# Chaitanya's Random Pages

## February 27, 2015

### Cross sections of a cube

Filed under: mathematics — ckrao @ 9:59 pm
When a plane intersects a cube there is a variety of shapes of the resulting cross section.
• a single point (a vertex of the cube)
• a line segment (an edge of the cube)
• a triangle (if three adjacent faces of the cube are intersected)
• a parallelogram (if two pairs of opposite faces are intersected – this includes a rhombus or rectangle)
• a trapezium (if two pairs of
• a pentagon (if the plane meets all but one face of the cube)
• a hexagon (if the plane meets all faces of the cube)

The last five of these (the non-degenerate cases) are illustrated below and at http://cococubed.asu.edu/images/raybox/five_shapes_800.png . Some are demonstrated in the video below too.

One can use  to experiment interactively with cross sections given points on the edges or faces, while  shows how to complete the cross section geometrically if one is given three points on the edges.

Let us be systematic in determining properties of the cross sections above. Firstly, if the plane is parallel to an edge (any of four parallel edges), the cross section can be seen to be a line or rectangle with the longer dimension of length at most $\sqrt{2}$ times the other. That rectangle becomes a square if the plane is parallel to a face.

If the plane is not parallel to a face, we may set up a coordinate system where a unit cube is placed in the first octant aligned with the coordinate axes and the normal to the plane has positive x, y and z coordinates. In other words, we may assume the plane has equation $ax + by + cz = 1$ and intercepts at $(1/a,0,0), (0,1/b,0)$ and $(0,0,1/c)$, where $a, b, c$ are positive. The cross section satisfies $ax + by + cz = 1$ and the inequalities $0 \leq x \leq 1$, $0 \leq y \leq 1$ and $0 \leq z \leq 1$. This can be considered the intersection of the two regions $\displaystyle ax + by + cz = 1, 0 \leq x, 0 \leq y, 0 \leq z,$ $\displaystyle ax + by + cz = 1, x \leq 1, y \leq 1, z \leq 1,$

each of which is an acute-angled triangle in the same plane (acute because one can show that the sum of the squares of any two sides is strictly greater than the square of the third side). Note that the triangles have parallel corresponding sides, being bounded by the pairs of parallel faces of the cube $x = 0, x=1, y = 0, y= 1, z=0, z=1$. Hence the two triangles are oppositely similar with a centre of similarity.

The following diagram shows the coordinates of the vertices of the two triangles, which in this case intersect in a hexagon. The centre of similarity of the two triangles is the intersection of two lines joining corresponding sides – this can be found to be the point $(1/(a+b+c), 1/(a+b+c), 1/(a+b+c))$, which is the intersection of the unit cube’s diagonal from the origin (to $(1,1,1)$) and the plane $ax + by + cz = 1$.

Side lengths of the triangles and distances between corresponding parallel sides may be found by Pythagoras’ theorem and are shown below for one pair of corresponding sides (the remaining lengths can be found by cyclically permuting $a,b,c$). To sum up, all of the possible cross sections of a cube where the plane is not parallel to an edge can be described by the intersection of two oppositely similar triangles with corresponding sides parallel.

The type of polygon obtained depends on which vertices of the figure below are selected, as determined by the values of $a,b,c$. In this figure a vertex for the cross-sectional polygon is chosen if the constraint associated with it is satisfied. A red vertex has a conflicting constraint with its neighbouring two blue vertices, so either a red point or one or more blue points in this area can be chosen. Note that for the plane to intersect the cube at all we require $(1,1,1)$ to be on the different side of the plane from the origin, or in other words, $a + b + c \geq 1$.

Let us look at a few examples. Firstly, if $a, b, c$ are all greater than 1 we choose the following triangle. Similarly if $a+b, b+c, c+a$ all are less than 1, the oppositely similar triangle on the red vertices would be chosen.

Next, if $c > 1, a < 1, a+b < 1$ we obtain the following parallelogram. If $c > 1, a < 1, a+b > 1$ we obtain either a pentagon (parallelogram truncated at a vertex) or a trapezium depending on whether $b < 1$ or $b \geq 1$ respectively. $b < 1$:  $b\geq 1$: Finally, if $a, b, c$ are less than 1 and $a+b, b+c, c+a$ are greater than 1, we obtain a hexagon. For details on calculating the areas of such polygons refer to , especially the method applying the area cosine principle that relates an area of a figure to its projection. For calculating volumes related to regions obtained by the cross section refer to .

#### References

 Cross Sections of a Cube: http://www.wou.edu/~burtonl/flash/sandbox.html

 Episode 16 – Cross sections of a cube: http://sectioneurosens.free.fr/docs/premiere/s02e16s.pdf

## 1 Comment »

1. […] this recent post, the following figure was formed by considering two triangular regions in three-dimensional space, […]

Pingback by A triangle centre arising from central projections | Chaitanya's Random Pages — March 28, 2015 @ 10:23 pm

Create a free website or blog at WordPress.com.