8
$\begingroup$

This was a challenge given to us in our Team Building exercise in an MBA course I took. Needless to say our team didnt do well. So I am asking the Stack community if there is a clear way.

We were in a Gym with NO equipment. We had 3 players on our team. The instructor gave us a long rope (may be 2 inches in diameter) and asked us to make a square out of the rope.

The catch: We were blind folded.

Also we were told that we cannot use any Gym related equipment including walls or windows etc. Just us, the blind folds and the damn long rope. No cutting either. Dont have to use all the rope We were given 10 minutes to do it.

We failed.

Any clear ideas?

The Team that won did a good job but not perfect one.

|improve this question|||||
$\endgroup$
  • 1
    $\begingroup$ Tie a square knot ;) $\endgroup$ – Grumpyllama59 Jul 21 '17 at 15:34
  • $\begingroup$ 3 players on your team, including or excluding you? $\endgroup$ – tuskiomi Jul 21 '17 at 20:23
  • 1
    $\begingroup$ Were you told you had to leave the blindfolds on? Much easier if you take the blindfolds off, make the shape, then put them back on. $\endgroup$ – RDFozz Jul 21 '17 at 23:50
  • $\begingroup$ With only 3 people, I think this would be difficult enough without the blindfold! $\endgroup$ – Strawberry Jul 22 '17 at 13:21

12 Answers 12

Active Oldest Votes
8
$\begingroup$

Use the shortest member of your group as a measuring tool and gravity for right angles.

enter image description here

|improve this answer|||||
$\endgroup$
  • $\begingroup$ In our class this was the idea that got most votes. $\endgroup$ – DEEM Jul 23 '17 at 12:41
  • $\begingroup$ Yeah, this is smart $\endgroup$ – Strawberry Jul 24 '17 at 6:34
  • $\begingroup$ Well, this makes a square shape in the 'air' ! How about making one on terra firma ? $\endgroup$ – Mea Culpa Nay Jul 24 '17 at 8:40
  • $\begingroup$ How can you be sure that you are holding the ropes perfectly vertical? What is the point of steps 5 and 6? Are they just re-verification that A and B haven't moved since last time they were measured? $\endgroup$ – Trenin Jul 25 '17 at 13:38
  • $\begingroup$ @Trenin the person holding the rope should let it fall naturally so that it is perpendicular to the ground. Step 5 shows person C handing the loose end of the rope to person A (since neither A or B can move) and Step 6 is the final position. (person C doesn't have to lie on the ground, but it's a way of making sure the distance hasn't changed since they got up) $\endgroup$ – as4s4hetic Jul 25 '17 at 14:02
5
$\begingroup$

Steps:

  1. Lie the rope on the ground and have Person $A$ hold both ends together.
  2. Have a Person $B$ pull the rope away until the apex is reached, creating two equal lengths of rope.
  3. While Person $B$ holds the apex on the floor firmly in place, Person $C$ brings one of the ends of the rope over to Person $B$, who now also holds it in place.
  4. Person $C$ brings that rope out to an apex, creating three lengths of rope: $x/2$, $x/4$, $x/4$.
  5. Person $A$ now brings the end of rope that Person $B$ is holding to meet up with Person $C$.
  6. Person $C$ walks off to the left perpendicular to the rope on the floor until the rope is taught.

They have now constructed a shape vaguely resembling $4$ on the floor using the rope. $4$ is, of course, a square.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ O C'mon Ian....... $\endgroup$ – DEEM Jul 21 '17 at 21:15
  • $\begingroup$ @DeepakMahulikar I like this answer; it demonstrates very nicely that the problem here is the question itself. :) What does "make a square using a rope" even mean. I can think of 10 different interpretations. (Ian, +1, this was not one of them :D ) $\endgroup$ – Tasos Papastylianou Jul 21 '17 at 23:23
  • 1
    $\begingroup$ LOL. so 1 is also a square of 1. $\endgroup$ – DEEM Jul 22 '17 at 20:37
  • 1
    $\begingroup$ Perhaps, but I wanted to demonstrate something with a few more steps than "just lie the rope on the ground in a straight line". ;) $\endgroup$ – Ian MacDonald Jul 24 '17 at 14:31
4
$\begingroup$

Person A picks up the rope in the middle of it and holds a segment taut between their two hands, using their thumb and pointer fingers they make approximate right angle

Person B picks up the end of the rope on Person A's left side, measures out a segment against the one in person A's hands and then aligns the thumb and pointer finger of his right hand with A's left hand.

Person C picks up the end of the rope to person A's right side, measures out two segments against the one in person A's hands by folding it. Person C passes the end of the second segment to person B's left hand. Person C holds the first segment and aligns the thumb and pointer finger of his left hand with Person A's right hand.

Voila, a square.

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ Perpendicular perlicues! $\endgroup$ – Tom Jul 22 '17 at 11:15
3
$\begingroup$

This will get very close.

Amy, Barb, and Catie pick up the rope. Amy holds one end of the rope while Barb holds the other end. They walk to each other and now Amy is holding two ends of the rope and both sides are the same. Barb holds the middle. Catie picks up the rope in something that feels like the middle of the folded 2 sides. Barb and Amy walk to each other while Catie holds her section. She adjusts her grip until she is holding the middle of the folded rope. The rope is now folded twice and there are 4 "equal" sections. Now that Catie has the folded middle, she holds it while Amy and Barb walk back and she feels a pull on each side that "feels about like a 90 degree pull." All 3 women set down their rope. Using The Pythagorean Theorem, Catie counts X steps heel-toe from her corner to Amy. Catie adds X^2 + X^2, then takes the square root of that sum and makes sure Amy and Barb are that distance apart by walking from one to the other heel-toe. Catie will then return to her corner, take the other folded corner of the rope and walk it heel-toe to the opposite 4th corner, dropping it at the length of the hypotenuse, ensuring it is the same distance as the distance from Amy to Barb. As the sides are equal and the hypotenuses are equal, this will be a nearly perfect square.

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ You can create a perfect 90 degree angle by making sure you can form a 3-4-5 Pythagorean Triangle, for example again in heel-toe steps. Creating two angles at the same rectangle side then straightening the ropes could be more stable at creating straight edges all around than taking the long walk through the middle. $\endgroup$ – HeroicKatora Jul 21 '17 at 15:50
  • $\begingroup$ You're suggesting dividing in 4 sides, using one side in a 345 triangle to get to 90 degree angle, then laying the rest out by step count? Interesting approach. I'll remove my statement of confidence. 🙂 $\endgroup$ – Forklift Jul 21 '17 at 15:53
3
$\begingroup$

This is supposed to be done on the floor. I consider a 6 meter rope "damn long rope" as you said. This probably the longest rope three persons can handle.

With a longer rope it probably still possible to do it with this recipe slightly modified and perhaps less perfect.

6 steps and 3 stick figures viewed from top (except step 7 :) ...

Steps with stick figures

    • Alice holds the ends of the rope and
    • Bart pulls the rope away from Alice.
    • Alice still holds the ends,
    • Caren grabs the middle of the rope half and holds it while
    • Bart pulls the end towards Alice.
    • Caren releases as much of the rope as necessary to bring the loop to Alice
    • Alice holds the first loop and one end in place while
    • Bart holds the second loop and the other end in place.
    • Caren then pulls the third loop 90° towards Bart to make an L-shape
    • Alice and Bart hold the ends and the first and third loop in place.
    • Caren pulls the second loop away from the two ends to the opposite corner
    • Everybody is happy
|improve this answer|||||
$\endgroup$
  • $\begingroup$ My problem with this is the 90 degrees bit $\endgroup$ – Strawberry Jul 24 '17 at 6:35
  • $\begingroup$ Agreed with Strawberry. How do you know when you've hit 90 degrees? You are blindfolded and can't eyball it. $\endgroup$ – Trenin Jul 25 '17 at 13:40
2
$\begingroup$

Have Teammate A, B each use their right hand to touch their own's left hand's elbow while their left hand pointing forward, forming a 90 degree angle. Now have them using their left hand to touch each other's right elbow thus forming a square, which is easy enough even blindfolded. Now have Teammate A and B, while maintaining the square, their hands on the ground.

This result in forming a square with their hands on the ground. Teammate C will now fill the inner square with the rope, with the 'frame' formed by Teammate A and B.

One can use their legs too, but as colleagues I won't recommend to do so...

|improve this answer|||||
$\endgroup$
  • $\begingroup$ This assumes that A and B's body dimensions are the same. :) You could do this but you need to align to whomever has the shortest dimensions. $\endgroup$ – LeppyR64 Jul 21 '17 at 14:17
  • $\begingroup$ @LeppyR64 probably! haha I would assume even with different body dimensions it wouldn't be too hard to adjust, in worst case a rectangle. Touching (sense) is not hard while blindfolded $\endgroup$ – Alex Jul 21 '17 at 14:22
1
$\begingroup$

Well, I have a way !

Assume the rope is of length 12m and it has two open ends (that means it is not a ring / closed one!)

Decide the five points which are equi-distant among them, one-after-another

This can be done by two persons(say A,B) holding each end and the third somewhere in between and then A and B move towards each other with C remaining fixed at a point (With this iteration, we got 2 x 6 m) Repeat the above step to arrive at five points which divide the 12m rope in four equal parts, each having 3m lenght and identified as shown below

|--------|--------|--------|--------|

Al-----Br-----Bl------Cr-----Cl

where Al represents A's left fist, Ar represnts A's right fist and so on.

Then comes the crucial part (traversing in semi-circular path)

Let B hold the rope tightly with both his fists and let A start walking with the lenth of the rope AlBr and keeping the rope tight so that he walks along a semi-circular path (and while walking let him count the steps) till he reaches Bl and thereby should walk-back half of the steps to arrive at a point which is one vertex of the required square (Br is first vertex) and

Repeat the same with C

But in this case the diagram changes as shown below:

|--------|--------|--------|--------|

Al-----Ar-----Bl------Br-----Cl

there by fixing the remaining vertices(Cr and another one found by the above operation) of the square.

And I bet this is the most precise way to arrive at a near square !

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Deciding upon 5 equidistant points seems non-trivial for 3 blindfolded people. The steps for that decision should be a part of your solution. $\endgroup$ – Forklift Jul 21 '17 at 15:02
  • $\begingroup$ @Forklift ... please look at the description / steps I mentioned. They are described. $\endgroup$ – Mea Culpa Nay Jul 21 '17 at 15:03
  • $\begingroup$ Ah, ok that part is same as mine. The next steps are different. $\endgroup$ – Forklift Jul 21 '17 at 15:04
1
$\begingroup$

This may take some time, but here is an idea. No complex math, but it does make use of the fact that the two diagonals of a square are the same length.

The end result will look something like this:

enter image description here

Obviously with smaller loops in the corners, and one corner will not have a loop, but simply the crossed ends using up the same amount of rope you would find in the loops (i.e. as if the rope was cut in the centre of one of the loops)

Our three people are Adam, Bill, and Catherine.

Find the Middle of the Rope

First off, all three stand within arms reach of each other and access to the rope. They first must split the rope in half. Adam will volunteer to take the two ends of the rope and Bill will hold onto a middle section. Adam will stay still, and Bill will pull the rope away letting it slide through his hands until both sides are taught. Bill now has found the middle of the rope and won't let go. He returns to where Adam and Catherine are and makes a tight loop as seen in the picture.

At this point, Bill will "measure" the length of the loop in his rope with his forearm or handspans. This is key to ensure Adam's sides don't end up too long. Adam will use this information to overlap his ends in the same manner so that an equivalent amount of rope is not part of the square at his corner.

Find the middle of a side

Adam will continue to hold the ends, and Bill hold the middle. Catherine will pick up both ropes on the ground and they will repeat the process. This time, Catherine will move away from Bill and Adam letting both ropes slide through her hands. When all 4 segments become taught, they have found the two middle sections.

Making a (rough) Right Angle Triangle

Catherine will make loops in the ropes as in the picture. Catherine will continue to hold the middle of both ropes tightly, one in each hand, with one hand directly on top of the other. Adam and Bill will begin to move away from each other and Catherine will tell them when to stop when she figures that they are approximately 90 degrees.

Making a (perfect) Rhombus

Now Catherine will drop the bottom rope (being careful to preserve the loop where the middle is) and walk forward with the top rope until both sides are taught (the other ends are held by Bill and Adam). Assuming they have done everything correct, they now have a perfect Rhombus. The only source of error is Catherine's judgement on 90 degrees, which we will correct in the next step.

Fixing the diagonals

Bill will drop his rope, careful to keep the loop in tact. He will walk to the corner not occupied by Catherine (i.e. where Catherine dropped her bottom rope). Catherine will now return to the far corning walking heel-to-toe keeping an exact count, using Bill's voice to guide her. She should be able to get a very accurate measure this way and can retry if she wanders too far off course.

She will now walk to Adam and Bill will return to his original corner. Catherine will repeat her measurement. Whichever diagonal is longer, they will fix by moving in a bit. Bill and Catherine will go to those corners, and Adam at a third. They will move in slightly and Adam will stretch to keep the rope taught. Adam will then walk to the other corner and tighten that as well.

The measurement process is repeated until both diagonals are within tolerance (one of Catherine's feet).

At this point, they should have a nearly perfect square.

|improve this answer|||||
$\endgroup$
1
$\begingroup$

Make the square hang vertically with one side on the ground.

  1. Tie the two ends together at some appropriate length to make a loop, with a circumference of about 4-6 feet.
  2. Person A and B can each grab one of the top two corners, if they stand close enough it should not be hard for them to make sure they are level with each other
  3. Person C can then approximate the distance between the person A and B and the floor by feeling where the top corners and the floor are.
  4. Also person C can feel around for slack in the rope that needs to be picked up.

This does depend on the exact rules off the task though and is easier with a smaller square.

|improve this answer|||||
$\endgroup$
1
$\begingroup$

I apologize in advance for the wall of text.

This solution will use the full length of the rope, assuming that one twelfth of the rope is within the distance of one individual's arm's reach. If the rope is longer, the remainder can be ignored, as only one end-point is needed and not using the entire rope is within the requirements of the puzzle.

For sake of notation, one end of the rope will be referred to as point A and the other end will be referred to as point B. The section of rope between two points will be referred to as the point names, e.g. section AB is the section of rope from point A to point B. A single unit will be one twelfth of the length of section AB. People will be referred to as P1, P2, and P3. For diagrams, each dash (-) represents one one unit of rope and the holders of the point will be marked.

1: measure one third of section AB. This can be done by having P1 tightly hold point A and loosely hold a bend in the rope close to point B, while P2 is tightly holding point B and loosely holding a bend in the rope close to point A. As they walk away from each other and pull the rope taut, the loosely held sections will slip through their hands to separate the rope into 3 parts. The loosely held points will be at 4 units and 8 units from point A. The point at 4 units from A will now be known as point C.

A(P1)----C(P2)----(P1)----B(P2)

2: P1 will drop their loosely held point. P2 will give point C to P1's free hand. P3 will loosely hold section AC and stretch the rope taut to discover the center point of section AC. This midpoint will be D.

A(P1)--D(P3)--C(P1)--------B(P2)

3: C will be given to P3's free hand and D will be given to P1's now free hand. Using his newly freed hand, P3 will repeat the above process to find the midpoint of section AD, which will be point E. P1 will drop D and take E from P3.

A(P1)-E(P1)---C(P3)--------B(P2)

4: P1 and P3 will pull the section CE taut. P2 will run section CB along the section CE and hold the point at which it meets point E, which will be point F. After F is found, P1 will drop E.

A(P1)----C(P3)---F(P2)-----B(P2)

5: At the point, section AB is separated into 3 sections, of lengths 4, 3 and 5 units. P2 will give P1 F and afterwards, receive A from P1.

A(P2)----C(P3)---F(P1)-----B(P2)

6: P2 will hold points A and B together and all three sections will be pulled taut. The only way for all three section to be taut is for the angle at point C be a right angle. At this point, the rope should be on the ground to prevent it from moving while not attended. P2 will abandon A in its current position.

A----C(P3,right)---F(P1)-----B(P2)

7: P2 will run section FB along section FC. Once P2 reaches C, P3 will hold the point where FB reaches C. This will be point G.

A----C(P3,right)---F(P1)---G(P3)--B(P2)

Note: C and G are at the same location

8: P2 will run section GB along section CF, holding the point along CF where B reaches. This will be point H. P3 can now drop point G and P1 will hold point point H, while making sure not to move point F (this is the reason for the one unit being within arm's reach). P2 will move back to A and hold onto it. Given that C and F have not moved, the right angle can be confirmed again.

A(P2)----C(P3,right)--H(P1)-F(P1)-----B(P2)

9: This is where it gets tricky. P2 will loosely hold a point at least one unit from point A along section AC to ensure that the angle remains unchanged. P3 will ensure that the rope does not leave point C, but will allow it to slide freely through that point. P1 will keep their hand on the ground where point F is. After this preparation, P1 will pull point H to where their hand marked point F on the floor and hold it there. This will cause AC to shorten by 1 unit (the length of section HF). P2 will adjust B to the new position of A.

A(P2)---C(P3, right)---H(P1)------B(P2)

10: P1 will use their free hand to swing the loose section HB in P3's direction. P3 will catch it, while making sure not to move C, and pull HB against point C, marking the point where section HB meets point C. This will be point I.

A(P2)---C(P3, right)---H(P1)---I(P3)---B(P2)

11: P3 will leave C and pull point I to the opposite corner until it becomes taut. This should result in a square.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Enthusiastic ^vote with a note: This looks promising and direct but could use a diagram or two to make it convincing. Probably not worth the effort though. $\endgroup$ – humn Jul 24 '17 at 7:26
1
$\begingroup$

I'm not sure how easy this would be to do blindfolded - but here's my thought - and it requires some fore-knowledge (or at least some arithmetic)

A square of sides 2.5 units has a hypotenuse of roughly 3.5 units (ok - not exactly, but near enough).

So...

  1. Divide the rope in half.
  2. Then divide one half into thirds (this is actually pretty easy to do - even blindfolded.
  3. Now exclude the first third (2 units). The remainder of the rope is 10 units. We need to take care to keep track of both the 10 unit mark and the 11 unit mark, so someone needs to grip both these positions, but they can't tie a knot - we still need to reference the overall length of the rope!
  4. Divide the 10 unit length in half and half again - 4 lengths of 2.5 units
  5. Select the first 2.5 unit length from the other end. The distance from this length to the middle of the whole rope is 3.5 units.
  6. Using any section of the other end of the rope (but still keeping track of the final unit position, form another length of 3.5 units to match. Using tension, we can now lay out a right angle isosceles triangle of sides 2.5, 2.5, 3.5.
  7. So we have two sides of the square, and 6 units until we reach the final unit length. From here, it's trivial to form two lengths of 2.5 units and 3.5 units, which we can use to position the far corner of the square, and finally close the square.
|improve this answer|||||
$\endgroup$
1
$\begingroup$

Initial analysis:

  • Don’t have to use all the rope.

    Extra rope may be used...

    ...for measurement (diagrams A and B below) and for a guide frame (diagrams B and C).

  • We had 3 players on our team.

    With three teammates, two teammates can...

    ...each hold 2 corners of a 4 - cornered guide frame (diagrams B and C).

        ...while the third teammate constructs the square.

  • Rope may be 2 inches in diameter.

    A rope with thickness can...

    ...form a rope-filled solid square, not just an outline.

Almost-self-explanatory diagrams:


(Click here to keep these diagrams visible.)


Recipe:

All of this is performed on the floor and can be done by feel and communication.

A. Make a “unit” segment of rope by tying two knots that are n thicknesses of rope apart, as in diagram A above.

B. Use that unit length to measure two overlapping 3 – 4 –5 triangles as in diagram B. Two teammates each hold two corners taut of what constitutes a guide frame that has two adjacent interior right angles separated by a gap of (3n  −1) × thickness.

C. The third teammate lays 3n  −1 back-and-forth stretches of rope on the floor, under the guide frame, as in diagram C. This forms a perfectly square area of rope.

  D. All other rope is moved aside, leaving just the square with sides  (3n  −1) × thickness  long.

(If anyone insists on an outlined square, just stretch rope around the perimeter of what you have and remove the insides.)

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.