bonanova

Agree.

Let's say that after k Martians have landed p(covering center) = .5. Would that mean expect number is k and the time is 12:(k-1)?

@gavinksong

Short and probably unneeded dissertation on color: Outdoor light comes from the sun and is essentially white. But as sunlight passes through the atmosphere it is scattered, especially in the shorter (blue) wavelengths. So the sky, which is not a light source per se, does not look black (as it does in space, away from our atmosphere.) Scattered simply means deviating from straight-line propagation from the source of light. This also explains reddish sunrises and sunsets: when the sun is near the horizon, direct sunlight must pass through more atmosphere to get to us, and more blue light is removed as it is scattered back into space. Hence the "blue planet." So direct sunlight is whitish overhead, producing ambient bluish light from the sky, and reddish when low in the sky, producing ambient reddish "sunset" light reflected from clouds. That's why ambient skylight (what's still present in a penumbral shadow of the sun) could preferentially augment the brightness of either the red post or the blue post.

You needn't choose until tomorrow. Why would that be?

Um... Is this your answer for some value of n? Do you have a general method?

Nice solve. There are cubic equations for sin, cos, or tan of the acute angle. But if God had wanted to impose on us the agony of solving cubic equations, he would not have invented computers. Your triangle is the unique alternative to the equilateral triangle. It is true that for acute triangles the squares relate to the length of their respective sides. The case of a right scalene triangle is a degenerate case in which the two (coalesced) squares on the two (unequal) legs have, necessarily, the same area. But they are both larger than the square that rests on the hypotenuse. Not until the triangle becomes (slightly) obtuse (apex slightly larger than 110o), and of course isosceles, does the square opposite the apex increase to equal the other two. What is surprising to me is that this triangle was apparently unknown (unpublished, at least) until less than 20 years ago. A fact that makes this, I think, an interesting puzzle question.

OK, I am willing to supply a huge clue, namely: for each n, the pair of coins that should be used. Is a clue desired?

plasmid's analysis is fairly complete. It begins with "depends on ... ," which is another way of saying the question can't be answered without more information than the OP gives. I will add to the information contained in the OP the assumption that the puzzle creator has fulfilled the burden of providing sufficient information for an answer to be reached.

Not a red herring; a demonstration for a simple case. I'll not say more until a proof is posted one way or another.

Well I found the image. But then what kind of puzzle to make of it?

I had thought of making this a different problem: Given the black square (only) construct a shape inside the square that is 1/10 of its area. It would entail making the circumcircle, the red square inscribed in the upper semicircle (the hard part) and finally the two red diagonals. I wonder [1] if anyone would have gotten that solution or [2] if there is a simpler way.

That was the referenced insight.

What fraction of the blue square does the orange square occupy? Lines join midpoints to opposite vertices. Gold star for most insightful solution.

What fraction of the black square does the red right triangle occupy?

Yes. You are allowed to use a different pair of coins for each n.

A fair coin permits the selection of the first two counting numbers 1, 2 each with the probability 1/2. A cubic die permits the selection of the first six counting numbers 1, 2, ..., 6 each with the probability of 1/6. What if we want to select equally from 1, 2, 3 with probability of 1/3? That's easy with a single roll of a die, counting modulo 3. But could it be done instead using two coins (that are not necessarily fair coins)? Yes. Use a coin with 1/3 - 2/3 probability, along with a fair coin. Mark the first coin with a "1" on the 1/3 side and "2 or 3" on the 2/3 side. Mark the fair coin with a "2" on one side and "3" on the other. Flip the uneven coin. If it comes up "1" we have the outcome of 1. If it comes up "2 or 3" flip the fair coin, and we have the outcome of 2 or 3. Each outcome has a 1/3 chance. Is there a general way to produce n outcomes, each with 1/n probability, using two coins and finite number of flips? You are allowed to use a different pair of coins for each n. If it is not always possible, find a value of n for which it is not possible and prove impossibility for it.