A very cute puzzle from a great book by Ian Stewart "Professor Stewart's Cabinet of Mathematical Curiosities."

The little known village of Apres-le-Ski is situated in a deep mountain valley with vertical cliffs on both sides. The cliffs are 600 meters high on one side and 400 meters high on the other. A cable car 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? To make your explanations easier I marked with "h" the height we are looking for and with "a" and "b" parts of the base that you may want to use in your calculations. Remember that "h", "a" and "b" are all unknowns and we want the answer for "h" as a number.

The little known village of Apres-le-Ski is situated in a deep mountain valley with vertical cliffs on both sides. The cliffs are 600 meters high on one side and 400 meters high on the other. A cable car 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? To make your explanations easier I marked with "h" the height we are looking for and with "a" and "b" parts of the base that you may want to use in your calculations. Remember that "h", "a" and "b" are all unknowns and we want the answer for "h" as a number.

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

## 6 comments:

Short answer h = 240.

I'm going to do this rather a long way. There are much shorter methods. The results are kind of amazing.

Let h1 = 600 m

Let h2 = 400 m

From similar triangles we get the following relationships for h1 and h2

h1/(a + b) = h/b . . . . . (1)

h2/(a + b) = h/a . . . . . (2)

Transform 1 into

h1/h = (a + b) / b (2) has a similar appearance.

h2/h = (a + b) / a

Divide the right hand side of both equations.

h1/h = 1 + a/b . . . . . (3)

h2/h = 1 + b/a . . . . . (4)

Transfer the 1 to the left hand side for both equations.

h1/h - 1 = a/b . . . . (5)

h2/h - 1 = b/a . . . . (6)

Use just 5 for the moment.

(h1 - h)/h = a/b Turn this into a reciprocal.

h/(h1 - h) = b/a which looks the same on the right as equation 6. So equate them.

(h2/h) - 1 = h/(h1 - h) Get a common denominator for the left side.

(h2 - h)/h = h / (h1 - h) Cross multiply. [ Can I use a term like that? ]

(h2 - h)(h1 - h) = h^2 . . . . . (7)

h2*h1 - h*h1 - h*h2 + h^2 = h^2 Cancel h^2

h2*h1 = h(h1 + h2) Divide by h and h2*h1

1/h = (h1 + h2)/h1*h2 = 1/h1 + 1/h2

The result is

1/h = 1/h1 + 1/h2

If you have suffered through high school physics, you will notice that that is the formula for resistors in parallel ie

1/Rt = 1/r1 + 1/r2.

The total distance (a + b) is not relevant. That distance can be anything from 1 inch to 40 miles. h will be the same as long as h1 and h2 remain the same throughout your calculations using various a's and b's.

h is equal to 240 meters. The classic "two-pole" geometry problem has a solution that is independent of the distance between the two cliffs. The solution is the product of the heights divided by the sum of the heights.

We could prove this with geometry and algebra - either using coordinates and equations of lines or with similar triangles and lots of ensuing algebra. Either proof gets messy for a comment board.

Tough one for me.(Probably the cables should not actually cross, not touch, not exactly.)

The diagram may be misdrawn, for the problem states the 600 meter cable should extend to the FOOT of the 400 cliff. The sum of a+b will the base of both right triangles, I hope. So the hypotenuse of the taller triangle will always be longer also, but the diagram makes that unclear, or less clear.

I think the sum of a+b probably doesn’t matter, but the ratio of a:b matters. And that base could 50 or 100 meters, or 2k, and I think h stays the same.

H >200 (half of 400), and <300 (half of 600). 250 is tempting, as is 225.

Maybe the right-left placement of h doesn’t really matter, or maybe it does, but h will always be at the same near-midpoint if the relative heights are the same 3:2.

The l/r position of the h line will be at the same ratio-point along ab, regardless of the sum length of ab.

I fear I need to use some trig (sin) in this problem, and have forgotten it all. Can’t do this one yet. Maybe tomorrow. Maybe not.

I labeled points: The line labeled 400 got 2 points, 'A' at the top and 'B' at the base. Line with length a+b got point 'C' on the left and 'B' on the right. Line with length h got points 'D' at the top and 'E' at the base. Line labeled 600 got points 'F' at the top and 'C' at the base. The diagram is broken into similar triangles ABC and CDE, as well as similar triangles BCF and BED. Therefore AB:DE=a+b:a and FC:DE=a+b:b. Then you substitute and get 400:h=a+b:a and 600:h=a+b:b. Then you can cross multiply and get 400a=ha+hb and 600b=ha+hb. I then solved for 'a' in terms of 'b' and got that 2/3*a=b. Then I need only one equation 400a=ha+h(2/3a) which equals 400a=ha+2/3ha which equals 400a=5/3ha. I can then divide by a and get 400=5/3h. Then multiply each side by 3/5 to get 240=h. The height at which the cables cross (h) is equal to 240 m.

h = 240

b/h = a +b/600 and a/h = a+b/400

a = 3/2b and b = 2/3a

b/h = 5/2b/600 h = 240

and

a/h = 5/3a/400 h = 240

240 it is.

I am repeating Annie's solution, just inserting some line breaks. Similar triangle rule states that if one right triangle is inside another, then their heights and bases are proportional. From that:

b/h = (a+b)/600

and

a/h = (a+b)/400

expressing h from both of these equations and making one equal another we find that:

a = 3/2b

b = 2/3a

substituting this back into first equation we get:

b/h = (5/2)b/600

solving for h:

h = 240

Puzzle point for everyone who dared to play with this!

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