8 ms·

Interesting math puzzle - harder than it sounds

20 points by mschireson 15y ago

byoung2 15y ago
40 feet?
- mschireson 15y ago
  Nicely done. Take the diagonal!
  - jonnycat 15y ago
    You're going to have to explain your math here, because I don't believe the diagonal is going to give you 40 (unless I'm visualizing the problem incorrectly). In fact, I believe the diagonal is longer than the naive solution.
    - mschireson 15y ago
      It depends which diagonal. I'll explain in more detail later but don't want to completely spoil it :) Stay tuned!
    - byoung2 15y ago
      The "obvious" solution is down 11 feet to the floor, 30 feet across the floor, and 1 foot up to the honey, for a total of 42 feet. Instead picture the room "unfolded" into a series of squares and rectangles and laid flat. Then use the pythagorean theorem to find the length of the diagonal that connects the ant and the honey while staying on a surface the whole way. You'll get (6+12+6)^2 + (1+30+1)^2 = c^2 or c = 40.
- akavi 15y ago
  I'm pretty sure this is the correct answer too. Visualization tip: Consider the possible nets of the rectangular prism that makes up the room, and find what the shortest path on one of them would look like.
  - mschireson 15y ago
    Its the best I've found. Another way of thinking about the visualization is to make a model of the surface that can fold up to form it. The shortest path (which will be a diagonal if not a straight line) on any of those is the answer.
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mschireson 15y ago
This one got my kids a bit interested but not enough to get them really engaged in trying to solve it. Anyone have fun puzzles for middle-schoolers?
dreamux 15y ago
31 feet. Ant jumps off the wall and onto the floor (no walking necessary, and easily survivable), walk straight across the 30 foot floor and 1 foot up the wall.
- noblethrasher 15y ago
  Less than 31 feet since, in the presence of gravity, the honey is moving towards the floor as well.
- mschireson 15y ago
  Lol. That's out of the box thinking!
dredmorbius 15y ago
0 feet. It's a flying ant.
Eduard 15y ago
Does the ant have wings?
- mschireson 15y ago
  No, it does not. And while someone suggested that it could drop to the floor without injury, you could consider it to be afraid to do so :)
mayoff 15y ago
The answer, with a diagram, can be found here: http://mathworld.wolfram.com/SpiderandFlyProblem.html http://mathworld.wolfram.com/SpiderandFlyProblem.html
- losvedir 15y ago
  Ha, well that was a lot easier than I made it. I ended up with the right answer by finding the minimum of the equation f(x,y) = 2(sqrt(1+x^2) + sqrt((6-x)^2 + y^2) + sqrt((15-y)^2 + 36)) which is the length of the path taken if the ant goes up to the ceiling at a point x to the East of straight up, and from there goes to the side wall at a point y to the south, and then goes from there to the center of that wall. (And then it's mirrored). I should have thought to unravel the room! Errr.
  - d0mine 15y ago
    for the unraveled room: sqrt(1+x^2) + sqrt((6-x)^2+y^2)+sqrt(12^2+z^2)+ sqrt((30-z-y)^2+u^2)+sqrt((6-u)^2+1) 40 at {x -> 0.75, y -> 7, z -> 16, u -> 5.25} http://www.wolframalpha.com/input/?i=minimize%5B%7Bsqrt%281%2Bx%5E2%29+%2B+sqrt%28%286-x%29%5E2%2By%5E2%29%2Bsqrt%2812%5E2%2Bz%5E2%29%2Bsqrt%28%2830-z-y%29%5E2%2Bu%5E2%29%2Bsqrt%28%286-u%29%5E2%2B1%29%2Cz%3E%3D0%2Cy%2Bz%3C%3D30%2C0%3C%3Dx%3C%3D6%2C0%3C%3Dy%3C%3D30%2Cu%3E%3D0%2Cu%3C%3D6%7D%2C%7Bx%2Cy%2Cz%2Cu%7D%5D http://www.wolframalpha.com/input/?i=minimize%5B%7Bsqrt%281%... your equation: http://www.wolframalpha.com/input/?i=minimize+2%28sqrt%281%2Bx%5E2%29+%2B+sqrt%28%286-x%29%5E2+%2B+y%5E2%29+%2B+sqrt%28%2815-y%29%5E2+%2B+36%29%29 http://www.wolframalpha.com/input/?i=minimize+2%28sqrt%281%2...
  - tzs 15y ago
    To make the "minimize the equation" approach complete, other paths should be considered, such as instead of going up to the ceiling, going to the side and taking a side wall. Some limits can be placed on the number of possible paths by considering some general properties of an optimal solution: 1. Assuming the start point is A, and the path leaves the start side at point B, the optimal path will take a straight line from A to B. Proof: if it were not straight, it could be replaced by a straight line which would shorten the path, contradicting the assumption the path was optimum. 2. Once the optimal path leaves a side, it will not return to that side. Proof. Suppose the path leaves the side at point B, then later enters at C, and then leaves again at D. The route from B to C could be replaced with a straight line from C to D, which would shorten the path, again contradicting the assumption that the path was optimum. This cuts down the number of possible path templates greatly, since a given surface can only appear once in the optimal path, and only contain a single straight line segment. I expect that a little more reasoning can deduce some more limits on possible optimal paths, to get it down to just two or three possible equations to maximize.

20 points by mschireson 15y ago

byoung2 15y ago

40 feet?

mschireson 15y ago
Nicely done. Take the diagonal!
- jonnycat 15y ago
  You're going to have to explain your math here, because I don't believe the diagonal is going to give you 40 (unless I'm visualizing the problem incorrectly). In fact, I believe the diagonal is longer than the naive solution.
  - mschireson 15y ago
    It depends which diagonal. I'll explain in more detail later but don't want to completely spoil it :) Stay tuned!
  - byoung2 15y ago
    The "obvious" solution is down 11 feet to the floor, 30 feet across the floor, and 1 foot up to the honey, for a total of 42 feet. Instead picture the room "unfolded" into a series of squares and rectangles and laid flat. Then use the pythagorean theorem to find the length of the diagonal that connects the ant and the honey while staying on a surface the whole way. You'll get (6+12+6)^2 + (1+30+1)^2 = c^2 or c = 40.
akavi 15y ago
I'm pretty sure this is the correct answer too. Visualization tip: Consider the possible nets of the rectangular prism that makes up the room, and find what the shortest path on one of them would look like.
- mschireson 15y ago
  Its the best I've found. Another way of thinking about the visualization is to make a model of the surface that can fold up to form it. The shortest path (which will be a diagonal if not a straight line) on any of those is the answer.
deleted 15y ago
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deleted 15y ago
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mschireson 15y ago

This one got my kids a bit interested but not enough to get them really engaged in trying to solve it. Anyone have fun puzzles for middle-schoolers?

dreamux 15y ago

31 feet. Ant jumps off the wall and onto the floor (no walking necessary, and easily survivable), walk straight across the 30 foot floor and 1 foot up the wall.

noblethrasher 15y ago
Less than 31 feet since, in the presence of gravity, the honey is moving towards the floor as well.
mschireson 15y ago
Lol. That's out of the box thinking!

dredmorbius 15y ago

0 feet. It's a flying ant.

Eduard 15y ago

Does the ant have wings?

mschireson 15y ago
No, it does not. And while someone suggested that it could drop to the floor without injury, you could consider it to be afraid to do so :)

mayoff 15y ago

The answer, with a diagram, can be found here: http://mathworld.wolfram.com/SpiderandFlyProblem.html http://mathworld.wolfram.com/SpiderandFlyProblem.html

losvedir 15y ago
Ha, well that was a lot easier than I made it. I ended up with the right answer by finding the minimum of the equation f(x,y) = 2(sqrt(1+x^2) + sqrt((6-x)^2 + y^2) + sqrt((15-y)^2 + 36)) which is the length of the path taken if the ant goes up to the ceiling at a point x to the East of straight up, and from there goes to the side wall at a point y to the south, and then goes from there to the center of that wall. (And then it's mirrored). I should have thought to unravel the room! Errr.
- d0mine 15y ago
  for the unraveled room: sqrt(1+x^2) + sqrt((6-x)^2+y^2)+sqrt(12^2+z^2)+ sqrt((30-z-y)^2+u^2)+sqrt((6-u)^2+1) 40 at {x -> 0.75, y -> 7, z -> 16, u -> 5.25} http://www.wolframalpha.com/input/?i=minimize%5B%7Bsqrt%281%2Bx%5E2%29+%2B+sqrt%28%286-x%29%5E2%2By%5E2%29%2Bsqrt%2812%5E2%2Bz%5E2%29%2Bsqrt%28%2830-z-y%29%5E2%2Bu%5E2%29%2Bsqrt%28%286-u%29%5E2%2B1%29%2Cz%3E%3D0%2Cy%2Bz%3C%3D30%2C0%3C%3Dx%3C%3D6%2C0%3C%3Dy%3C%3D30%2Cu%3E%3D0%2Cu%3C%3D6%7D%2C%7Bx%2Cy%2Cz%2Cu%7D%5D http://www.wolframalpha.com/input/?i=minimize%5B%7Bsqrt%281%... your equation: http://www.wolframalpha.com/input/?i=minimize+2%28sqrt%281%2Bx%5E2%29+%2B+sqrt%28%286-x%29%5E2+%2B+y%5E2%29+%2B+sqrt%28%2815-y%29%5E2+%2B+36%29%29 http://www.wolframalpha.com/input/?i=minimize+2%28sqrt%281%2...
- tzs 15y ago
  To make the "minimize the equation" approach complete, other paths should be considered, such as instead of going up to the ceiling, going to the side and taking a side wall. Some limits can be placed on the number of possible paths by considering some general properties of an optimal solution: 1. Assuming the start point is A, and the path leaves the start side at point B, the optimal path will take a straight line from A to B. Proof: if it were not straight, it could be replaced by a straight line which would shorten the path, contradicting the assumption the path was optimum. 2. Once the optimal path leaves a side, it will not return to that side. Proof. Suppose the path leaves the side at point B, then later enters at C, and then leaves again at D. The route from B to C could be replaced with a straight line from C to D, which would shorten the path, again contradicting the assumption that the path was optimum. This cuts down the number of possible path templates greatly, since a given surface can only appear once in the optimal path, and only contain a single straight line segment. I expect that a little more reasoning can deduce some more limits on possible optimal paths, to get it down to just two or three possible equations to maximize.