# Ernie and the Alchemist's Gift

For Ernie, the task of choosing Christmas presents for his friends is always a struggle. I think that is because, while he finds it easy to solve problems relating to machines, mathematics, and materials, he has difficulty with minds. He is never sure exactly what presents his friends will like and as a result, agonizes for weeks until he can convince himself that he has chosen something appropriate. I always try to reassure him that "it's the thought that counts" (and Ernie certainly puts plenty of thought into his choices of gift), but that still doesn't reassure him. This year he had started earlier than usual (back in June I think), and had finally sorted out gifts for everyone except his old school friend Aristotle.

Aristotle was everything that Ernie wasn't. Emotional, where Ernie was logical; intuitive, where Ernie was rational; innumerate, where Ernie was mathematically inclined. And... an Alchemist by profession. Aristotle had never actually managed to transmute an element, and his experimental elixirs of life were anything but life-enhancing. But after every failed attempt to isolate phlogiston or dig up the philosophers stone, he would just smile a little wistfully and begin planning for his next experiment. Surprisingly, considering Ernie's usual views on "witch-doctors, charlatans and anti-scientists", they remained the best of friends, although their gifts to each other often had a somewhat ironic aspect.

To be honest, I had no idea how to work out which piece would be best from first principles, so I cheated a little. Ernie had said that both pieces would be just big enough, so I measured the side-lengths, worked out which one had the smallest area and (after lots of practice with sheets of newspaper) finally managed to wrap the cube in a single uncut sheet of gold according to Ernie's requirements. Ernie was right, with the appropriate orientation and correct folding the cube was perfectly covered in gold. Oddly enough, it must be the Christmas rush that has made me even more absent-minded than usual, I have already forgotten the measurements I made and can't even remember if I used the square or triangle. Can you help me to remember which piece I used, and the lengths I measured?

• Thanks for the compliment. The puzzles, words and graphics are all my own work. The main aim is to make puzzles that entertain in at least three different ways. – Penguino Dec 22 '14 at 8:08
• Success, then. I'd recommend to compile your Ernie's (plus answers) in PDFs fore a later "Ernie puzzle collection". I very much like you artwork and humor in the puzzles - something to aspire to for myself. (Btw, I'm going to delete some of my comments later to keep it uncluttered.) – BmyGuest Dec 22 '14 at 8:12
• +1 (and hence a new bronze badge for you) since the first few sentences could be a description of myself ... I mean, since it's a great puzzle :-) – Rand al'Thor Dec 22 '14 at 12:47
• The answer can be found online, but I don't know how they figured out the equilateral triangle case. Frankly it's beyond me. – McMagister Dec 23 '14 at 15:18
• Well, I googled. I will not reveal which one is better, the square or the triangle, but the difference in area is below 1 percent. – Florian F Dec 24 '14 at 0:16

The optimal wrapping for a square has been shown to be $8\ dm^2$.

For the triangle, the almost best wrapping I could find is the following:

In the picture above, the grid has $5 \ cm$ squares. The triangle is constrained by 3 grid points that must lie on the 3 sides and the fact that the right vertex lies on the grid line where it seems to be. The shape that is cut out wraps exactly a cube of size $1\ dm$. The leftovers can be folded nicely.

The side is $4.215123\ dm$ as I found numerically. This gives an area of $7.693450 \ dm^2$.

So, in the end, it is still the triangle that wraps a cube best.

PS: The solution can be slightly improved by turning the 3 sides a bit more, counterclockwise around the 3 pivot points. The bottom edge would start eating in the tip at the right, but that could be covered by the leftover at the bottom left.

I'm pretty sure you used a square with a length of $200\sqrt{2} \approx 282.84 mm$ (total area $80000 mm^2$). The optimal wrapping of a cube in a square looks like this. I imagine the triangle will end up with more wasted surface area due to the dissimilarity of the shapes, but I haven't figured out the optimal wrapping strategy.