bonanova

BMAD, for generating-function result, see

Nice. At first blush it looks solid. Intuitively I'm expecting something closer to 1/13, but it is what the process says it is. Having just learned generating functions, I'm thinking the gf for this approach might be reasonably straightforward.

I misread the order of the students.

The answer is even closer so if no one can find an even more precise answer then this will be the answer. And before I get some guessess with a bunch of nines, the answer that i found is not consisting solely of running nines after the decimal. It would repeat, being rational. I just tried to investigate its closeness to 1/13. I'm still thinking about the small excess count for multiples.

the first student used a 30 degree angle and the second student used a 45 degree angle. The op did not give the third person's angle. I guess I don't understand. Was that not the question?

x\y| 1 2 3 4 5 6 ---+--------------------- 0 | 1 0 -3 -8 -15 -24 1 | 3 3 1 -3 -9 -17 2 | 7 8 7 4 -1 -8 3 | 13 15 15 13 9 3 4 | 21 24 25 24 21 16 5 | 31 35 37 37 35 31 6 | 43 48 51 52 51 48

Hi w10, and welcome to the Den. Hope you enjoy solving puzzles here and posting some of your own. You have a certain "outside the box" approach to solving them.

By switching positions, I meant as in the 15-square puzzle. These two positions are in separate solution spaces. You can't start from one and after a series of moves get to the other. 1 2 3 4 2 1 3 4 5 6 7 8 5 6 7 8 9 10 11 12 9 10 11 12 13 14 15 __ 13 14 15 __ But two switches changes the parity back again. So these two positions are in the same solution space: You can start from one and after a series of moves get to the other. 1 2 3 4 3 2 1 4 5 6 7 8 5 6 7 8 9 10 11 12 9 10 11 12 13 14 15 __ 13 14 15 __ So I was wondering whether there are similar sets of disjoint configurations present here. It's not simple to visualize what happens even after two or three moves. It's very interesting.

Chain the dog in the middle of a 100-foot wall. In the front yard he has an 80-foot-radius semicircle of access. When he gets to the ends of the wall, he has 30 feet of chain left. As he walks into the back yard he gains access to a 30-foot-radius semicircle on each side. Area [three semicircles] = pi/2 [802 + 302 + 302] = pi/2 [6400 + 900 + 900] = pi [3200 + 900] = 4100 pi ft2 General area formula (x is distance from end of wall to the chain point): 0<x<20:.. A = pi/2 [802 + (80-x)2] 20<x<50: .A = pi/2 [802 + (80-x)2 + (x-20)2]

Suppose I conjure a function f(x,y). I give you a matrix of f values for x = {0 1 2 3 4 5 6} and y = {1 2 3 4 5 6}. It looks like this: x\y| 1 2 3 4 5 6 ---+------------------ 0 | 3 6 9 12 15 18 1 | 1 4 7 10 13 16 2 | -1 2 5 8 11 14 3 | -3 0 3 6 9 12 4 | -5 -2 1 4 7 10 5 | -7 -4 -1 2 5 8 6 | -9 -6 -3 0 3 6 From the first row you can see that f(0,y) = 3y From the first column you can see that f(x,1) = 3-2x From the second column it's clear that f(x,2) = 6-2x, and so forth. At some point it's evident, and the other values so confirm, that f(x,y) = 3y-2x. OK that's the idea. A 2-diminsional "what's the next term in this sequence" puzzle. Only we don't ask the next term, we ask for the simplest function that generates the given terms. See if this one is too easy (there are plenty of clues in the table): Same range of values for x and y as above. x\y| 1 2 3 4 5 6 ---+-------------------------- 0 | 1 1 1 1 1 1 1 | 0 1 2 3 4 5 2 | -1 0 1 0 -7 -28 3 | -2 -1 0 -17 -118 -513 4 | -3 0 17 0 -399 -2800 5 | -4 7 118 399 0 -7849 6 | -5 28 513 2800 7849 0 What is f(x,y)? If you like this puzzle type, then let the solver make one, and we'll keep the thread going. Aligning a table like this is easy: just use Courier font.

OK I see what you did. It seems more natural to assume the same thing for the dog that one assumes for the men. The men march at a constant speed; the dog trots at a constant speed. The men's feet are in contact with the ground; the dog's feet are in contact with the ground. If we were talking about an airplane, then both airspeed and ground speed are of importance. If we were talking about a boat, then both water speed and "geographical" speed are of importance. If the dog's feet were somehow in contact with the men, a "soldier speed" would have import. That might happen as follows: Imagine the soldiers are carrying a 100' x 100' platform on their shoulders. A dog runs at constant speed around the perimeter of the platform. Now constant speed would mean with respect to the soldiers. What is the distance traveled by the dog? With respect to the platform, exactly 400' With respect to the ground, something like your answer.

Circles aren't bigger or smaller than squares until constraints are added. If the constraint is a given perimeter, circles are bigger (area wise) If the constraint is maximum and minimum values of x and y, then squares are bigger. The present constraints are of the second type.

If we let the house assume a parallelogram shape, then the above numbers change. If the tie point is closest to an angle that becomes acute, the accessible area increases. If the closest corner becomes obtuse, the accessible area decreases. The extreme case is that the house collapses into a 100 foot wall. The accessible area depends on the distance x the dog is tethered from the end of the wall If x = 0 (end) the maximum accessible area is the 80-foot radius circle: 6400 pi square feet. If x = 50 (midpoint) the accessible area hits a minimum: 4100 pi square feet. The previously found areas occur at intermediate values of x: x. . . Area/pi -------------- 0 .....6400 15.2 ..5300 20 ... 5000 27.6 ..4600 30 . ..4500 50 .. 4100

assures the detector is inside a cube.

If the solution has not already been posted, the only other assumption I can imagine making is that the tether is linked to a railing that allows it to slide along the perimeter of the house. The the accessible portion of the yard in that case is four rectangles and a complete circle. I won't bother to draw it, but the area is clearly 80p + pi802 where p is the perimeter of the house. I won't spoiler this, because don't see how the OP could be construed to describe that condition. I don't see what else to try. If we're still missing something, a clue would be nice.

Was that the assumption you made in your graphical solution?

Very nice problem. I verified TSLF's solution numerically, but then realized the optimal condition can be calculated directly. Which is probably what TSLF did. Since I went through the calculation I thought I'd share it.