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Geometry problem |
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| Nov10-04, 11:21 AM | #1 |
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Geometry problem
Consider an alley with two ladders placed crosswise.
Kinda like this sucky diagram below: [tex] \begin{array}{cccccccc} |&* & & & & & &| \\ |& &* & & & & &| \\ |& & &* & & &&/| \\ |& & & &* &/& &| \\ |& & &/& &* & &| \\ |&/& & & & &* &| \end{array} [/tex] ( Ugh, I hope you get the idea)One ladder has a length of 2 m. The other is 3 m long. They cross each other at a height of 1 m. What is the width of the alley? It's not hard to see there's a unique solution to this question. Good luck
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| Nov10-04, 12:15 PM | #2 |
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[tex]\sqrt{9-w^2} + \sqrt{4-w^2} = \sqrt{(9-w^2)(4-w^2)} [/tex]
This might look pretty, but it's both messy and inelegant...there's got to be a nicer way. PS : That gives me w = about 1.231m |
| Nov10-04, 12:36 PM | #3 |
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The solution to these 'crossed ladder' problems, except in trivial cases, is a quartic. They are almost always messy:
spoiler - highlight or Ctrl-A to view w^8 - 22w^6 + 163w^4 - 454w^2 + 385 = 0 w is about 1.2312 m. Best found (IMO) by successive approximation, ie. guessing. |
| Nov10-04, 03:41 PM | #4 |
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Geometry problemOn the left side I see the height of the 3m ladder plus the height of the 2m ladder and on the right side the product. Why are these equal? |
| Nov10-04, 06:03 PM | #5 |
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I don't know how they did things, but there are lots of similar triangles to pick from. It seems like that should be enough to work things out.
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| Nov23-04, 09:20 PM | #6 |
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could someone explain this one for me? this intrigues me
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| Nov24-04, 04:14 AM | #7 |
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Me too. C'mon Gokul43201!
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| Nov24-04, 03:54 PM | #8 |
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Galileo,
this is sort of a hint and i am not sure whether Gokul did it this way... Let the ladders be L1 and L2. Let H_L1 be the height at which ladder L1 meets the wall Let H_L2 be the height at which ladder L2 meets the wall can u show that, 1/H_L1 + 1/H_L2 = 1 -- AI |
| Nov24-04, 03:59 PM | #9 |
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Here's a hint: highlight to see
Use trigonometry (cos, sin, etc.......). It's not a difficult problem as you will see, the solution is quite "simple." |
| Nov24-04, 04:08 PM | #10 |
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| Dec18-04, 11:32 AM | #11 |
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Eureka !! I have discovered a truly unremarkable Al Gore Rythym to solve this teaser. *.102m Unfortunatley, it will not fit in this small margin. Get the idea? Thanx*encrypted so not to spoil it for others>12 clock arithmetic. |
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