The electrician's plight - a riddle?

Question

An electrician has to connect 2 points in a room with wire. The wire cannot be suspended, but must run along the walls, ceiling or floor. The room is 8ft wide, 15 ft long and 12 ft high. The two points are on the two opposite smaller sides of the room, one is a foot above the ground in the centre of the wall, and the other a foot from the ceiling in the centre of the wall. What is the shortest piece of wire he needs?

Homework - I gave up homework years ago.

Eric D - try again

This is a riddle you guys - IT"S NOT HOMEWORK. And the answer is not 27ft.

Greeneyedevil - not quite

greeneddevil - you got the right method - just the wrong answer.

albert - it doesn't look like a riddle. It's deceptively simple. That's the beauty of it. Anyway greeneyedevil practically solved it, just a little calculation mistake on his part

Alright I accept the challenge. But let's just wait 5 minutes to give anyone else a chance

Allright time's up. When you unfold the sides, do it so that one of the shorter walls has a common edge with the floor. And the other of the shorter walls has a common edge with the longer side wall. Then the height of the triangle is 4ft + 11ft = 15ft. The base is 1ft + 15ft + 4 ft = 20ft. So the hypotenuese is 25ft exactly (multiple of 3,4,5 triangle).

Pascal - greeneyeddevil was closer to the actual answer.

Pascal - bad luck

Answer 1

25.08ft. The way you do this problem is to imagine that you can lay the walls out flat like you were unfolding a box. You create a triangle, where the base runs the length of the paper, from the center of one room to the center of the the other. The height is the difference between the location of the first point in the center of the wall (1 ft high) and the location of the second (11 ft high). That makes the height of the triangle 10 feet. In your example, both points are in the center of the far walls, so that's 4 ft of length for each, plus 15 for the long wall, making the base of the triangle 23 feet. The hypotenuse of this triangle represents the "straight" line shortest distance between both points. You can derive this using pythagorean theory: a^2 + b^ = c^, where c is the hypotenuse. 23^2 + 10^ = 629. The square root of 629 is 25.08 feet.

Addendum:
Unless this "riddle" is a play on words, the answer indicated is the shortest possible distance, of course he could cut the wire up into any arbitrary length he wanted, so the riddle is invalid anyway.

Correction: The author's method is more efficient. Sometimes you have to go down to go up.

Apology: Next time I'll get out a pen and paper and draw a diagram.

Have fun!

Postscript: I strongly suspected the answer was actually 25 feet, due to the commonality of the 3-4-5 multiples of the pythagorean theorem. My error was finding a "straight line" and forgetting that I was drawing on a folded surface. All I had to do was flip my triangle upside down and rotate it 45 degrees to get a shorter "straight" line.

Answer 2

The shortest length of wire possible is √689 ft, which is approximately 26.24881. Here's how: unfold the walls of the room like so:

+--
..--
..--+
..--

Where the --'s represent the floor, ceiling, and long sides of the room (with the -- on the top being the floor, and the -- second from the bottom being the ceiling), and the +'s represent the two short sides of the room (with the one on the left being the one with the point close to the floor). The shortest distance between those points along the walls is given by the straight line between them on this unfolding, which spans a horizontal distance of 17 ft (15 along the side + 1ft from the first point to the floor + 1 ft from the ceiling to the second point), and a vertical distance of 20 ft (12 ft up the side + 4 ft from the middle of the wall to the side and 4 ft back). This gives a right triangle, whose hypotenuse has length √(17²+20²)=√(289+400)=√689 ft. Therefore the shortest wire possible is √689 ft.

Edit: this is a simple variant of the spider and fly problem. You can read more about it at http://mathworld.wolfram.com/SpiderandFlyProblem.html

Edit 2: And that's what I get for assuming that this is the same problem without actually checking all possible permutations of the sides.

Answer 3

Over the top or under the floor should be 1+15+7=23'
I think that is the shortest. Around either of the sides would be 4+4+sqrt(15^2+4^2) which would be larger (the length of the wire needed for the 15' wall is found using the Pythagorean Theorem since it would be sloped to make up for the difference in height on the wall of the two points.

(If this was your homework I hope you get in trouble! If not, hope this helps!)

Answer 4

Since both points are at the wall centers, when you
view the room from the side the wire does not have to go
forward or back, which makes it one dimensional easier.
And you didn't say that it can't run thru the center of the floor. So it's 1 ft down to the floor, across 15 ft and up
11 ft. Which gives us 1+15+11=27ft.
Did I miss something. That's not a riddle. It's a simple problem.

Answer 5

I think the answer is 27 (11+15+1)
what you do is you connect the wire along the ceiling of the room to and down to the other side, it is the most direct rout and therefore the shortest distance.
it really helps to draw a diagram.

Answer 6

I think 27 feet.

Answer 7

Do your own homework!

Answer 8

Is this your homework???!!!

Answer 9

what?