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Two Pyramid Puzzle Solutions

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Edited by Richard Walker, Friday 23 January 2026 at 21:55

I asked here what ratio between the height of a pyramid and the length of its sides would give the greatest volume for a given surface area. Below I will give two solutions, the first I think quite insightful and intuitive, and doesn't need much maths at all, but you have to take a particular fact on trust. The second is more fully worked out but does need some maths.

As well as the side length and height I've added the slant height to the sketch, because we shall use it later, and because it is often referred to in discussion about the proportions of pyramids. The Great Pyramid of Giza originally stood about 147 m tall and its side length at the base was about 230 m. If we work out the slant height and compute its ratio to half the base length it comes to about 1.623, which is sometimes said to be strikingly close to the Golden Ration phi almost equals 1.618 .

And now: the solution to the puzzle. The height should be <drum roll>

one divided by Square root of two almost equals 0.707

times the side length. It's worth noting that this makes the slant height to half side length ratio Square root of three times s almost equals 1.732 times s , so Pharaoh's new pyramid will be a bit pointier than that at Giza with its ratio of 1.623.

Explanation 1 (intuitive)

Imagine the pyramid is just the top half of a solid whose other half is an inverted copy of the pyramid. What would it look like? Well, it would have eight faces and it's not hard to see it would be an octahedron.

If you like you an imagine the bottom half being buried in the sand!

Our goal is to maximise the volume for a given surface and the octahedron that does that is the regular octahedron. It's also not too difficult to workout that if the octahedrons sides have a length of s its height will be the diameter of a square of side length s , which is Square root of two times s and the height of the top half one half of that, which is one divided by Square root of two times s .

This is a very pleasing explanation which feels intuitively correct  but of course I have just claimed that the regular octagon is optimal without proof. However I'm not going to attempt a proof here, because I think making it watertight would be quite lengthy. If I can find or come up with a nice simple proof I'll write about it a separate post.

Explanation 2 (mathy)

Let the combined area of the four faces be cap a , the slant height script l , the vertical height h , as shown. Let the volume of the pyramid be cap v .

Then firstly, using the formula for the area of a triangle and Pythagoras gives cap a equals four full stop times one divided by two times s times script l and script l squared equals left parenthesis s divided by two right parenthesis squared plus h squared respectively.

To avoid square roots later we square the first equation so we have cap a squared equals four times s squared times script l squared . By substituting for script l squared and making h squared the subject we obtain

h squared equals cap a squared minus s super four divided by four times s squared

Next the formula for the volume of a pyramid tells us that cap v equals one divided by three times s squared times h and it is convenien to square this also, giving cap v squared equals one divided by nine times s super four times h squared . Substituting the expression for h squared found earlier we obtain

equation sequence part 1 cap v squared equals part 2 one divided by nine times s super four times cap a squared minus s super four divided by four times s squared equals part 3 one divided by 36 times left parenthesis cap a squared times s squared minus s super six right parenthesis

We want to maximise cap v rather the cap v squared but they will both be maximised at the same value of s . So we differentiate this expression and find the non-zero root. The derivative is

one divided by 36 times left parenthesis two times cap a squared times s minus six times s super five right parenthesis

which is zero when s equals Square root of cap a divided by Square root of three . Plugging this into our expression for h squared and tidying up we get h equals Square root of cap a divided by two times Square root of three .

Finally we get h divided by s equals one divided by Square root of two and that is the ratio of the optimal height to the side length.

PS This has bee a great project to practice my LaTeX!

Picture credit for octahedron: Wikimedia https://commons.wikimedia.org/wiki/File:Octahedron-MKL4.png
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