May 18, 2009

Maths Challenge #5 – Après–le–Ski

The little-known Alpine village of Après-le-Ski is situated in a deep mountain valley with vertical cliffs on both sides. The cliffs are 600 metres high on one side and 400 metres high on the other. A cable runs from the foot of each cliff to the top of the other cliff, and the cables are perfectly straight.

Apres-le-Ski
At what height above the ground do the two cables cross?

(Picture not to scale)


- 9 comments by 3 or more people Not publicly viewable

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  1. y = 400x
    y = 600 – 600x

    Solve for y.

    18 May 2009, 12:15

  2. @healthmed

    Worth patenting the cabling materials.
    Answer a factorial sequence/2.

    18 May 2009, 14:00

  3. Eleanor Lovell

    The answer is 240m.

    It’s simpler to tackle a more general problem, where the lengths are as shown. By similar triangles,

    \frac{x + y}{a} = \frac{y}{c} and \frac{x + y}{b} = \frac{x}{c}

    Adding, we get

    (x + y)(\frac{1}{a}+\frac{1}{b}) = \frac{x + y}{c}

    Dividing by x+y, we obtain

    \frac{1}{a}+\frac{1}{b}=\frac{1}{c}

    leading to

    c = \frac{ab}{a + b}

    We notice that c does not depend on x or y, which is a good job since the puzzle didnt tell us those. We know that a = 600, and b = 400, so

    c = \frac{600 \times 400}{1000} = 240

    So, the point at which the cables cross is at a height of 240m

    19 May 2009, 11:47

  4. Iain

    This is where I was trying to go with my solution but forgot about the neat ratio of sides idea. Jim’s simultaneous linear equations solution is by far the easiest method though – you could pretty much do it in your head from that.

    19 May 2009, 13:13

  5. Simon Whitehouse

    My working was a bit more spurious.

    The point where the cable crosses is a 6:4 ratio along the horizontal (spurious bit #1 – I’m sure it’s true but didn’t justify it). The horizontal distance is irrelevant so make it easy and use 1000m (spurious bit #2 as it doesn’t give you a more general answer to the problem.

    So, x = 600m and y = 400m

    Then I used similar triangles, using the 600m vertical on the left

    a/(x+y) = 600/1000 = 3/5

    Apply that to the smaller triangle :-

    (a-c)/x = 3/5

    (600 – c)/600 = 3/5

    600 – c = 360

    c = 240m

    19 May 2009, 16:37

  6. I do like all this triangle stuff, but I also like to think that the straight line geometry that I used just makes a mockery of it all, even if the short cuts of y=mx+c are equivalent to a less Cartesian approach.

    Oh, and:

    19 May 2009, 22:09

  7. Mathew Mannion

    For those who like left-field maths challenges, I highly recommend Jim Miles’ (above) Wednesday Maths Problems: http://blogs.warwick.ac.uk/wednesdaymaths/entry/links_to_all/

    20 May 2009, 09:51

  8. I sooooooooooo need to update that and do some more.

    I love that you call them “left-field” – I don’t think of them as such, but I guess the style is pretty loony.

    20 May 2009, 17:18

  9. David

    although it’s not the easiest resolution, that was the one that sprang in my head.

    Take a referential using the 600m cliff.
    Consider the common base A

    The equation of its cable would be given by:
    Y1= -600x/a + 600
    The equation of the 400m cliff would then be given similarly by:
    Y2=400x/a

    solving the equation Y1=Y2 comes
    x=6a/10

    using that value in any of the above expressions, comes that

    Y2= 400/a * 6a/10 = 400*6a/10a = 240 m.

    the height is thus 240m.

    30 May 2009, 23:53


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