Trick question #1

I forget to add that two poles are perpendicular to the ground all the time.
Thanx,
xl

50

I agree with vn77, the distance is zero.

Can we assume that the "ground" is flat?  Or might the two poles situated on, and be perpendicular to, the surface of a sphere?  If so, what is the diameter of that sphere?

Hi vn77,
> right triangle (???)
I'd rather opt for a parabola.

Cheers!

Stefan

It never touches the ground. Poles 50 ft high. String 100ft long. When the poles are adjacent, the distance from top pole a to ground to top pole b = 100ft. Because some of the string will have been used to tie the knots, the slack will be less than 100ft.

vn77 nailed it.The string itself is only 100 ft long. For it to touch the ground in the middle while suspended by two 50' poles means that is basically folded in half (50' down to the ground, and 50' back up again; there is no slack left for creating parabolic curves). To achieve that, the poles have to be moved together, so that it's 50' up Pole A, 50' down the string to the ground, 50' back up the string to Pole B, 50' back down to the ground. Your experiment winds up being completely vertical. :)

>> I am prepare for a interview.

The whole point of these questions (if there is a point at all) is that you CAN'T prepare for them.  If you know the answer in advance, it proves nothing.  If you have never heard of a problem anything like it, but are able to come up with an outside-the-box solution, it could indicate something about your problem-solving ability.

So here is the secret.... if you have heard the answer to any of these screwball querstions, don't *DO NOT* immediately write it down.  Instead take a long time with it... draw a diagram.  Show a matrix of possibilities... THEN pretend to think of the answer with a flash of intuition.  

Before going to the interview, practice saying "ahhh HA!" in the mirror.  Maybe the occcasional "Eureka!" would be ok, but I'd opt for spending the most time practicing the "puzzled-expresssion-turns-to-quiet-smile-as-you-start-writing" facial configuration...  

A perfectly-executed "squenched-up-brow-resolves-to-twinkle-in-the-eye" could EASILY be worth $40 per hour.

DanRollins,
It is a very interesting skill :). Thanks for sharing with us,
Cheers,
vn

Sounds like old HW prob.
I agree with Dan.
Any quick answer could also indicate lack of additional experience that could give one pause to remember what the answer was, leading to conclusion of being too green.

Personally, I distaste such question any interview.

Yo! Rockingham!!

Winchester!!

Wrong Thread !!!
Oops!

Just want to give hint.. Software companies do not ask geometric questions until and unless they need some geometric answer. But fefinately they will ask question asked in previous thread( bulb ).

There is book called" How do you move Mount Fiji?".
Puzzle for superBrain
Mind teasers are good books.

Hope this help

Good luck

do not have a f****** clue

I know the question is closed, but assuming the ground is flat, the tops of the two poles are 50 ft apart when the string touches the ground.  Simple geometry.

vn77, congratulations. These are the easiest points I've ever seen for a wrong answer. The distance between the tops of the poles when the string touches the ground is 50 ft.

You start with this situation - a rectange 50 ft high and 100 ft wide.
---------------
|                |
|                | 50 ft
|                |
---------------
  100 ft

After pulling the string, assuming the poles do not bend and the string does not stretch. You have two equilateral (not right angled) triangles.
  /\    /\
 /  \  /  \
/    \/    \
-----------
The poles are still 50 ft long. The string touches the ground mid way between the poles, so it is 50 ft from the base of each pole. The length of string from the ground to the top of each pole is 50 ft (half the length of the original string on each side).
So you have two equilateral triangles with sides of 50 ft touching each other. From here it's pretty easy to see that the tops of the triangles are 50 ft apart.

jhshukla, you would have a catenary if the string was hanging freely, but the question says you pull it, so it effectively becomes a straight line (unless it’s really heavy string).

Guess I won't be hiring at EE just yet. :)

vn77

Guess I owe you an apology. I've just noticed the author's addendum about the poles remaining perpendicular. It's a strange constraint, but legal in the puzzle universe, and in that case your answer is dead on. That will teach me to read all the posts properly before responding in future.

Oops, I guess I made the same oversight.

It said the poles remain perpendicular to the ground, it didn't say that the base of the poles remained 100 ft from each other.

---------------
|                |
|                | 50 ft
|                |
---------------
  100 ft

Becomes:
---------------
       ||
       || 50 ft
       ||
---------------
  100 ft

yes, very tricky.

             

Why do we let the asker decide who gets the points?   By definition, he does not know the answer and is apparently completely unable to figure it out for himself.

Dan, the asker would be the one who has to corroborate the answer and see if it gets him the points, or the job, or whatever.  Wow, but you do make a good point - he/she may never know if the answer was deemed "right" by the tester, potential employer, etc.

Actually i believe the question is impossible from the start and here's why.  If all you had was a 100' piece of string then technically would wouldn't even be able to tie it to the two poles.  you need atleast a piece of string 100' 1" long allowing for 1/2" on each side to be tied to the poles at a bare minimum.  That is where the trick part comes in.  This is just my 2 cents.

Was browsing thru some puzzles for fun....
saw this one .....which made me join this grp!!
bcoz the solution is very simple

Essentially the poles on both end would bend as a part of circle as both the distance of base of pole and it's tip from the  centre would be 50 ft .
using some basic trignometry
the distance = 2 x 50 Cos(180/pi)

Ans =  53.98 ft

jatze,

Welcome to the group.

You didn't read the whole thread.  Look at the first comment by xl2003 after asking the question.  Both poles remain perpendicular.  Go back 3-4 posts for the entire explanantion.

> jhshukla, you would have a catenary if the string was hanging freely, but the question says you pull it, so it effectively becomes a straight line (unless it’s really heavy string)

read the question again. it asks you to bring the poles together, not to pull the string.
1) There are two poles with same height of 50 ft. The two poles are separated from each other 100ft, and there is a string 100ft long whose ends are tied to tops of the poles. Now try to bring the two poles together until the string touches the ground. The question is: when the string touches the ground, how far apart the two poles from each other?

> you need atleast a piece of string 100' 1"
let's use glue.

actually, it may be impossible due to another reason. the rope has some non-zero thickness. so two halves of the rope would be some distance apart and the bend at the bottom has some finite curvature. depending on the stiffness of the rope, it could look like:
  ||
  U
____

If we get into another "stiffness" argument, I am going to scream!  It's as stiff as it's gunna get, now let's move on.