# Tag Info

3

Three pieces of information are needed to answer this: So the number of nails should be In this case, the information about would be red herrings, and the only real knowledge we need would be

5

The five-passenger boat is obviously possible, as you can simply bring every animal in one trip. The four-passenger boat is The three-passenger boat is The two-passenger boat is The one-passenger boat is Thus, the least expensive option is

-2

I think that if you look at the arrangement you can see a: So the answer is the:

9

It spells:

2

Congratz to @Anon for finding the intended solution. I'm posting this just to show off a to scale picture I happened to have lying around:

4

At first glance, I have ONE very simple answer, but I might have miscounted something. Let's start by analysing the range of possible turns and r values thanks to the equation. The equation is simplified to: so: knowing that r has to be one digit to reach area 1, it also: We can then calculate a bunch of possible T, and search for any paths in the maze ...

9

TL;DR First, and easiest, we can do it in: Using the principle of the: The method is as follows: You then just need to rearrange your pieces. Why does this work? Now that we understand that method, how can we do it better? We now need to prove that we cannot do better. However, this is impossible. The following is a sketch of a proof:

2

Left Guardian is ... Right Guardian is ...

2

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: Why do I suspect this is optimal? Why am I not 100% sure? Total cost: What could be the second best? The difference would then be a mere

7

I think the solution is This thing is a poem, a riddle containing:    Cosmic Grandeur’s Clarion; A statement about one who, camels sustaining,    Makes disciples carrion; For none, yet for all, ‘tis a tale involving    Men who will defeat us; It sings for celestial dawn, for evolving    Ape and nascent foetus. Credit where due:

2

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

4

Partial Answer This is better as a comment to the existing answer, as it's more observations that might HELP solve (not an actual solution), but I can't comment yet. I think @Stiv is on the right track, and have come up with several other names Furthermore...

13

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 ...