### 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.

At what height above the ground do the two cables cross?

(Picture not to scale)

## 9 comments by 3 or more people

[Skip to the latest comment]y = 400x

y = 600 – 600x

Solve for y.

18 May 2009, 12:15

## @healthmed

Worth patenting the cabling materials.

Answer a factorial sequence/2.

18 May 2009, 14:00

## Eleanor Lovell

The answer is 240m.

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

andAdding, we get

Dividing by , we obtain

leading to

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

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

19 May 2009, 11:47

## 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

## 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

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

Cartesianapproach.Oh, and:

19 May 2009, 22:09

## 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

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

## 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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