»  Solutions to puzzles in my VDARE.com monthly Diary

  October 2017


My October diary included the following brainteaser.

I have a circular paper disk. I draw two radii, the lesser angle between them being α (the greater angle is of course 360 degrees minus α).

Then I cut along these radii, giving me two pieces of paper, each of which can be rolled up, radius joining radius, to make a circular cone.

What value of α should I use to maximize the total volume of these two cones?


• Solution

The friend who sent me this problem assured me that it "can be solved using standard high school real functions that appear on every scientific calculator."

Oh yeah? I don't know about that … but judge for yourself. It's certainly a very cute problem, though. (Defined to mean: It'll waste hours of your time.)

One's first guess is that the maximum total volume is attained when α = 180°, i.e. when you just cut the disk in half.

Surprisingly, not only does α = 180° not give the maximum volume, it actually gives a local minimum! That is, if you vary α very slightly away from 180°, the total volume increases.

OK, let's get formal. Wolog we can assume that the original disk has unit radius. And since plainly arc lengths of a circle are going to be involved, let's switch from degrees to radians. A radian is 57.29578 degrees; 180° is π radians. These will simplify things considerably.

For any value of α we shall get two cones. If they are not identical (α ≠ π), one is taller and skinnier, the other more short and squat.

The circular bases of the cones have circumferences that are the arc lengths of the original circular disk cut off by the two radii we drew. These arc lengths are α and 2π − α. It follows that the radii of the two cone bases are α/2π and 1 − α/2π.

Change of variable: Set r = α/2π. Then the cone-bases have radii r, 1 − r.

The volume of a cone is one-third the area of its base times its vertical height. What are the vertical heights of our two cones?

Well, the slant heights are both 1, the radius of the original disk; so by Pythagoras' Theorem the vertical heights are  √(1 − r²) and  √(1 − (1 − r)²).

Now we can write down an expression for the total volume of the two cones.

V = (π/3)[r² √(1 − r²) + (1 − r)² √(1 − (1 − r)²)]

Well, that's straightforward enough. Time for another change of variable.

If you squint at that expression for the total volume you'll see that it is symmetrical about r = ½. That is, if you substitute ½ + δ for r, you get the same expression you'd have gotten by substituting ½ − δ. (Well, not strictly typographically the same; but the value will be the same.)

So we can improve the symmetry of that expression for V by switching to the variable u = ½ − r. Then, after cleaning up some fractions, our expression for V looks like this:

V = (π/24)[(1 − 2u)² √(3 + 4u − 4u²) + (1 + 2u)² √(3 − 4u − 4u²)]

You can see the symmetry more clearly there. If you substitute −u for u, the value of the expression is unchanged.

We just need to find a maximum for that, then track back through our changes of variable to get α. Piece of cake.

Total volume

Before proceeding, though, to guide our further inquiries, let's graph that expression.

I'll let α range over all possible values from 0 to 2π; that is, dropping the assumption that α is the lesser angle. There's no harm in this, it just captures the full symmetry.

So 0 ≤ α ≤ 2π. That means −½ ≤ u ≤½.

Here at right is a graph. As promised, there is a local minimum at u = 0, i.e. α = π. (Although it's hard to make out on the scale as shown. Changing the u-scale to squinch the graph centrally makes the local minimum more plain.)

For a fast check at this point we can compare the value of V on the graph when u = 0 (that is, when the original disk was cut precisely in half) with the total volume of the two equal cones we then get. From our original expression for V this is easily shown to be one-twelfth of π√3, which is a tad less than 0.45345. Sure enough.

That's a local minimum, though. What is the maximum? To what value of α does it correspond?

Total volume

Subdividing the coordinates on GraphSketch.com (sorry, my subscription to Mathematica has expired) until the resolution starts to break up, I get the maximum around u = ±0.17598615, the maximum of V then being about 0.456640590999588, which is 0.7 percent more than the volume at the u = 0 local minimum. Hey.

(The u-axis in this graph at the left goes from 0.1759861 to 0.1759862; the V-axis from 0.45664059099958 to 0.45664059099959. I couldn't get GraphSketch.com to show gridlines at these scales.)

Working back through my changes of coordinates, the positive value of u corresponds to r = 0.3240138, for which α = 2.035839 radians, or 116.644968 degrees. If you want to get Babylonian about it, that's 116° 38′ 41.95″. (The negative value of u corresponds to 360 degrees minus that angle.)

All that is of course merely graphical. Can we get a closed-form solution by algebra and calcuus?

We-ell. We can differentiate that expression for V in terms of u and find the zeros of the derivative. You have to square and multiply some polynomials, ending up (when you have eliminated the u = 0 solution) with a cubic equation in u². Since cubic equations with rational coefficients admit of closed-form solutions in terms of rational numbers and roots, we should be home and dry …

 … Except that you have to extract cube roots of complex numbers. And that's assuming you got all the polynomial manipulations correct. This is what we math geeks call "breaking rocks."

My rock-breaking skills are much decayed. I had two tries at this, working the polynomials by hand. I got two different results, neither of them corresponding to the graphical solution.

A friend who is handier with pick and hammer than I am came to the rescue. The cubic can, he says, be solved exactly: "But as with many exact cubic root solutions, it's ugly." Over to him.

Let this mess of the cube root of a complex number be A. The cubic complement is the cube root of the same complex number, but with a negative sign on the imaginary part as well. So B = (((-9i*sqrt(191)-1121)/4)^(1/3).

Let C = A+B, which is a real number. It is equivalent to A + (43/A), or if a and b are the real and imaginary parts of A, then C = 2a.

Then let y = (C - 5) / 9 = 0.2190289

Then the answer in radians is pi*(1 - sqrt(1 - 4y)) = 2.035839 = 116.645 degrees.

Thanks, pal.

[Added later: The friend who sent me this puzzle reminds me of the trigonometric solution to a cubic with three real roots.

Remember that a cubic may have one, two, or three real roots. Examples: (x − 1)(x² + 2), (x − 1)(x − 2)², (x − 1)(x − 2)(x − 3).

For the one-solution case you only need square and cube roots. For the three-solution case (of which the two-solution case is just a special instance) you can get the solutions using a formula involving cosines and inverse cosines. It's all explained here.

Which, he says, validates the claim that the parent problem here "can be solved using standard high school real functions that appear on every scientific calculator." Yes it does. Thanks again!]