Ernie and the Artificial Emmental

Question

I dropped in on Ernie last week to help him prepare for his annual hosting of the Wine and Cheese Club party. This was a special occasion because the invited guest was the famous oenophile and turophile Jean-Claude la-Grange. Ernie’s party had been planned to the last detail. “The components will be based on the appreciation of Antipodean contrasts”, Ernie had explained, “I am mating a New Zealand Ata Rangi Pinot Noir (2014) with a Swiss Rohmilch Emmentaler”. “Isn’t that wine a little young for tasting?” I asked. But Ernie explained that he had invented a Wine Vintage Accelerator which, by a combination of heat, vibration, and other secret factors, could add 10 years of cellaring to a bottle in minutes – this was to be its first test.

At that moment the door-bell rang. “Answer that can you? That will be the cheese shipment from my affineur“, said Ernie. But disaster was in wait! Instead of a refrigerated box of Swiss cheese I was handed a scrap of crushed packaging stamped “Damaged in Transit”. Ernie was distraught. “There isn’t another cheese of that quality within miles!”, he cried. Ernie looked carefully through the wreckage of the delivery. With a pair of tweezers he extracted the barest crumb of cheese that had somehow remained lodged in the packaging. “Maybe all is not lost”, he muttered. “I can determine the exact composition with my Acme Cheese Analyser and then recreate it using milk from my mechanical dairy cow”. While the analyser was at work Ernie sketched a rapid diagram onto his chalk-board, explaining that he could suitably age the cheese by adapting his Vintage Accelerator.

I rushed out to buy suitable feed-stock while Ernie assembled and augmented the components of his Instant Dairy Factory. When I returned everything was ready to go. I fed in the turnips and hay and minutes later a block of cheese was deposited on a waiting plate. The cheese certainly looked, felt and smelled right and when I took a tentative bite the nutty taste was perfect. But something was not quite right. “Ernie, it doesn’t have any holes! Jean-Claude will not be fooled for a moment”, I cried. Ernie agreed and altered his diagram. “I can use my new Acme 3-D Printer. But there is no time to scan a cheese to make a suitable model – I will have to hack directly into its analogue controls”.

Ernie rapidly assembled a complicated network of capacitors, inductors, relays and electromechanical components to programme the printer. He explained how it would work:

1. The system prints a perfect cube of cheese precisely 100 mm on each side (with interior and/or exterior holes).
2. The thumb-wheel determines the exact number of holes in the block of cheese (all of identical diameter).
3. The bridge resistor determines the size of the holes. Each 1.0 kΩ of resistance will ‘add’ precisely 1% holes by volume (in the range 0-100%) in the block.
4. The cheese is printed with the smallest diameter holes consistent with the required number of holes and percentage volume of void space.

“Where will the holes be positioned in the block?” I asked. “Well, depending on the number of holes and the volume to be removed by them they could overlap each other. They could also be completely internal to the cube of cheese or could overlap its edges, so long as their volume and diameters are consistent with rules 1-4” Ernie replied after a moments thought. Ernie asked me to solder in a 10 kΩ resistor across the bridge so we could get a cheese with exactly 10% void volume. I hastily complied while he made the final attachments between the cheese machine and the 3D printer.

When all was ready, Ernie set a 1 on the thumb-wheel and pressed the go button. The printer clicked and buzzed for a few minutes and spat out a short spiral of printed paper, but when the output hopper swung open it was empty. Ernie scratched his head, and tried again with 2 set on the thumb-wheel – the result was the same, as it was when he tried a third time with a 3 on the thumb-wheel. “Maybe there is a blockage in the print jet” he exclaimed. “While I’m looking for that, can you read aloud what is printed on the paper?” I tore off the sheet and glanced over it. It described the setting data for each of the three cheese-less 3D-print runs we had made. I began reading the first aloud – “Run 1:: Feedstock: cheese. Cubical block: side length = 100.000 mm. Spherical voids: number = 1; diameter = 173.205 mm …”

“Stop right there. The problem is obvious! – some fool...” (Ernie was obviously more than a little stressed) “...has soldered in the wrong valued resistor”. He de-soldered my resistor, threw it into the rubbish bin and soldered in a new one – this time 10 kΩ. “…and I think 7 holes will be perfect” he said as he adjusted the thumb-wheel and hit the go button. A few seconds later a perfect cube of cheese rested in the output hopper, and when I cut it into wedges the beautiful spherical holes were revealed. And not a moment too soon - the door-bell rang, Ernie’s guests arrived and I spent the next hour opening bottles of Pinot and (secretly) printing more cheese to Ernie’s specifications.

At the end of the evening Jean-Claude commented that the wine was ‘perfectly’ matured and its quality was “…only exceeded by the most exquisite cheese I have ever tasted”, so the party was a success. I never did get to work out exactly what value of resistance I had mistakenly used and have no recollection of the diameters of the spherical voids printed out for the 2-hole and 3-hole cheese-less cheeses. Can you help me understand what they were (As usual, accuracy to the nearest Ohm and micron should be enough)?

Answer 1

173.205 mm is $\sqrt3$ times 100 mm, so a 100 mm cube fits exactly into a sphere with diameter 173.205 mm. The resistor must have been 100 kΩ, which explains why nothing was printed!

The rest of the question is to determine the smallest possible diameter of two identical spheres that cover a cube, and the smallest possible diameter of three identical spheres that cover a cube.

If we want to cover a cube with two identical spheres that are smaller than 173.205 mm, there must be two vertices on the same edge that are covered by different spheres. Then the midpoint of the opposite edge is 150 mm away from both, so the spheres must have a diameter of at least 150 mm. This is achieved by cutting the cube in half parallel to one of its faces and circumscribing a sphere around each half.

For three spheres, we can divide the cube into a 100 mm by 100 mm by 12.5 mm block and two 100 mm by 50 mm by 87.5 mm blocks, and surround each with a sphere with a diameter of 141.973 mm ($\frac{\sqrt{129}}{8}*100$ mm). This is optimal:

Suppose that three spheres with a smaller diameter could cover the cube. By the Pigeonhole Principle, one must cover at least three of the eight vertices. This is only possible if covers three vertices of a face. Define a coordinate system so that the three vertices it covers are (0,0,0), (1,0,0), and (0,1,0). The sphere cannot cover (1,0,1/8), (0,1,1/8), (1,1,1/8), or any point on the edges of the opposite face. In addition, it cannot cover (0,0,7/8) because the triangle with vertices (1,0,0) (0,1,0) (0,0,7/8) will not fit in a circle of diameter 141.973 mm.

Whichever sphere covers (0,0,7/8) cannot cover (1,1,1/8) or (1,1,1), so (1,1,1/8) and (1,1,1) must be covered by the same sphere. (1/2,0,1) is 141.973 mm away from (0,1,1/8) and (1,1,1/8), so it cannot be in the same sphere as either of those. Therefore (0,1,1/8) and (1,1,1/8) are in the same sphere. Similarly, (1,0,1/8) and (1,1,1/8) must be in the same sphere. But then (1,0,1/8), (0,1,1/8), and (1,1,1) are all in the same sphere, which is impossible because the triangle with those three vertices does not fit in a circle with diameter 141.973 mm.

Therefore the spheres must have a diameter of at least 141.973 mm.

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In summary: the resistor was 100 kΩ, the diameter for 2 holes was 150.000 mm, and the diameter for 3 holes was 141.973 mm.