bonanova

This puzzle not mine, but I believe I have found the minimum answer. If goes like this: You are given a square pane of glass 1 meter on a side. You are required to cut the pane into 100 pieces, all of the same area. There is no requirement what shape the pieces are, nor do they have to all be the same shape. Just so all 100 pieces have the same area. The instrument you use is a cutting laser, and you may assume it will cut through glass of arbitrary thickness. However, the laser consumes vast amounts of energy as it cuts. For both cost and "green" reasons, you want to accomplish the cutting using as little energy as possible. If you make multiple cuts - i.e. when the cut reaches the edge of the glass - you may turn the laser off immediately. So the problem is: minimize the total length of your cuts. Enjoy

Starbreaker has the approach, k-man showed the derivation and has the right answer CR and horia were close with the numbers [horia suggested it] imran had a close guess. Kudos to all.

Yup.

Sounds plausible. I've made a sketch but not done the math yet.

Nice!

You have a cube of cheese; each edge is a cm long. You make a plane cut through the cube that creates a face having six sides. That's cool, you think, let's see if I can do that again. So you make a second plane cut, parallel to the first; it creates another face with six sides. The cube is now in three pieces; one of them has two 6-sided faces. How thick can that piece be?

Great channeling...

See if it works.

They say a magician never reveals his/her secrets. Well this is your lucky day. I'm "giving" you a cool trick today, absolutely free. Here's how it goes ... play along as a spectator. Please select a whole number between 51 and 100. Got it? Good. Don't reveal it. Please quiet your mind for ten seconds while I channel your thoughts. Fine. Now I will write a different number on this paper, then I will seal it in this envelope. Now I want you to add 76 to your number, cross out the first digit of the sum and add that digit to the result. Finally, subtract that sum from your original number. Do you have your final result? Good. I now hand you the sealed envelope. Isn't that cool? Now you can play this trick on your friends. And just to vary things, you don't have to use 76. You could use other numbers. Like 62. Oh, but then you'd have to hide a different number in the envelope. What number should you hide?

A friend gives you the game of 15 squares that slide in a 4x4 container. After a little manipulation, expert that you are, you solve the puzzle, obtaining the following arrangement: ..1..2..3..4.. ..5..6..7..8.. ..9.10.11.12.. .13.14.15.__.. where __ shows the empty space. If you're unfamiliar with this puzzle, you have two moves you could make from this position, namely slide 12 down or slide 15 to the right. And so on. Note a move can be described by giving the number on the square that is moved. Here's the puzzle, in two parts. Using only legal moves, can you make a 4x4 magic square? . In a magic square, numbers in each of 4 rows, 4 columns and 2 diagonals sum to the same number. Count the empty space as zero. . Hint: not all configurations can be achieved. For example, two adjacent numbers cannot exchange positions. For example, you could not get the top row to be 2 1 3 4 with the other 3 rows unchanged. So if write out a magic square of 0 -15 on paper, there's a 50% chance it's not reachable from the above starting position. . If your answer is yes, how can it be achieved with the fewest moves? The solution is the 4x4 array of numbers and the list of moves: e.g., 12, 8, 7, 6, ... etc. Above all, have fun.

What if the headwind equaled the plane's maximum airspeed?

CR and k-man have it. imran, right answer with incorrect math - translating words to equations was the real task. JarZe, probably answered who drove the car but didn't say which he specified. As they used to tell me in school ... show your work. preflop, Prime, took the road less traveled [pun intended], as always.

Imram has it. Even left the final step as an exercise. Kudos.

Earth to Prime. The solution was given 6 hours after OP, in post #10, and 2 of the previous posts were yours. . If I felt like tweaking you, I'd claim that I just got impatient waiting for you. But that's not my style. btw, you don't have to posture so much. you're well recognized for your insightful analysis already. The fact that you were included in to OP proves that.

It's algebra night, me lads, Alex intoned as he set an empty mug on his fav table at Morty's last night. And you might want to bring a pencil along, this one is a wee bit complicated. Ian was home sick, so that left Davey and Jamie to humor Alex. Pencils are ready, they said, shoot! So, Alex Jr. and I took a trip the other day, one of us on our horse, the other in our car. After a while, I observed that if I had gone three times as far as I had, I would have half as far to go as I had; and if Jr. had gone half as far as he had, he would have three times as far to go as he had. By the way, we started at the same time and place, and were headed for the same destination. Did you get all that? They nodded, but Davey was stroking his beard just a little more vigorously than usual. OK, they said, but ... what's the question? The question, lads is this: which one of us rode the horse? It took Davey only a minute to answer. I think I know... Do you?

I think that's the best finite-speed solution so far.

Thanks, both. Curious... is it the song or the book? Or are they related? [foth i mean]

Here is a related question. Suppose Prime and I hold opposite ends of a taut string. What is the area swept out by the string? As Y-san would say, and so would I, please show your proof.

You have a fuse that burns at an uneven rate, but it's symmetric about its midpoint. Starting from either end, it burns completely in one hour. What's the shortest time interval you can measure using the fuse?

Really? Hmmmm.... maybe the answers are ... also ... riddles?