c^2}=1$ with two negative signs.

Volume calculation is done usual way of integration using meridian curve equation given above.

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### Vim

Updated on March 18, 2020

• Vim over 2 years

We have a cube with edge length $L$, now rotate it around its major diagonal (a complete turn, that is to say, the angle is 360 degrees), which object are we gonna get?
Astoundingly the answer is D. And here is a demonstration:
Well now I'm required to calculate the volume of this monster. It's far beyond my capability. I don't know how to analytically describe the curve between the two cones (although I believe it is a hyperbola). And I'm still not sure why it should be a curve rather than a segment or something. Could you help me? Thanks in advance.

• Vim over 7 years
This is a problem in my 3d analytical geometry test and really drives me mad. I been thinking on it for a week and still can't even interpret the demo. 😔
• krirkrirk over 7 years
Is the angle of the rotation given ?
• Vim over 7 years
@krirkrirk of course it's a whole rev. (360 deg)
• Vim over 7 years
@krirkrirk. Are my two pictures properly displayed? Perhaps there are some problems with picture display on this page. But on my cellphone it is ok.
• krirkrirk over 7 years
It's perfectly clear, good job :)
• Andrew D. Hwang over 7 years
(+1) If you have reason to edit the question in the future, you might mention explicitly that the solid in question is swept out, i.e., is the union of all positions of the cube during one full turn. As is, the wording could be interpreted to mean the solid is the intersection of all the cube's positions, which would be a convex solid. (This point isn't worth a special edit, however.)
• Vim over 7 years
@user86418 thank u for good advice. Well I guess it'd be even harder in the "intersection" case than this one ;)
• String over 7 years
@Vim: For the intersection case, it becomes surprisingly much simpler, since then the curve is composed by the lines $f(x)=x/\sqrt2$ and $g(x)=-(x-\sqrt3)/\sqrt2$ which if desired can be combined to $$h(x)=(\sqrt6-\sqrt 2|2x-\sqrt3|)/4$$ in which case the solid of revolution is simply composed of two cones.
• robjohn over 7 years
This seems to be a duplicate of this question.
• String over 7 years
@robjohn: Nice answer and animation you have there!
• Vim over 7 years
@String. Sorry but I don't understand how to determine the curve for the intersection case. Seems like it can't be done the same way as the union case is.
• String over 7 years
@Vim: I added that case to my animated diagram in my answer. Have a look and let me know if that helps. Edges describe paths furthest away from a given diagonal of the cube, but a diagonal of a side describes a path closest to a diagonal of the cube.
• Vim over 7 years
Thank you a lot. But could you briefly explain why the surface should be a Catenoid if it isn't too deep? I just don't know how to analytically start.
• Frieder over 7 years
The picture (D) is a good advise for me. The edges of the cube (the longer ones) are generating cones. The smaller ones "vanish" inside. Seem's very regular to me. Do Carmo may help (Differentialgeometry) Chapter 4.2
• Frieder over 7 years
As far as I know, one can generate the Catenoid also with straight-lines in $\mathbb{R}^3$. But for me, it's far in the past. Now geometry is not subject, but I love it.
• Vim over 7 years
Well differential geometry is far ahead of me since I'm still learning elementary math analysis.. So at least one thing I'm sure is that this problem is far too difficult to appear in that test, then it should be my prof's fault rather than mine that I couldn't solve it :). And thank you again for all your time
• Frieder over 7 years
I'm with you! It is too difficult within a test.
• Vim over 7 years
.......but the other answer says it's a Catenoid. Do you think that answer is wrong?
• String over 7 years
@Vim: I will think about it a little more and eventually return. I just plugged the setup into my knowledge of 3D-geometry and this popped out.
• String over 7 years
@Vim: It appears several people agree that this should be the curve, not the catenoid. How do you like my 3D-model :)
• Steven Stadnicki over 7 years
This answer is just wrong - the catenoid is not a ruled surface. The correct answer is that it's a section of a hyperboloid (specifically, a hyperboloid of one sheet).
• abel over 7 years
string, really neat. i just made a paper model. i am cutting the petrie polygon into half. i have a hexagon and equilateral triangle for the top and bottom. i need to figure out how to parametric the intermediate irregular hexagons.
• abel over 7 years
nice. it will take sometime for me to fully understand. will try.
• Frieder over 7 years
@Steven: My first view on picture (D) above made me thought, that could be a catenoid. Now it's very clear. Great work done! Thank you all together.
• Frieder over 7 years
@String: Excellent animation!
• Andrew D. Hwang over 7 years
@String: (+1) It might be easier to calculate the orthogonal distance from an edge to the axis using linear algebra rather than minimizing a quadratic? Anyway, nice solution. :)
• String over 7 years
@user86418: Thank you! Yes, that might be easier - although pojections are also tedious when working with variable coordinates. I have not considered that very carefully, though.
• String over 7 years
@Frieder: I am glad to hear you like it! I enjoyed adding that animation part very much myself. It took some extra work! Animations in GeoGebra has become kind of "my thing", as can be seen several places in my answering history.
• Frieder over 7 years
@String: I like to build animations with mathematica. I didn't know much about GeoBra, just that it exists. And once more: Very nice!
• Vim over 7 years
Brilliant demo! @String
• Vim over 7 years
I'm sorry that it was too late yesterday and I went to bed. I'm just getting up and will take a close look at your post. Thank you so [email protected]
• Vim over 7 years
@String incredible! I believe that's the best answer a freshman can understand.
• Vim over 7 years
1 million thanks!