bonanova

This question gets much more interesting if we go, say, 1.5 block-lengths outside of the original footprint--shape of tower differs dramatically. The balance is precarious at the 1-2 edge and the 4-5 edge, but some of the 1/24 can be spent to make them stable. If the middle blocks are rotated, hmmm... might not matter. Nice job.

Hopefully, this solves all tiebreak situations and removes the uncertainty. Reading from the OP: Given that it [Cole's strategy] is uncertain, what can we determine regarding Bobby's shooting accuracy? From your reasoning regarding Cole's strategy, what is the answer?

This puzzle was discussed at length a few years ago in this forum. It was rich enough that I thought it deserved another airing, with a largely different set of puzzle solvers. I can't find it using search, to know for sure, but I'm wondering whether Prime was one of the solvers last time. For those to whom this is a new puzzle, I hope you will enjoy it. It's simple to state, and challenging to solve. I'm pretty poor at darts, but I managed to put three of them on the board. Assuming my fourth dart also hits the board [circular], then what is the probability it will lie inside the triangle formed by the first three?what is the probability the four darts will not form a convex quadrilateral?

Almost time to harvest the oranges. Anyone still working on this, or are there millions wanting to know how it's possible? More time if you like, otherwise I'll post the layout at week's end.

Earlier this year I noticed that the date could be written as 1.13.13 [January 13, 2013] and the day and month multiplied gave the year [modulo 100] Thirteen being both prime and greater then the number of months, this is the only such date in 2013. What year this century contains the most such dates?

You are stacking children's building blocks on a level table. They are perfect, identical cubes, evenly weighted, and you're not fastening them together in any way. The stack is not very straight. In fact at one point you notice the topmost block (viewed from overhead) is completely outside the footprint of the bottom block. What is the minimum number of blocks in your tower? Or is that even possible?

That's not what paragraph 2 says in the OP. You're right. OP is corrected. I was presumptuous to teach you how to solve ... Apologies.

Someone posed this problem to me, and I tried several ways to count cases and finally gave up. This prize is a cooperative sharing of k-man and Prime, but I can mark only one post. So I marked the post that got the final count. Good job both!

Hi PL, it's nice to have you back! If you remember your old username, we can probably fix it up for you to use again.

Yes, that is the answer. Nice job.

Very nice approach. Nice categories and elimination of symmetric cases. Never mind, I think I see two cases after all. I think that's it.

A twist on the "truel" puzzle - a duel among three participants, where shots are taken in turn. Alex always hits his man. Bobby is not perfect, but he shoots better than Cole does. The referee thus gives Cole the first shot, followed by Bobby and Alex in that order. When a man is hit, he drops out [or drops dead]. The shooting continues in sequence until only one man is unhit. Cole does not want to take Bobby out of the competition, then Alex will hit Cole. Game over. But it may not be wise for Cole to hit Alex either. Cole is disadvantaged in an ordinary duel. So Cole's first-shot strategy is uncertain. Given that it is uncertain, what can we determine regarding Bobby's shooting accuracy? Assume each knows the others' accuracy level.

Very nice, and I thought you had me on that point.But I read the OP again, and sure enough, it says ... You disagree and ask her to replace the spoons and repeat the process, But you get honorable mention anyway. Your solution works except for the red letter. However, if you didn't know how many spoons were taken, would you ask please replace the spoon? The question of implication is an interesting one. A previous Forum discussion centered on whether the statement "All my cars are Fords" implied that I had at least one car, and that it was a Ford. What do you think? My take was that it meant "For any object that is a car and that I own, that object is also a Ford." There may not be any such objects, but the statement nonetheless has meaning: If there is such an object then that object is a ford. The premise does not have to be true on a conditional. The discussion continued until I said that I once saw an empty field that bore a sign "All trespassers will be prosecuted." The sign was meaningful even tho the field clearly was empty. That is, the sign did not imply the existence of at least one trespasser. I'll concede your point as it applies to "some number of spoons." That number clearly could be one. Or zero, for that matter. But if she had drawn just one spoon, I could not say "replace the spoons". I would have to say "replace all the spoons that you drew." Well, that is the appropriate statement to have made, I guess. I didn't know the number; the OP simply presupposes that I knew is was plural. So your point is better taken than I first thought. Let's say, though, that had she drawn only one spoon, then to be responsive to what I asked, she would have to have said "I can't do as you ask." or "I don't understand what you are referring to." or something other than just doing the replacement. In that sense, the OP (by careful reading, and way too much over-thinking) rules out the one-spoon case. Kind of. In any case it was only a fortuitous accident that the OP covered the case of one spoon at all. I'm not clever enough to have done that by plan. I simply didn't think of the case. Good solve!

Correct. Good job. I wonder if there is a proof of this that is not overly complex? Edit: Well, No. I just found the proof, and it's not beautiful for its simplicity. You start with the Law of Sines, and 2 1/2 pages later you have a symmetrical expression for one side. "Do not try this at home."

Nice.

Very nice approach. Nice categories and elimination of symmetric cases.

I was going to add the caveat that I cannot verify the usefulness of google translate. But yes, I thought all monkeys played football. Nice. Whereas one of the two Google founders was born in Moscow. You must know Russian to appreciate and fully enjoy the beauty of the translation. The first sentence of the translation actually is: The rope [to] pulley had ran up to ... (As if the rope or pulley had legs and actually ran. Hard to say which.) Working off the OP in English with some aid of Russian translation I get a very large negative weight for the rope. 0 for 2 is not bad. It's positive, and not all that long. The tricky part is ascribing the right time frames to the numbers.

Is there a consensus? Can you construct the possible configurations in groups, using words or a sketch? My first try to solve this was to take 12 things 4 at a time, then remove the symmetrically equivalent solutions. But I kept removing too many cases. Next try was to group the edges and distribute 4 blues among the groups. I'm not convinced yet of my answer.

You're right about the bisector case. But for the trisector case there is more than one point. In fact there are three places where a trisector of one angle first intersects a trisector of one of the other angles. And there is something special about those three points. Actually they are not collinear. That being the case, they form a triangle. So the OP really asks: what is special about the triangle formed by these three points?

Good solve ... although I got the "basic" nature of the puzzle, I did not see the second "?" in the OP.

Very nice, and I thought you had me on that point. But I read the OP again, and sure enough, it says ... You disagree and ask her to replace the spoons and repeat the process, But you get honorable mention anyway. Your solution works except for the red letter.