When passing Ernie's letterbox this morning, I found a courier bag about the size of a shoe-box, resting inside (along with a smaller unlabelled bag). I immediately guessed that it was the special delivery from Acme Industries that Ernie had been muttering about for days - as the contents were listed as Acme Industries 1x $Un_2Cl_{14}(H_2O)_7$ single crystal (6 $dm^3$ cuboidal block, irrational edge lengths). Unobtanium septa-chlorohydrate, according to Ernie, was the most amazing material known to science. "It has a remarkable crystal structure such that the six faces of a perfect crystal are 'optically asymmetric'". And sure enough, when Ernie slit the bag open and deposited the contents on the breakfast table, it was in the form of a rectangular brick with top, bottom, front, back, left, and right faces respectively coloured silver, bronze, aqua, lavender, gold, and vermillion. But all was not well - Ernie let out a somewhat rude (for him) exclamation and moaned that "the dimensions are all wrong, it's a $1:2:3$ edge-length ratio cuboidal block... and I was promised a $\sqrt{2}:\sqrt{3}:\sqrt{6}$ edge-length ratio block!!!", and simultaneously I noticed that on the packaging there were a number of crudely applied stickers covering some of the letters. I peeled them off to expose the lettering underneath: Acne Imdustries 1x $Un_2F_{14}(H_2O)_7$ single crystal (6 $dm^3$ cuboidal block, rational edge lengths). Once again, Ernie had mail-ordered a bogus product from Acne Imdustries.

But after spending a bit of time with a set of Vernier calipers, followed by some browsing on a number of online chemical databases, he announced that things were not quite as bad as he had thought. The fluoride (after a quick check with his pocket spectrometer) had exactly the same unit cell dimensions as the chloride, and was also optically asymmetric - perfect for his experiment. The only problem was that the block, whilst the correct volume, had the wrong dimensions. "...and I need those exact dimensions", he explained, "because of reasons".

Fortunately, he thought there was a solution at hand. Ernie had ordered some 'universal crystal glue' at the same time and the smaller bag contained a tiny bottle of exactly that - the labelling confirmed that it was the genuine Acme product. All that was required, Ernie explained, was to cut the $1:2:3$ block up into a number of pieces, and re-assemble it (with the aid of the glue) into a $\sqrt{2}:\sqrt{3}:\sqrt{6}$ cuboid, and then he could run his experiment. "So why don't you do that", he said to me "while I draft a note to the Mail-order Fraud Investigation Authority". enter image description here

Ernie told me I would need to use the particle-beam cutter (as most Unobtanium compounds were very hard) but, "not to worry", he had realised that Louisette was a bit of a hazard, so had made a few changes in how the guillotine now worked. So while drafting his letter of complaint he called out a number of instructions:

  1. Ernie had improved the beam focus and mounted the cutter on an an x-y table with a beam-proof base, so now it could make vanishingly narrow, vertically oriented, straight cuts through any material. You could program the coordinates of where each cut started and finished (in the x-y plane), accurate to the nearest micron.
  2. It was advisable to cut through only one piece of single crystal material at a time. The beam could cut through any thickness of material, but if you stacked layers there was a risk of 'beam refraction' at the interface. (I asked what that meant, and Ernie replied that the technical details were a bit complicated, but the implication was that the cutting beam could be deflected at right angles. As the table was approximately at waist height I decided to follow that instruction carefully).
  3. If the glue was painted carefully over two appropriately oriented surfaces and the surfaces were pressed together, the glue would bond the two pieces and a catalytic process would convert them almost instantaneously back into a single crystal - which could now be safely cut according to rules 1 and 2 if required. (When I asked what 'appropriately oriented' meant, he suggested that so long as any two pieces being joined retained their original orientations - i.e. were not rotated about any axis relative to each other - it shouldn't be a problem. But best not to get it wrong or the $Un_2F_{14}(H_2O)_7$ might spontaneously decompose into a superheated cloud of unobtanium-oxide and hydrofluoric acid. I decided to follow that instruction carefully too).
  4. As the glue was fairly expensive (it cost precisely \$10.00 for enough glue to join two 1 $dm^2$ surfaces), and Louisette used a lot of electricity (it cost precisely \$10.00 for the electricity to make a single cut 1 dm long - regardless of the thickness of the cut). Ernie would appreciate it if I could make the cutting/joining process as economical as possible.

To be honest, I wasn't even sure if it was possible to cut a rational-sided cuboid into an irrational-sided one, and I was so concerned by the fear of disembowelment and/or deflagration that I didn't take any notes of how many cuts I made, or joints I glued, or what order I did them in. But finally I ended up with a single brick of material that looked about the right shape. I handed it to Ernie who, after a few measurements confirmed that he was "fairly pleased" as the block was a single crystal cuboid with the desired $\sqrt{2}$ dm x $\sqrt{3}$ dm x $\sqrt{6}$ dm edge lengths. But he couldn't quite understand why I had chosen only the second-most efficient solution (in terms of cost) rather than the most efficient.

Now I am feeling guilty that I have cost Ernie an unnecessary expense and I am considering buying him something in compensation for the financial loss caused by my clumsy cutting process. Can anybody help me work out how much I owe him?

Postscript 1: I still haven't worked out how much I owe Ernie for my error. It's affecting my sleep and I have started having recurring dreams of the conversation I had with him after I "re-formatted" the crystal. In last night's dream I remembered a little more of the conversation and am sure that Ernie definitely stated "It's curious that both your solution and my better one, consisted two cuts, followed by gluing the all the pieces back into one piece, followed by two more cuts, followed by gluing those pieces back into the final shape"

Postscript 2: After eating a large plate of cheese last night before retiring to sleep, I had an even more vivid dream of my conversation with Ernie. I'm not sure it will help, but now I recollect that Ernie also said "It is nice that the orientations of the crystal lattice planes, relative to the surfaces of the irrational cuboid you have cut, remain the same as the orientations of the crystal lattice planes relative to the surfaces of the original rational cuboid. That results from you keeping as much of the original silver, bronze, aqua, lavender, gold, and vermillion surfaces of the rational cuboid at the surface of your irrational cuboid."

  • $\begingroup$ +1, and hoping that the puzzle is as high quality as the art :) $\endgroup$
    – Avi
    Oct 9, 2021 at 4:39
  • $\begingroup$ Are you allowed to cut the crystal so that one or more of the cut faces are not parallel to the table (i.e. perpendicular to the cut)? $\endgroup$ Nov 1, 2021 at 4:47
  • $\begingroup$ @2012rcampion The cutting beam is mounted so it aims vertically downwards (in the -z direction), and moves horizontally in teh xy plane when making a cut. So cuts are inherently perpendicular to Louisette's base-plate. If I recollect correctly, when I asked Ernie (while carrying out the original task) if I could tilt a block, so as to make a cut that wasn't perpendicular to the bottom face of the block, he replied in his usual blunt manner that "I suppose you could, but why ever would you want to?". ... so I didn't. $\endgroup$
    – Penguino
    Nov 1, 2021 at 21:52
  • $\begingroup$ Guess I must have mistakenly misremembered Ernie's comment (see above). I will leave it there for posterity (historical infamy or whatever). But to confirm the correct answer to 2010rcampion's question - yes you can tilt the crystal if you believe it might help $\endgroup$
    – Penguino
    Nov 11, 2021 at 22:36

2 Answers 2


I have no proof that this solution is optimal, but it is better than the previous ones.

Step 1:

  • Rearrange the $2\times 3$ face into a $\sqrt{3}\times 2\sqrt{3}$ face.
  • Make the long diagonal cut from the side for a length of $1$.
  • Make the short cut from the top for a length of $2\sqrt{3}-3\approx 0.464$.
  • The glued surfaces have a total area of $2\sqrt{3}\times 1+(2-\sqrt{3})\times 1=2+\sqrt{3}\approx 3.732$.

Step 2:

  • Rearrange the $1\times 2\sqrt{3}$ face into a $\sqrt{2}\times\sqrt{6}$ face.
  • Again, make the long diagonal cut from the side for a length of $\sqrt{3}\approx 1.732$.
  • Make the short cut from the top for a length of $\sqrt{2}-1\approx 0.414$.
  • The glued surfaces have a total area of $\sqrt{7}\times \sqrt{3}+(2\sqrt{3}-\sqrt{6})\times \sqrt{3}=6-3\sqrt{2}+\sqrt{21}\approx 6.340$.

The total cost is:

$$\\\$10\times\left(1+(2\sqrt{3}-3)+\sqrt{3}+(\sqrt{2}-1)\right) + \\\$10\times\left((2+\sqrt{3})+(6-3\sqrt{2}+\sqrt{21})\right) \\ =\\\$10\times(5-2\sqrt{2}+4\sqrt{3}+\sqrt{21})\approx \\\$136.83$$

We can decrease the efficiency very slightly by:

Making the first small cut from the side instead of the top,

which increases the cost by:


As a bonus, here is the second most efficient dissection.

Step 1:

Step 2:

Total cost:

$\\\$142.33$, an increase of $\\\$5.50$ from the optimal.

  • $\begingroup$ For some reason the equations involving dollar signs are being split over two lines when rendered, but not in the preview. Anyone seen that before? $\endgroup$ Nov 3, 2021 at 2:31
  • $\begingroup$ @2012rchampion, what a beautifully rendered solution! From my frustrated experience, dollar signs misbehave on multiply spoilered lines. They just do. $\endgroup$
    – rotta
    Nov 3, 2021 at 5:35
  • $\begingroup$ Love the sketches. Will see Ernie this weekend and try and get to the bottom of this. $\endgroup$
    – Penguino
    Nov 4, 2021 at 1:14
  • $\begingroup$ I suspect this is still not the correct answer, as both solutions require tilting the blocks before cutting. $\endgroup$ Nov 5, 2021 at 3:09
  • $\begingroup$ That is much better than mine! and, yes, $ sgins are a mess for me, too. $\endgroup$
    – loopy walt
    Nov 5, 2021 at 7:35


With the new hint one mght suspect what Ernie has in mind are essentially two

2D steps like
enter image description here

This procedure transforms

a rectangle with sides a,b and diagonal d into a rectangle with sides a',b' and diagonal d'

and is possible as long as

(1) ab = a'b' and
(2) $a,b \le d'$ and $a',b' \le d$.

I can, however, only find one way of applying this strategy. So I seem still to be missing something.

Anyway, let's go through all possible permutations.

No matter which order we settle for we will always have to glue all the original faces (cost in units of 10\$:$1\times 2 + 1\times 3 + 2\times 3 = 11$) and cut all the new edges (cost:$\sqrt 2 + \sqrt 3 + \sqrt 6 \approx 5.6$.

We'll leave out this shared cost from all calculations.

An additional, order specific cost comes from the fact that

each 2D step creates 2 new edges for a total of 4 where we only need 3. This translates to cutting 1 excess edge and glueing 1 excess pair of faces.

Case 1: start with 2x3 face

Case 1a: creating $\sqrt 6$ edge

Other edge will also be $\sqrt 6$. Therefore second step would have to be $\sqrt 6 \times 1 \rightarrow \sqrt 2 \times \sqrt 3$ This is

impossible because of (2) as the new diagonal $d'=\sqrt 5$ is too short

Case 1b: creating $\sqrt 2$ edge This is

impossible because the new side $\sqrt 18$ is longer than the original diagonal $\sqrt 13$

Case 1c: creating $\sqrt 3$ edge This is

impossible, because the second step would have to be $\sqrt 12 \times 1 \rightarrow \sqrt 2 \times \sqrt 6$ with too short new diagonal $\sqrt 8$.

Case 2: Start with 1x3 face

Case 2a: creating $\sqrt 6$ side: The other new side would be $\sqrt {3/2}$ and the diagonal $\sqrt{15/2}$ which is too short.

This also rules out

Cases 2b and c: because they produce even shorter diagonals.

This leaves as only viable starting point

Case 3: start with side 2x1

Case 3a: producing new side $\sqrt 6$. Can be immediately dismissed because original diagonal is only $\sqrt 5$

Case 3b: producing two new sides $\sqrt 2$. Excess cost: Cutting: $\sqrt 2$, Glue: $3\times \sqrt 2$, Total: $4 \times \sqrt 2 \approx 5.657

Case 3c: producing new sides $\sqrt 3$ and $2/\sqrt 3$. This would require as second step $2\sqrt 3 \times 3 \rightarrow \sqrt 2 \times \sqrt 6$ and one last time we observe that the new diagonal $\sqrt 8$ is too short.


Here is what I believe to be the cheapest. No proof, though, and only a vague idea what the second best might be.

(Almost) without words:

enter image description here
Fig 1: 1x2x3
enter image description here
Fig 2: first cut, creates $\sqrt 2$ (teal shade)
enter image description here
Fig 3: rearrange bits
enter image description here
Fig 4: glue and rotate 45°
enter image description here
Fig 5: second cut, creates $\sqrt 3$ (blue) and $\sqrt 6$ (red)
enter image description here
Fig 6: rearrange and glue

Why do I suspect this is optimal?

Because glued area is precisely half the surface of the original box (1x2+1x3+2x3)

Why am I not 100% sure?

Because there are ways doing it with fewer cuts

Total cost:

$(11 + \sqrt 2 + 2\sqrt 3 + 2 \sqrt 6) \times 10\\\$ \approx 207.77\\\$ $

What could be the second best?

Perhaps similar to the above but with a $\sqrt 2 \times \sqrt 2 \times 3$ cube intermediate. That doubles the cuts needed in step 1 ($+ \sqrt 2 \times 10\\\$ $) but halves those in step 2 ($-(\sqrt 3 +\sqrt 6) \times 10\\\$ $). What makes it more expensive is that we also have to glue an additional two $\sqrt 2 \times 3$ surfaces together.

The difference would then be a mere

$(4\times \sqrt 2-(\sqrt 3 + \sqrt 6))\times 10\\\$ \approx 14.75\\\$ $

  • $\begingroup$ I can se where your $\sqrt 2$ length cut comes from, but I am having difficulties working which lines represent the $\sqrt 3$ and $\sqrt 6$ cuts. in your diagrams (probably my fault, as my eyes are not as good as they used to be - perhaps colour coding would help) But regardless of that, I have a vague recollection that the total glued area in my second-best solution was less than in your suggested best one. $\endgroup$
    – Penguino
    Oct 11, 2021 at 2:29
  • $\begingroup$ @Penguino. Now that you mention it I realise that I made a silly accounting mistake. The glued area is not equal to the area of surface glued (as I calculated it) but half of that;.it is fixed now. Funnily enough, I already had done it right in the difference calculation. Re the sqrts of 3 and 6 they can best be seen in the top right panels of figs 5 and 6. The top right panel is the view from the top in both you can see the sqrt3xsqrt6 face. It is rotated relative to the standard axes. Its sides are hypothenues of right triangles with short sides (sqrt2,1) and (sqrt2,2) $\endgroup$
    – loopy walt
    Oct 11, 2021 at 4:58
  • $\begingroup$ @Penguino I redid the figures, Not sure the colours help making it clearer, though. Let me know what you think. $\endgroup$
    – loopy walt
    Oct 11, 2021 at 7:10

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