Friday, June 1, 2012

The Man Who Walked Between The Towers

You have probably heard about the daredevil Philippe Petit who sneaked on top of the World Trade Center and walked between the towers on a tightrope in  the August of 1974. I learned about him from a wonderful picture book by Mordecai Gerstein: The man who walked between the towers.  Here is one of the illustrations:

Some of you also likely saw the documentary made about Philippe Petit: Man on Wire and the animated film based on the above book.

A book, a documentary, an animation. Now it is time for a puzzle.

When Philippe Petit stepped onto the tightrope it was horizontally stretched between the towers. Under his weight, the cable hang a bit down. When Philippe reached the midpoint, the cable under his feet was 10 feet below its original horizontal position and the overall length of the cable got extended by 1 foot. What is the original length of the cable between the towers?

Your answers are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.


SteveGoodman18 said...

The original cable was 199.5 feet long.

If we let x = half of the length of the original wire, then x and 10 are the legs of a right triangle, with hypotenuse x + 1/2. (The 1/2 comes from the extra foot when the cable is stretched - 1/2 a foot on each side).

Our good friend Pythagoras shows that the solution to x^2 + 10^2 = (x + 1/2)^2 is x = 99.75. Since this is half the length of the original cable, the overall length is 99.75 * 2 or 199.5 feet.

Jerome said...

Let the original length be x. In the middle, you will create two right angle triangles from 1 isosceles triangle.

Let the hypotenuse be (1/2)x + 1/2 feet.
Let the original length of the horizontal leg = (1/2)x
Let the third side be the vertical increase in dip = 10 feet.

((1/2)x + 1/2)^2 = 10^2 + ((1/2) x )^2
Multiply through by 4. (Get rid of unnecessary fractions).

(x + 1)^2 = 400 + x^2 Expand (x + 1)^2 and cancel out the x^2s.
2x + 1 = 400
2x = 399
x = 199.5

Seems like I'm getting that the distance between the towers is 199.5 feet.

The distance given in this article agrees with my answer.

Anonymous said...

This can be solved (at least that's how I worked on it) using the Pythagorean theorem that states that the squares of the sides of a right triangle add up to the square of the hypotenuse: a^2 + b^2 = c^2

My answer: the original cable between the towers was 199 1/2 feet in length.

x is the original length of the cable between the towers (in feet)
x+1 is the stretched out length of the cable (in feet)
x/2 is how far Petit has gone out on the original (pre-stretched) cable (feet)
10 feet is how far vertically the cable has moved from its original position because of Petit's weight

a = x/2
b = 10
c = x/2 + 1/2 = (x+1)/2

a^2 + b^2 = c^2
(x/2)^2 + 10^2 = ((x+1)/2)^2
(x^2)/4 + 100 = (x^2 + 2x + 1)/4
x^2 + 400 = x^2 + 2x + 1
400 = 2x + 1
2x = 399
x = 199 1/2 feet


Ilya said...

The original cable and the stretched one with Philippe in the middle comprise an isosceles triangle. The original length is X, and the equal sides each are X/2+1/2. The altitude is 10. The right triangles formed by the altitude have sides of X/2, 10 and X/2+1/2. With the help of dear old Pythagoras, we have X^2/4 + 100 = (X/2 + 1/2)^2 or X^2/4 + 100 = X^2/4 + X/2 + 1/4. From this we have X/2 = 99.75, so X=199.5

kalyan said...


Anonymous said...

I know the original length was 140 between the towers. However, I have no idea on how to do this mathematically. A thought experiment is by using rope or even string and the rope or string would sag more but it seems like one would have to know the tensile strength of the cable to know the distance between the buildings since different tensile strengths of different "fabrics" would be different.amount of "sag" and extended length. I am interested in how one would figure this out.


Ryan said...

Let the distance be d. Half way across he has travelled d/2 feet (horizontally). Now the rope has stretched 1 foot, so he has (d+1)/2 feet between the bottom of his foot and the opposite tower. The bottom of his foot is 10 feet below where it would have been had the rope not stretching making a right triangle. (This is much easier to see with a diagram). By Pythagorean Thm this gives us:

(d/2)^2+10^2=((d+1)/2)^2 -expand and multiply by 4

d^2+400=d^2+2d+1 - solving for d


Thad said...

Using the Pythagorean Theorem, with x as the original length of the cable, we have:
(x/2)^2+10^2 = ((x+1)/2)^2
which has solution 399/2.

The cable was originally 199' 6" feet long.

anne-marie said...

I will use the Pythagorean theorem.
The tangent being (L+1)/2
We have,
(L/2)^2+ (10^2) = ((L+1)/2)^2
I solved for L and found 199.5 feet

Annie said...

The length of the tightrope was 199.5 ft.

Imagine an isosceles triangle with the base being 2x, representing the length of the tightrope before the walk, each side being x+ 1/2 ( to account for the increase in length of 1 ft.) and the height from the midpoint of the base being 10 ft. If we use the right triangle to solve for x we can use: a2 + b2 = c2 or x2 + (10)2 = (x+ 1/2)2. If we solve for x, the length to the midpoint was 99.75 ft.

X2 + 100 = x2 + x + 1/4
X = 99.75
2x = 199.5

Therefore the entire length of the tightrope before Phillippe's walk was 199.5 ft.
(If you are receiving this twice, I apologize. My new iPad is not doing what I expected so I'm trying it again!)

Laura W said...

199.5 feet

Let x=the original length of the cable. The midpoint of the cable is x/2. The depth at the midpoint, with Phillippe on it would be 10 feet. The hypotenuse of the right triangle would be x/2 + 1/2. (x/2)^2 + 10^2 = (x/2 + 1/2)^2

anne-marie said...

I did not mean the tangent but the hypotenuse.

Maria said...

You all are absolutely right. It is a geometrical puzzle. Below is a link to the image. Remember to click "back" button after viewing it. Google doesn't allow to open it in a new window. A puzzle point for everyone who answered.
Click to see the sketch

Laura W said...

Actually, we are all probably wrong, as the cable would have had to have been somewhat longer to anchor it at each end.

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