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Topic: Two ladders (Read 3266 times) |
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NickH
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Two ladders are placed cross-wise in an alley to form a lopsided X-shape. Both walls of the alley are perpendicular to the ground. The top of the longer ladder touches the alley wall 5 feet higher than the top of the shorter ladder touches the opposite wall, which in turn is 4 feet higher than the intersection of the two ladders. How high above the ground is that intersection?
Nick
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| « Last Edit: Sep 20th, 2003, 8:41pm by Icarus » |
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Icarus
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Re: NEW PUZZLE: Two ladders
« Reply #1 on: Oct 12th, 2002, 8:00pm » |
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A little playing with similar triangles gives me an answer (which I won't mention to avoid spoiling), but I should point out that it requires an additional assumption: the bottom of both ladders are against the opposite walls of the alley. Otherwise, it is only the maximum possible height. To see this, cut both ladders off the same distance above the ground, and slide the assembly down until they are on the ground again. All the conditions of the original puzzle are still intact, but the intersection height is now lower.
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NickH
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Re: NEW PUZZLE: Two ladders
« Reply #2 on: Oct 13th, 2002, 3:35am » |
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Good point, Icarus, I should have made that clear.
The nice thing about this puzzle is that the height of the intersection is independent of the width of the alley!
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steven_s
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Re: NEW PUZZLE: Two ladders
« Reply #3 on: Oct 23rd, 2002, 1:28am » |
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is it 6 ft?
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| « Last Edit: Oct 23rd, 2002, 2:56am by steven_s » |
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S. Owen
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Re: NEW PUZZLE: Two ladders
« Reply #4 on: Oct 23rd, 2002, 8:12am » |
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That's it. One kind of interesting generalization you can make is that if the ladders touch the walls at heights y and z in the manner described in the problem, then the height x of their intersection satisfies: 1/x = 1/y + 1/z. That still works if the walls are slanted, but still parallel... but then x is the "slanted" height of the intersection.
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Anonymous
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Re: NEW PUZZLE: Two ladders
« Reply #5 on: Nov 12th, 2002, 4:51pm » |
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I still don't get it. Can someone please post a solution? 
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Icarus
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Re: NEW PUZZLE: Two ladders
« Reply #6 on: Nov 12th, 2002, 5:34pm » |
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A couple of hints:
Hint 1: By adding a few horizontal and vertical lines you can form two pairs of similar triangles.
Hint 2: Just because you have more unknowns than equations does not mean you can't solve completely for some of the unknowns.
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Anonymous
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Re: NEW PUZZLE: Two ladders
« Reply #7 on: Nov 13th, 2002, 9:19am » |
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I examined every similar-triangle relationship in the diagram, but I still have x as a function of u and v (where v is the distance from the left wall to the intersection, and u is the distance from the right wall to the intersection). If I knew the ratio of u:v then I could solve it, but there's nothing in the problem to give that value. 
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S. Owen
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Re: NEW PUZZLE: Two ladders
« Reply #8 on: Nov 13th, 2002, 11:35am » |
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Let y be the height where one ladder touches the right wall, let z be the height where the other ladder touches the left wall, and let x be the height of the intersection. Let u be the horizontal distance from the intersection to the right wall, and let v be the horizontal distance from the intersection to the left wall.
First...
u/x = (u+v)/z, and
v/x = (u+v)/y
by similar triangles.
and then...
Add those equations together:
(u+v)/x = (u+v)(1/z + 1/y), or
1/x = 1/y + 1/z.
Since y = x+4 and z=x+9 you can solve this.
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James Fingas
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Re: NEW PUZZLE: Two ladders
« Reply #9 on: Nov 13th, 2002, 12:03pm » |
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An alternate solution (maybe a little simpler?)
Label the points at the bottom of the walls A and C. Label the point above A as B, and the point above C as D. Label the intersection E. AB is longer than CD.
Notice that ABE and DCE are similar triangles. Call the height of point E x. Now notice that 4 on the small triangle corresponds to x on the large triangle. Also notice that 9 on the large triangle corresponds to x on the small triangle.
AB/CD = 9/x
AB/CD = x/4
Now combine to get 9/x = x/4, so x is the geometric mean of 9 and 4. That is to say,
x = 6
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| « Last Edit: Nov 13th, 2002, 1:42pm by James Fingas » |
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Re: NEW PUZZLE: Two ladders
« Reply #10 on: Nov 13th, 2002, 7:35pm » |
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Ah that clears it up. Thanks S. Owen! 
To think, adding equations together yields new information... cool.
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