bonanova

Bravo. Not an easy solve. Yes, that is the coveted bonanova gold star.

The second of your words is the one I was looking for.

Yes indeed. But there are enough clues to say which one. Hint: read everything.

With a nod to plasmid's I am not ... series. I am a place where rich men are lain, but add a night, and I'm a hero's bain bane. What am I?

What word has the opposite meaning of "if"?

I drew five letters from the Scrabble pile. Amazingly, they were the five vowels, a e i o u. I discarded two, at random, and drew two different consonants. Playing with these five letters, I arranged them into a 1-syllable word I rearranged them into a 2-syllable word I rearranged them again, into a 3-syllable word. What were my five letters, and what were the words?

A book on the shelf, sandwiched among several others, reads "How to Jog." What is the book about?

You find four snips of paper on which are written these numbers: 13, 14, 95 and 62. You see a fifth snip of paper. What number is written on it?

I think your second pour is not right.

Barcallica - nice work. OK we have a solution for 11 pourings. Can anyone get it down to 10? Or 9?

I thought of that, too. But it scales down to 120/25 = 4.8 squares. Not quite 5. So I don't think it would apply, unless that turns out to be harder than the original problem.

@Phil, I am thinking along these same lines. The task I'm working on is how to determine a provable worst case for the squares placement. It's the nature of such a proof that is puzzling me for the moment.

You have two 10-liter beakers filled with water, and two empty beakers, of capacity 4 and 5 liters, respectively. You may pour from one beaker to another, in such a way that either the source beaker is emptied, or the receiving beaker is filled. The object is for the two smaller beakers each to contain 2 liters of water. No water is to be wasted, and no more than nine pourings is allowed.

A solid in the shape of a regular tetrahedron had uniform density and a mass of 1 kg. It was mutilated by the removal of its vertices, each made by a planar cut, parallel to its opposite face. The solid now has eight faces, whose areas are 1, 2, 3, 4, 5, 6, 7 and 8 square units, in some order. What is the area of its original faces? What is the mass of the resulting solid?

@gavinksong No, I can't. For solids it just seems intuitive. For points and lines, I am ready to change my answer. Points and lines have no surfaces to illuminate. In that sense they need 0 lights. Shining 0 lights on a point or line suffices to leave no surface un-illuminated.

@gs, your solution is better. Which square did you start on?