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Cube standing on a corner


This question has arisen from this post and the picture and the insights have been taken from the answers mentioned there.


$\quad\qquad\qquad\qquad\qquad\qquad\qquad$Cone and the cube

In the above picture, a cube is standing on one of its corners. Let the cone lie on a $xz \text{ plane}$ and assume $y$-axis (the red line) passes through the centre of the cone and through one of the body diagonals of the cube. Now for the cube, imagine a plane passing though the points $QS$ and parallel to the $xz \text{ plane}$.

What I thought:

I thought, after looking at a dice for some time, that the $3$ vertices $QST$ would indeed form a triangle but the $3^{rd}$ vertex will be at a different height (not in the same plane), like the triangle $QST'$ below. So when a $xz \text{ plane}$ will cut the cube, what we'll get will be some polygon which with my level of understanding and imagination wasn't possible to fathom then.

Front view $\qquad$ Top view
Image on the Left: Front View; $\qquad\qquad\qquad$Image on the Right: Top View

What it turned out to be:

It turned out that all the $3$ vertices, $QST$ indeed lie on single $xz \text{ plane}$, like $QST$ above, and thus form a triangle. Similarly it's true for the $3$ vertices above these (the vertices $U,V,W$). And add to that all the $4$ $xz \text{ planes}$ upon which all these vertices lie are at intervals of equal height!

I want to know that:

  1. How can we know whether a triangle forms and not some other polygon for the vertices $Q,S,T$ on the $xz \text{ plane}$ parallel to cone's base and passing through $1$ or more of the $3$ vertices (which may be an unfamiliar figure, maybe a pentagon or some quadrilateral, as below)?
    xz plane capture
  2. Whether all the 3 vertices of the cube will lie on a plane parallel to $xz \text{ plane}$ and not on 2 planes parallel to $xz \text{ plane}$
  3. How to determine the height of these $xz \text{ planes}$ from one another?
  4. How will one calculate the centre and the radius for the obtained polygon in point 1?
    [Note: Point 4 has a bit to do with the linked post (in case you don't get what I'm saying)]
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    $\begingroup$ 1. For the 3 vertices $Q, S, T$? Maybe I am misunderstanding, but three vertices (that are not collinear) would form a $3$-gon, not some other polygons. $\endgroup$
    – peterwhy
    Aug 10, 2022 at 21:51
  • $\begingroup$ @peterwhy thanks a lot! made the correction in my question. Its xz plane instead of xy. $\endgroup$ Aug 11, 2022 at 4:25
  • $\begingroup$ @peterwhy 3 vertices will form a triangle as in front view (and top view) but when the xz plane cuts the cube, we'll have something else on the xy plane if all 3 vertices are not on the same height. $\endgroup$ Aug 11, 2022 at 4:27
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    $\begingroup$ Kindly don't "close vote" for "needs more focus" as these seemingly multiple questions are actually a single question itself, regarding cube standing on a corner, but so that the answerer may know the confusion I'm having, I have broken it down. Example: one may solve the question with addressing 1. or 2. or 3. or 4.. $\endgroup$ Aug 11, 2022 at 12:01
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    $\begingroup$ @hardmath great observation! Thank you! Made the edit. Luckily saw a way to correct the question and so that peterwhy don't need to change anything in their answer. $\endgroup$ Aug 12, 2022 at 18:25

1 Answer 1

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To find lengths, one may work with another cube rotated to a more convenient orientation.

Consider a cube with side length $a$, with two neighbouring vertices at $O(0,0,0)$ and $E(a, 0, 0)$, and a body diagonal from $O$ to $D(a,a,a)$.

Question 3

In your diagram, you are looking for the height of point $Q$ (or $S$ or $T$) along the $y$-direction, relative to point $P$ in your diagram.

In my rotated cube, this would be to find the projected length of $OE$ along the $OD$ direction. One may perform this projection by dot product:

  • The dot product of $\overrightarrow{OD}$ and $\overrightarrow{OE}$ is

    $$\overrightarrow{OD}\cdot \overrightarrow{OE} = (a,a,a)\cdot (a,0,0) = a^2$$

  • Dividing this by the length of the body diagonal $OD$, and calling the result $s$:

    $$\begin{align*} \overrightarrow{OD}\cdot \overrightarrow{OE} &= \left\|\overrightarrow{OD}\right\| \left\|\overrightarrow{OE}\right\| \cos\theta = \left\|\overrightarrow{OD}\right\|s\\ s &= \frac{\overrightarrow{OD}\cdot \overrightarrow{OE}} {\left\|\overrightarrow{OD}\right\|} = \frac{a^2}{\sqrt{3a^2}} = \frac{a}{\sqrt 3} \end{align*}$$

    where $\theta$ is the angle between $\overrightarrow{OD}$ and $\overrightarrow{OE}$, as in the linked Wikipedia page.

$s$ is the projected length of $OE$ along the $OD$ direction, which in your orientation is the height of $Q$ along the $y$-direction relative to $P$. Repeat the same process for other vertices of my cube, e.g. $(0,a,0)$, $(a,a,0)$, etc.

Question 4

From Q3 above we obtained $s$, the projected length of $\overrightarrow{OE}$ along the $\overrightarrow{OD}$ direction. If $r$ is the perpendicular distance from $E$ to line $OD$, then

$$\begin{align*} OE^2 &= s^2+r^2\\ r^2 &= OE^2 - s^2\\ &= a^2- \left(\frac{a}{\sqrt3}\right)^2\\ r&= a\sqrt{\frac23} \end{align*}$$

In your orientation, $r = a\sqrt{\frac23}$ would be the radial component of $Q$ away from the $y$-axis (or from $P$).

(If we find the exact "projected" vector with signed length $s$ and direction $\overrightarrow{OD}$, this will be the vector projection of $\overrightarrow{OE}$ onto $\overrightarrow{OD}$. Then subtract this vector from $\overrightarrow{OE}$, this will return a vector with length $r$ and perpendicular from $OD$. This is also the first steps of the Gram-Schmidt process.)

Question 1

From the same calculation as in Q3 above, points $Q, S,T$ all have the same $y$-coordinates, i.e. the same "height" from $P$. So they lie on the same plane parallel to the $xz$-plane.

The face diagonals $QS$, $ST$ and $TQ$ are all on this plane, and all have length $a\sqrt 2$. So $\triangle QST$ is an equilateral triangle on this plane.

Question 2

The cube has 3-fold rotational symmetry centred at a body diagonal.

For cross sections of a cube on a plane perpendicular to the body diagonal, the cross section shape maintains the same symmetry and may be either

  • an equilateral triangle, or
  • a convex hexagon.
  • (or just a single point at the two ends $P$ and $V$)

One interactive demo of cube cross sections from Google search is https://www.geogebra.org/m/X6GYjh2U (not by me).

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    $\begingroup$ Wow! Amazing! +1. Too clear and understandable explanation. Thanks a lot! Also, the links were a great help. $\endgroup$ Aug 11, 2022 at 15:57

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