### A math puzzle that’s harder than it sounds

A room is 30 feet long, 12 feet wide, and 12 feet high.

An ant is one foot from the ceiling in the middle of the 12 foot long wall on the north end of the room.

A drop of honey is in the middle of the 12 foot long wall on the south end of the room, one foot from the floor.

How far does the ant have to walk to get the honey.

Hint: if the obvious path were the shortest, I wouldn’t have bothered to post the problem.

Have fun!

— Max

Honey drops to the floor. Ant is releasing from the wall and also falls to the floor. Now the little ant has only to walk 12 feet for getting that delicious honey.

Another “creative” answer. It’s 30 feet the long way, but I suppose I should have nailed the honey to the wall (or come up with a slightly different metaphor) and asked how far the ant has to travel, not walk :)

31.6227766

How? Pretty sure it’s a bit further than that, but love to hear how the ant does it?

Yes you are right, 31.623… is the ‘as-the-crow-flies’ distance thru the volume. Possible if the ant is actually a termite and has wings ;)

The ‘straight-line’ distance is 42. So we are looking for a way to ‘cut a corner’ somewhere to lessen the distance.

Unrolling the walls into a plane we find a path that goes along the hypotenuse that takes the ant across three sides of the length from ceiling to floor. That way the bottom of the triangle is only 32 feet!

The side is 6+6 from each side wall and an extra 12 from the length.

So the total is sqrt(24*24+32*32) = 40.

Nice problem ;)

Here’s a picture of the 40 feet:

You forgot that the ant needs to travel two times 6 (center of the wall remember). So it would be 12+31.622776 = 43.622776. It’s walking, not flying across the room.

It seems pretty straightforward: Unfold the room to create a plane, have the ant walk straight to the honey. Unfold at the northwest vertical and the southwest vertical (or the northeast and southeast). In other words, either the west wall or the east wall is the main part of the plane, unfold the north and south walls to lie in this same plane.

The ant is now 10 vertical feet above the honey and 42 horizontal feet from the honey. The ant has to walk sqrt(42**2+10**2), which is roughly 43.2 feet.

This is longer than just walking straight there (down 11, across 30, up 1, 42 total). But your thinking is on a good path…

with 3 walls wolfram agrees on 43.1741 though it is worse than a straightforward (1+30+11)=42 and optimal 40 with 5 walls

http://www.wolframalpha.com/input/?i=minimize%5B%7Bsqrt%286%5E2%2Bx%5E2%29%2Bsqrt%2830%5E2%2By%5E2%29%2Bsqrt%286%5E2%2B%2810-%28x%2By%29%29%5E2%29%2Cx%3E%3D0%2Cx%3C%3D10%2Cy%3E%3D0%2Cx%2By%3C%3D10%7D%2C+%7Bx%2Cy%7D%5D

Funny they have this puzzle on their site I think.

36

Explain? Shorter than the answer I know…

sqrt((1+30+1)^2+(12/2+12+12/2)^2))

About 40.717 feet or is this beer stronger than I thought?

You have saved some distance but you can do better…

http://mathworld.wolfram.com/SpiderandFlyProblem.html

I know you’ve refined the problem to indicate travelled distance but:

1ft up to the ceiling and 30 ft along the ceiling. Then drop the 11ft to the honey. 31ft walked using the original specification. The honey should be able to stop the ant as he drops.

My girlfriend came up with 40 after I got 43.2 just like Peter did.

I have no idea. If the ant was a spider and the drop of honey a fly I know the answer would be 40′ but this ant/honey problem has me stumped

http://mathworld.wolfram.com/SpiderandFlyProblem.html

Well if I kept it as a spider and a fly it would have been way too easy to find on google :)

zero if to twist relity and blackhole him, is that creative enough

30′ “walk”. He can fall to the floor and climb up at the other end

That wild make it 31’… and that’s what I’ve came up with as my best stab at it — and this is only assuming that the ant is allowed to leave the surface and drop to the floor.

Max seems to imply in his reply to freakconomics below that it can be done in less than 23′, though, so I’m stumped…

No, not less than 23.

You can flatten the walls to get a path from the ant on the N wall to the ceiling, then along a side wall to the dot of honey on the S wall. It forms the hypotenuse of a right-triangle with sides 37 ft and 17 ft, so the hypotenuse is sqrt(17*17+37*37) = approx. 40.7 ft.

I don’t see how Qua Fibo gets a 24′ by 32′ right triangle

37. Up to the ceiling, across the room and down. Ants can walk upside down. Thought there must be a trick since this is so obvious…

Its 11 feet down from the ceiling to the honey, so that’s 42 feet.

Of course. I read “middle of the wall” for the south wall. Time to learn to read.

Or time for me to write more clearly!

Is the room an ellipse? I can’t be bothered calculating how long the path would be if so…

Actually that is dumb, read the question!!! Forget this.

Rectangle.

23 feet – down 11′, across 12′ – honey dripped to floor, hence 23′ to a tasty morsel. (or to complete journey to honeypot – 24′).

ant and honey aligned on same plane center of a 12′ wall so life is simple.

top down view

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ant/honey

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face

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Sorry if I wasn’t clear but its 30 ft n to s, and there’s a shorter way.

The ant travels 40 feet.

The general approach is to head up to ceiling and diagonally across the ceiling to a point along the wall, along the diagonal to the floor at a point before the corner, across the floor to the wall, and diagonally across to the honey.

The path can be described as three segments, and the first and the last are the same:

2*sqrt((1+x)**2 + 6**2) + sqrt((30 – 2x)**2 + 12**2)

x is the distance the ant saves on the 30 foot wall by cutting the corner.

The distance =

2*sqrt((1+x)**2 + 6**2) + sqrt((30 – 2x)**2 + 12**2)

2*sqrt((1+x)**2 + 6**2) + 2*sqrt((15 – x)**2 + 6**2)

The distance is minimized when

2*sqrt((1+x)**2 + 6**2) = 2*sqrt((15 – x)**2 + 6**2)

((1+x)**2 + 6**2) = ((15 – x)**2 + 6**2)

(1+x)**2 = (15 – x)**2

x = 8

So the ant initially heads 1 foot up the wall plus a further 7 feet along while travelling 6 feet towards the edge, travelling along the hypotenuse of a 6×8 foot triangle, for a total of 10 feet. This puts him on the top of the 30 foot long wall 7 feet from the corner. He travels down the 12 of the wall and aims for a spot 7 feet from the corner, so the hypotenuse of a 16×12 triangle,or 20 feet. From this spot on the floor 7 feet from the corner, he heads across the floor and up the wall, travelling along the hypotenuse of a 6×8 foot triangle, for a total of 10 feet.

In all, he travels 10 + 20 + 10 feet, for a total of 40 feet.

Can you confirm that it’s definitely over 30? Just so we know quite how creative we need to be.

Yes, definitely over 30. It doesn’t need any of that type of creativity. Its a math puzzle not a tricky wording puzzle.

I got 4 simple ways to do that because unlike humans, ants can walk wherever they want but I may be wrong! In my answers, ant took only the straight paths but there are also diagonal paths to take. I just skip those because that’s be too much; I only had oatmeal for breakfast :P

1ft from ceiling means 11ft because 12 – 1 = 11 ,so, ant has to walk 11ft to reach the ground. He then walks 30 ft to reach the south end. Now the honey is just 1 ft away. That means 11 + 30 + 1 = 42 ft

Ant can also take the way up, so, 1 ft up, 30 ft to the south and 11ft down. Which also = 42.

Now, ant is in the middle of the 12ft wall and it’s 12 ft wide right? Because she’s an ant, she can also walk to the side of the wall. So, she walks 6ft left, 30 ft right to reach the south and another 6 ft right to reach the honey. Also equals 42ft.

Now, she cal also take the 6ft right path, walk 30 ft south and 6ft left to reach the honey and that also = 42.

Now Max, you tell me!

Oops, looks like I was wrong. If ant was to use the side road, it would have to do another 10ft, so, the answer would be 6 + 30 + 6 + 10 = 52ft. If it were to use the side road that is!

40 ft.

Something I haven’t seen elsewhere: if the room is a cube, the direct route (vertically down, the across the floor and vertically up) is the shortest route. As the length of the room increased to longer than 18 feet, then the diagonal route becomes shorter.

‘Unfold’ the cuboid to reveal its plan. Then the solution is clearly (using pythagorus) sqrt (24×24 + 32×32) = 40 feet.

what if to put room down on north side since its cube and all what ant has to do to w8 til honey drops right on him him, not like its glued to the wall, so it gonna move down the wall.

The correct answer is 31.62. if you unfold the room sideways you determine the following

12′ high – 1′ foot from ceiling – 1′ from floor = height

30′ feet long side wall = length

Because ants can walk in any direction you can use pythagorean theorem to find the hypotenuse which is 31.62

I forgot to add 12 + 30 which the distance from the centers of the room plus the sidewall for the length(6+6+30)’ long

10′ high so the hypotenuse is 43.17

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10′

— 6′ 30′ 6′ *

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