ParaLogic

Hah! Now we bring out the monopoly money!

Oops. I accidentally (mentally) switched Eric/Adam's names when I read the first few lines.

The question 'yes?' is rather vague. I have no idea what he's trying to ask.

uno>tres>cuatro<=>seis dos>tres>cuatro<=>seis cinco<=>cinco siete>cinco<=>cinco ocho>cuatro<=>seis nueve>cinco<=>seis diez>cuatro<=>seis once, doce, trece, catorce, quince, dieciseis, diecisiete ... So it's nowhere as neat. Numbers either end up in the 4=6 loop or at 5. Now how about Roman Numerals?

At first, I thought this 35 page long argument was ridiculous, but after reading bona/syon's posts above, I can understand some of the trouble surrounding the ambiguity of the question. This is the way I see it: There are four possibilities for boy/girl combinations, as mentioned several times by now (I imagine): BB BG GB GG Note that this categorization does take into account the order of the children. We are shown that one of the children is a girl, which of course eliminates the first combination.This is the point where people jump to the answer of 1/3, since there only seem to be three combinations remaining. However, as I stated before, order does matter if you look at the problem this way. Therefore, there are still four possibilities: G1B2 G2B1 G1G2 G2G1 Where the number indicates order. The girl we see can be either the elder (G1) or younger (G2) sister, so we must look at both possibilities equally. Thus, the probability stands at 1/2, regardless of the order of birth. Another way to look at this is to completely disregard order in all of the cases. The resulting combinations are: BB GB (same as BG!) GG When BB is eliminated, the probability of GG remains at 1/2.

Wow! I was actually thinking about this exact situation a couple of weeks ago, and I noticed the same convergence. But I didn't consider proving it...

I think that's the best solution if you don't re-arrange any pieces until the end, but I didn't say you couldn't..? I'll just share my solutions for #7 now, since this thread has passed on.

Bah, just eat them all!

I don't like the implications of negative angles, but if you use them, then As for irrational numbers...well I don't (want to) know.

Looking at bonanova's post, maybe I did take the question too far... Either way, I just realized that for numbers like 7/4 and 11/6:

All right. I've cluttered up this thread very much, but this particular topic intrigued me very much. I believe my conclusion is correct, although I didn't actually prove it. (Someone else can handle that. ) So we've got that behind us. Edit: Did I take the question in the OP too far? It was pretty vague...

Eugh editing time ran out just when I noticed my mistake in Edit2. So ignore that one. Need to test more! Edit: Yep, it seems to work with 5/2 and 9/2 as well! Edit2: I think I have reached a conclusion!!. (post coming up)

Just picturing it in my head, I'm guessing that the resulting shape will be a Edit: Edit2:

I meant for #7 :

Sorry for the long wait with no reply; I've been very busy recently. @TSLF: Your answer to #1 is correct, although there are really an infinite number of solutions (to any of these) with the same number of cuts. And your second answer to #2 is also right. (Much simpler than your first, no? ) BobbyGo has the best solution for #5. I should also add a clarification: I have not seen any solutions that require you to flip or even rotate the pieces (yet). I realized that if you are allowed to rotate the pieces, you can cut down (heh) the number of moves by using this in conjunction with the Note provided in the OP. For now, let's avoid anything other than translations. That means that TSLF's answer to #4 is also good. I could do it with one fewer cut, but that would require some rotating. After some consideration, I think I've found the best solutions for #6 and #7. There is an especially creative solution to #7 that I found while doodling on graph paper. Unfortunately, I doubt that there is such a neat solution to #3. If it were to be done, I don't think any of the cuts would line up with the grid lines provided in my example picture.

It's been a looong time since I last visited this forum. (I forgot the username/password for my old account.) But now I'm capable of making my own puzzles (I hope)! Find the minimum number of straight lines needed to cut a greek cross (example above) into pieces that can be re-assembled to make: One square Two congruent squares Three congruent squares (I don't actually know the answer--or if it's even feasible--but maybe you'll surprise me!) Four congruent squares Five congruent squares (It's not quite as obvious as it looks!) Four congruent greek crosses (I have a solution, but it's probably not optimal) Five congruent greek crosses (Same as #6) I'm fairly confident that I have the optimal solutions for 1, 2, 4, and 5. I just threw in the others for an extra challenge.