# Ernie and the Equi-area Tetrahedra

I enjoy spending time with Ernie when he is 'pottering' in his workshop. It is a large open-plan area filled with soldering-stations and oscilloscopes, lathes and milling machines, band-saws and drill-presses, vacuum pumps and x-ray spectrometers... If you can name a tool or piece of equipment, Ernie most probably has one (always of the top quality) and it is almost always a pleasure to use them. But there is a small annex at the back of the workshop - which I prefer not to visit. It is crammed with experimental 'alpha' prototypes of equipment considered to be "not quite ready for commercial production". They are leant to Ernie by a range of manufacturers for testing purposes. He benefits from early access to exotic technologies, and they benefit from shake-down tests, stress tests, and comments Ernie provides in the form of annotations to their provisional 'Application Notes'. Some of the prototypes can be more than a little disconcerting - tools like the Plasma-discharge Paint Stripper, the Magneto-hydrodynamic Glue-gun, and the 4D Printer (even Ernie hasn't got his head around that one yet). But worst of all in my opinion, is Acne Imdustry's Heavy Hadron Beam Auto Guillotine (Ernie has nicknamed it Louisette for some reason or other) - that thing absolutely terrifies me!

• In each of them, the areas of the four individual faces were in the exact ratio 1:2:3:4.
• Each of them had a surface area totaling exactly 100 cm^2.
• The first one had (very very nearly) the largest surface area for any 1:2:3:4 tetrahedron that could possibly have been cut from the first Unobtanium sphere.
• The second one had the largest volume for any 1:2:3:4 tetrahedron that could possibly have been cut from the second Unobtanium sphere.

And then he told me that from the information he had just provided (and if I wished to indulge in a little "back of the envelope maths"), it would be possible to calculate the exact diameter of each of the spheres. I thought he was pulling my leg in that last statement, but do you think he is correct? If so, I would love to surprise him by telling him exactly how big each of the spheres was. Can anybody help me with that?

• Nice, Ernie is back! =D Sep 25, 2021 at 4:01
• I may be misunderstanding the requirements, but it seems that the first bullet and the last bullet contradict each other because (rot13) gur ynetrfg grgenurqeba gung pna or phg sebz n fcurer vf n erthyne bar. Sep 29, 2021 at 23:24
• @DqwertyC I can see the potential confusion. I have updated the bullet points to confirm the requirement for 1:2:3:4 surface area ratios in the 'largest' bullets. Sep 29, 2021 at 23:36

We can solve this problem by optimizing the surface area or volume of a tetrahedron in a unit sphere and then scaling the solution so the surface area matches the given value.

Let the vectors $$\vec x_i,\ i\in\{1\ldots 4\}$$ be the four vertices of the tetrahedron. Our first constraint is that the vertices must lie in the (unit) sphere; this can be written as simply:

$$\left\Vert \vec x_i\right\Vert^2\le 1$$

The next constraint is the ratios of the areas of the faces. The areas can be calculated by:

$$A_i = \frac{1}{2}\left\Vert(\vec x_{i+1}-\vec x_i)\times(\vec x_{i+2}-\vec x_i)\right\Vert$$

(Note the indices wrap around, e.g. $$\vec x_5=\vec x_1$$.) This is a complex nonlinear constraint but we can convert it to two quadratic constraints by defining a set of auxiliary vectors $$\vec c_i,\ i\in\{1\ldots 4\}$$:

$$\vec c_i=(\vec x_{i+1}-\vec x_i)\times(\vec x_{i+2}-\vec x_i) \\ (2A_i)^2 = \left\Vert\vec c_i\right\Vert^2$$

The defining equation for $$\vec c_i$$ is quadratic since, when the cross product is expanded, no terms have more than two $$x_i$$ multiplied together.

The area ratio constraint is then just:

$$A_i = i A_1$$

Note that although we have chosen a particular permutation of the faces, due to the symmetries of the tetrahedron any other face permutation is equivalent by permuting the vertices.

The volume can be calculated by:

$$V = \frac{1}{6}\left|(\vec x_2-\vec x_1)\times(\vec x_3-\vec x_1)\cdot(\vec x_4-\vec x_1)\right|$$

We can rewrite this to use $$\vec c_i$$, turning it in to a quadratic constraint:

$$6V = \vec c_1\cdot(\vec x_4-\vec x_1)$$

Note that the absolute value has disappeared; this is OK since we can convert a negative-signed-volume tetrahedron to a positive one by swapping to vertices.

Writing these constraints in ZIMPL looks like this:

# helper functions
defnumb mod4(i) := ((i - 1) mod 4) + 1;
defnumb mod3(i) := ((i - 1) mod 3) + 1;

# variables
var x[{1..4}*{1..3}] >= -1 <= 1;
var a[{1..4}] >= 0;
var c[{1..4}*{1..3}] >= -infinity <= infinity;
var v >= 0;

# objective; pick one
# maximize total_area: sum <i> in {1..4}: a[i];
maximize volume: v;

# constraints
subto in_sphere: forall <i> in {1..4}: sum <k> in {1..3}: x[i,k]^2 <= 1;
subto cross_prod: forall <i,k> in {1..4}*{1..3}:
c[i,k] == (x[mod4(i+1),mod3(k+1)]-x[i,mod3(k+1)])*(x[mod4(i+2),mod3(k+2)]-x[i,mod3(k+2)]) -
(x[mod4(i+1),mod3(k+2)]-x[i,mod3(k+2)])*(x[mod4(i+2),mod3(k+1)]-x[i,mod3(k+1)]);
subto area: forall <i> in {1..4}: (2*a[i])^2 == sum <k> in {1..3}: c[i,k]^2;
subto area_ratio: forall <i> in {2..4}: a[i] == i * a[1];
subto volume: 6*v == sum <k> in {1..3}: (x[4,k]-x[1,k])*c[1,k];
# constrain orientation to force a unique solution
# the first vertex must lie along the positive x-axis
# the second vertex must lie in the upper x-y plane
subto fix_vertex: x[1,1] >= 0 and x[1,2] == 0 and x[1,3] == 0 and x[2,2] >= 0 and x[2,3] == 0;


I solved this this using SCIP.

The solution for maximal surface area is:

solution status: optimal solution found
objective value:                     3.24759692195701
x#1#1                                               1   (obj:0)
x#2#1                              -0.499648917450054   (obj:0)
x#2#2                               0.866228124582392   (obj:0)
x#3#1                               0.250330301953486   (obj:0)
x#3#2                               0.866138389075211   (obj:0)
x#3#3                            -0.000750267933309599  (obj:0)
x#4#1                              -0.500270727125412   (obj:0)
x#4#2                              -0.865869295355317   (obj:0)
x#4#3                            -0.000453442238582569  (obj:0)
a#1                                 0.324759692195701   (obj:1)
a#2                                 0.649519384391403   (obj:1)
a#3                                 0.974279076587105   (obj:1)
a#4                                  1.29903876878281   (obj:1)
c#1#1                            -0.000648946627302448  (obj:0)
c#1#2                            -0.00112514724035245   (obj:0)
c#1#3                              -0.649518255693092   (obj:0)
c#2#1                            -0.00129950154812839   (obj:0)
c#2#2                            0.000340548156145226   (obj:0)
c#2#3                               -1.29903796871455   (obj:0)
c#3#1                            -0.00104238022931135   (obj:0)
c#3#2                             0.00078613317424887   (obj:0)
c#3#3                                1.94855751247677   (obj:0)
c#4#1                            -0.000391889620475885  (obj:0)
c#4#2                            -0.000680001769731886  (obj:0)
c#4#3                                2.59807741529706   (obj:0)
v                                0.000373725035811096   (obj:0)


Note that $$V$$ is very small, and the z-coordinates of all the points are zero or nearly zero. The tetrahedron is degenerate (flat), and looks something like this:

We can compute the radius at which the tetrahedron would have an area of 100 cm2 as $$r=\sqrt{100/A}$$, which is:

5.5491 cm

The solution for maximal volume is:

solution status: optimal solution found
objective value:                    0.168616181170499
x#1#1                                               1   (obj:0)
x#2#1                               0.103368738387049   (obj:0)
x#2#2                               0.994643103806537   (obj:0)
x#3#1                               0.446415095883521   (obj:0)
x#3#2                               0.678786929600625   (obj:0)
x#3#3                              -0.433510583267715   (obj:0)
x#4#1                              -0.734126599748252   (obj:0)
x#4#2                               -0.67901272007481   (obj:0)
x#4#3                            -0.000515264692628134  (obj:0)
a#1                                 0.291707753384547   (obj:0)
a#2                                 0.583415506769094   (obj:0)
a#3                                 0.875123260153641   (obj:0)
a#4                                  1.16683101353819   (obj:0)
c#1#1                              -0.431188370044785   (obj:0)
c#1#2                              -0.388698218976949   (obj:0)
c#1#3                             -0.0580016270153901   (obj:0)
c#2#1                              -0.725384940303185   (obj:0)
c#2#2                                0.36324047987551   (obj:0)
c#2#3                                -0.8386700819649   (obj:0)
c#3#1                              -0.294709100335685   (obj:0)
c#3#2                               0.751477471806906   (obj:0)
c#3#3                                1.55299391553877   (obj:0)
c#4#1                            -0.000512554591328825  (obj:0)
c#4#2                            -0.000462485871227904  (obj:0)
c#4#3                                2.33366155968202   (obj:0)
v                                   0.168616181170499   (obj:1)


The tetrahedron looks something like:

The sphere has a radius of:

5.8550 cm

And the volume of the (scaled) tetrahedron is

33.844 cm3

• It would be nice to see your rationale since the ‘aha’ moment is the most interesting part of puzzles. Sep 30, 2021 at 4:13
• @Lawrence Added. Unfortunately the "aha" moment looks like SCIP Status : problem is solved [optimal solution found], which is probably not as interesting as you hoped. Sep 30, 2021 at 16:05
• For the greatest area solution, loopy walt's (hidden/deleted) answer demonstrated the 'aha' moment and provided a nice analytic solution. For the greatest volume solution, I am still waiting on Ernie to let me know if there is an 'aha'moment. Oct 3, 2021 at 22:25

Partial result. I can't get the volume bit done.

Let us first solve the 2D equivalent: What is the largest triangle (by perimeter or area) that fits in a given circle? A plausible guess would be the inscribed regular triangle

and this guess can be confirmed using the concavity of the sine between 0° and 180° and Jensen's inequality.

We can use this to greedily solve the 3D maximise-the-surface-area problem. The largest possible face of any tetrahedron fitting inside a given sphere is

a regular triangle inscribed on a great circle.

The largest total surface area under the 1:2:3:4 constraint is therefore 10/4 times this triangle's surface area if we can demonstrate that we can extend the triangle to an 1:2:3:4 tetrahedron inside the sphere. For that we need to add a 4th (complanar) vertex such that the diagonals intersect at ratios 1:4 and 2:3 and verify it lies inside the great circle.

In units of the sphere's radius the triangle's edge length is $$\sqrt 3$$. For ease of reference let us assign $$x,y,z$$-coordinates to the vertices: $$(-1,0,0);(1/2¸-\sqrt 3 / 2,0);(1/2,\sqrt 3 / 2,0)$$ As the centre intersects the height in the ratio 2:1, the 4th vertex would have to be at $$(7/8,\pm\sqrt 3/8,0)$$ This point is comfortably within the circle. As the area of the inscribed regular triangle is $$3 \sqrt 3/4$$ the sphere's radius is

$$\sqrt \frac{40 \mathrm{cm}^2}{3 \sqrt 3/4} \approx 5.549055267 \mathrm{cm}$$

which, reassuringly, coincides with @2012rcampion's numerical result.

Technically, we would also want to show that this result is unique but I can't find a good way of doing that.

Official part ends here, below is an idea that may be useful in attacking the volume-maximisation. I can't finish it but leave it here for reference, just in case someone smarter wants to pick up from here.

If we cut the tetrahedron along the 4 edges that separate the 1:4 and 2:3 faces and straigthen the two pairs of faces out we obtain two quadtrilaterals with the same area (because 1+4 = 2+3) and the same edge lengths. It follows, then, from Bretschneider's formula the product of the lengths of diagonals must also be the same, and from that it follows that the angle between the diagonals must also be the same.