Let’s call the length of one edge of the cube x (x > 0). So the volume of the full cube is x^3 and the area of one full face is x^2.

Intuitively I know that in order to truncate the corner of a cube and create an equilateral triangle, you must remove the same distance along each of the 3 edges. As a proof, I guess I could say, lets call the distances removed along each edge a, b, and c (a,b,c > 0). So the 3 edges of the new equilateral triangle are:

Sqrt(a^2 + b^2)

Sqrt(a^2 + c^2)

Sqrt(b^2 + c^2)

Since this is an equilateral triangle,

Sqrt(a^2 + b^2) = Sqrt(a^2 + c^2)

a^2 + b^2 = a^2 + c^2

b^2 = c^2

since a,b,c > 0

b = c

Similarly we can also find that a = b or a = c, leading us to a = b = c.

Moving on, let’s find out what that distance is. Truncating a corner removes an area of (a^2)/2 from the area of the face (x^2). So (a^2)/2 = 1/8 * (x^2)

4 * a^2 = x^2

2 * a = x

a = x/2

This makes sense, because you can picture a square being made up of 8 right triangles (divide the square in half horizontally, then vertically, and then cut each of those 4 quadrants in half diagonally).

Now we have to calculate the volume of the removed corner as a function of a. The formula for the volume of a pyramid is 1/3*base*height. Well, since all the faces of the pyramid are triangles, I can call the "base" one of the right triangle sides, making the height = a. So…

V = 1/3*(1/2*a^2)*a = 1/6*a^3 = 1/6*(1/2*x)^3 = 1/6*1/8*x^3

Since we remove 8 corners, the total volume removed from the original cube is 8*V = 8*1/6*1/8*x^3 = 1/6*x^3. So the volume of the new solid is x^3 – 1/6*x^3 = 5/6*x^3.

So the new volume would be 5/6 the original volume.

Later!