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22 October 2005

AskMeCha— MATH Ok. I got challenged by a friend to figure this out, and even with the combined efforts of my incredibly large brain and the large brain of my pal the math major (now an actuary), we couldn't get the solution.

There are two poles, one 15 feet high, the other 9 feet high. If you draw a line from the top of one to the base of the other, and vice versa, the lines intersect six feet above the ground. How far apart are the poles?[More:]

I've already been told the solution, but I think I'll let some of you cast about for a moment. What I'm really interested in is how to get to the solution, which even the guy who gave me the puzzle couldn't accurately explain. The best we could figure out was by making similar triangles out of the 6:9 or 6:15, but we weren't even sure that told us anything all that useful.
i'm afraid you or your friend must have something wrong. the two lines always cross 5 feet above the ground, no matter far apart they are. at least, if you assume that they're so close that the earth is flat etc etc.

you can see that the height doesn't change with separation by drawing the diagram and then looking at the paper side-on so that the the poles look closer together - the heights don't change.

to get the 5 feet you need to do the maths with similar trangles. it's just a couple of simultaneous equations.
posted by andrew cooke 22 October | 11:16
ooops. that sounds rather absolute. the above is my understanding - i may, of course, be wrong.
posted by andrew cooke 22 October | 11:16
Hmm. He told us that it was six feet no matter how far apart they were. And while that may be why we couldn't figure it out (which suddenly makes the question kind of cruel), could you give me the equations that you used? I'm still a blank on how to figure out the other sides of the triangle (that's what you had to do, right?)
Or is it a sine/cosine thing?
posted by klangklangston 22 October | 11:20
(Unless, of course, I'm remembering it wrong... I'll have to ask him again... Surreptitiously)
posted by klangklangston 22 October | 11:22
oops. it's 5.625 feet above the ground! i was lazy and used a computer to divide two numbers, but it did integer division... :o)
posted by andrew cooke 22 October | 11:23
ah, maybe i've made a mistake. maybe it's 6 feet. i'm much more sure that it is the same height whatever the separation, rather than the absolute value.

sorry, didn't realise it was a trick question.

maths will follow next post...
posted by andrew cooke 22 October | 11:24
ok, i suspect there's a more elegant way to do this, but here's what i did. first draw a diagram, with 15 on the left and 9 on the right. cross-over is at height x and the distance from base of 15 to base of cross over is a. distance from base of crossover to base of 9 is b.

ascii art doesn't work here unfortunately.

from similar triangles (or tangent of bottom corner angles - sme thing):

15 / a+b = x / b
and
9 / a+b = x / a

rearranging:

15 . b = x(a+b)
9 . a = x(a+b) [A]

so equating x(a+b):

15 . b = 9 . a
a = 15/9 . b

substituting back into [A]:

9 . 15/9 . b = x(15/9 . b + b)

cancelling b from both sides:

9 . 15/9 = x (15/9 + 1)
15 = x . 24/9
x = 9 . 15/24
x = 9 . 5/8
x = 45/8
x = 5 5/8
x = 5.625
posted by andrew cooke 22 October | 11:34
and since you can get x indpendently of a and b (ie the separation), the height doesn't depend on the separation. that becomes more obvious if you try to solve for a and b - you can't.
posted by andrew cooke 22 October | 11:35
but to just see that the separation doesn't change the height, the best thing is to look sideways at the drawing, as i said earlier.

ok, back to work...
posted by andrew cooke 22 October | 11:41
Thanks so much!
posted by klangklangston 22 October | 12:19
NERDS!
posted by Eideteker 22 October | 13:58
it has been a good day. || I wanna buy some clothes

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