94 of 100: Cross Sections

It is possible to create all of the listed shapes by taking a cross-section of the cube except an octagon.

Equilateral triangle: Isosceles Triangle that is not equilateral: Scalene Triangle: Rectangle that is not a square: Square: Pentagon: Hexagon:

However, since 3 points make a plane, we cannot have more than 2 points on each face of the cube, therefore we can only have a maximum of 6 points on the cube (which makes a hexagon). Thus, it is impossible to create an octagon by taking a cross-section of a cube and the answer is $\boxed{7}$ .

The cross-section is a polygon, and each side of that polygon is the result of intersecting a face of the cube. Since there are only six cube faces, the cross-section can have no more than six sides. This rules out the octagon . The $\boxed{\text{seven}}$ other shapes mentioned here are possible:

Let the vertices of the cube have coordinates $(\pm 1, \pm 1, \pm1)$ . Consider the plane $x + y + z = 3 - \epsilon$ , where $\epsilon < 1$ is a positive number. This plane intersects the cube in the points $(1, 1, 1-\epsilon$ ) and permutations thereof; symmetry shows that this is an equilateral triangle .

Now tilt the plane a bit, e.g. $x + y + \tfrac43 z = 3-\epsilon$ . The intersections are $(1,1,\tfrac34-\tfrac34\epsilon)$ , $(1-\epsilon,1,\tfrac34)$ and $(1,1-\epsilon,\tfrac34)$ . This is symmetric under reflection around $x = y$ , so that the triangle is isosceles . It is not hard to see that it is not equilateral.

A further twist gives e.g. $x + \tfrac54 y + \tfrac43 z = 3-\epsilon$ , destroying all symmetry. Now we have a scalene triangle.

Move the plane closer to the center: $x + y + z = \epsilon$ , where $\epsilon < 1$ is positive. Intersections are $(\mp1,\pm1,\epsilon)$ and all permutations thereof, for a total of six intersection points. This is a hexagon .

Now consider the extreme case $x + y + z = 1$ . The intersection is now a large equilateral triangle, whose vertices are all permutations of $(1,1,-1)$ . Now tilt the plane, keeping one of the intersection points in place; e.g. $x + y + \frac65 z = \frac 45$ . The intersection points $(1,1,-1);\ (1,-1,\frac 23);\ (\frac35,-1,1);\ (-1,\frac35,1);\ (-1,1,\frac 23)$ form a pentagon .

The square is easy: take e.g. the plane $x = 0$ , and the intersection has points $(0,\pm 1,\pm1)$ .

For the rectangle , intersect with $x + 5y = 0$ , and get the intersection points $(\mp\frac15,\pm1,\pm 1)$ , where the signs of $x$ and $y$ are opposites.

Note : It is not possible to make any kind of polygon with six sides of less. For instance, polygons with 4, 5, 6 sides will have one, two, or three pairs of parallel sides, respectively.

Proof that you cannot make an octagon:

Assume there is a cross section that creates an octagon. There are 8 edges on an octagon. Spread among 6 faces, at least 1 face will have 2 edges of the octagon (by the Pigeonhole Principle ). However, it is impossible to have a plane intersect a single face in more than one line, by simple 3D geometry. Thus any shape with >6 edges cannot be created.

Showing that up to hexagon can be created is an exercise left to the reader :)

An octagon is not possible because an octagon has 8 sides and every side has to be in a face of the cube, but a cube has only 6 faces.

The other 7 shapes are possible.

• Equilateral Triangle
• Isosceles Triangle that is not equilateral
• Rectangle that is not a square
• Hexagon
• Scalene Triangle
• Square
• Pentagon

All of the possible shapes are possible except for Octagon. Here are the pictures to prove it:

Interestingly, the answer would be 8 in a 4-d cube and more generally 2*n in n-d cube.

I think it is impossible to make octagon

But it is easy to make other shapes