Now trying to remember my O Grade maths but I would assume its along the lines of the ladder becomes the hypoteneuse of a right angle triangle, (isosceles?) so its square equals the sums of the other two which I should be able to calculate knowing that each side is 6 + x...........
out with the calculator now....
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15ft from the base of the wall, 9ft from top of box
If you have any two dimensions of a triangle (side length or angle) you can work out any of the other side lengths or angles. You are also working with similar triangles.
Edited by daveyjp on 18/09/2008 at 13:48
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15ft from the base of the wall 9ft from top of box
What formula did you use?
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Two answers, it could also be 10ft from the base of the wall and 4ft from the top of the box - I wouldn't climb a ladder at that angle though!
EDIT: Wouldn't rather than would.
Edited by cheddar on 18/09/2008 at 14:00
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You are missing some information!
Firstly a box needs 3 dimensions. (unless you define it as a square (ie regular cube) box, in which case you only need 1 dimension!).
"rests against a corner" - do you mean edge or corner?
Is the box placed "square" agaist the wall?
pmh
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Is the answer 12.72 ?
If it is I will then try and explain how I did it!
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Sorry, the box is sat in the corner between the wall and the ground. The box is six feet out from the wall and six feet above the ground.
The ladder is resting against a point six feet out and up from the corner, with its botttom on the ground and its top against the wall.
I'm sure there's plenty of pythagorus in this and possibly even some trig. The angle of inclination, however, isn't 45° so the top part of the ladder above the box isn't 9'.
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I'm sure there's plenty of pythagorus in this and possibly even some trig.
Plenty of that. The answer is an infinite range roughly between 10.3 and 14.7 feet [depending on what angle you place the ladder ].
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There is only one solution where the ladder can touch the edge of the box, the ground and the wall.
Sorry, Bobby, the answer isn't what you calculated.
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I believe daveyjp and cheddar have the correct answers, but how?
You only have the 6" dimension as one of the sides of a triangle?
(Maths was never my strongpoint. I cheated and had to draw it on CAD.)
Edited by Rich 9-3 on 18/09/2008 at 14:18
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If I have understood the question correctly, then your following two statements seem incompatible:
"There is only one solution where the ladder can touch the edge of the box, the ground and the wall."
"The angle of inclination, however, isn't 45°"
So the computer says no, and I conclude that I must be reading your question wrong.
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There is only one solution where the ladder can touch the edge of the box the ground and the wall.
Nope, two solutions.
There would only be one solution if the ladder was angled at 45deg though it is not so there is not.
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Cheddar, if you imagine this in real life, you have the ladder standing against the side of the box. You slowly pull the bottom of the ladder across the ground, creating an angle. Surely only at one point will the top of the ladder touch the wall whilst the bottom is on the ground, and for it to be touching the corner of the box?
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Cheddar if you imagine this in real life >>
Bobby G, the imagine tipping the whole caboodle, wall, floor ladder and box on its side, so the floor becomes the wall and visa versa, it would still be a 6ft x 6ft box, the ladder would still touch it though the ladder angle would now be less than 45deg from the floor rather than more than 45deg from the floor.
EDIT: "from the floor" added.
Edited by cheddar on 18/09/2008 at 14:29
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Owing to the symmetry of the problem about a 45 degree line from the base of the wall, there must be an even number of solutions - most probably 2, and so, I wouldn't trust any answer that doesn't come via the solution of a quadratic (or quartic, etc, etc) equation.
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the ladder is the hypoteneuese of a right angle. It is 18, so squared is 324.
the other two sides are (6+x) and (6+y) but its finding out what x and y are
oh its lunchtime..... i give in...
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well, if you take the view that the ladder can be thought of as a straight line having
eqn y=ax +b
we know three points on the line
when x=0, y = a
when x=6, y=6 => b=6-6a
when y=0 x=-b/a
we also know that the distance (0,a) to (-b/a, 0) is 18 whichis where pythagorus comes in.
18^2 = a^2 + (b/a)^2
I think that leads to a quadratic eqn
2a^4 -4a^3 + 3a^2 -4a +36 = 0
but its sooooo long since I did those that I leave solving it as an exercise to the reader :)
just seen the link reply - my a and b include the 6
Edited by adverse camber on 18/09/2008 at 14:34
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Not as easy as it looks at first glance but quite a well known problem. This link gives you 2 methods to do the calculation:
www.mathematische-basteleien.de/ladder.htm
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If you call the distance from the edge of the box to the foot of the ladder x
and the distance from the top of the box to the top of the ladder y
you can write
(x+6)^2 + (y+6)^2 = 18^2
By similar triangles, you can also write;
6 / y = (6+x) / (6+y)
This is saying that the triangle formed by the upper edge of the box, the upper part of the wall, and the upper part of the ladder is the same shape as the triangle formed by the floor, the wall, and the entire ladder.
So, you have 2 equations, and 2 unknowns.
The answer drops out that
x=4.021269735
and
y=8.952396228
meaning that the tip of the ladder touches the wall either at 10.02 units high, or at 14.95 units high.
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This is saying that the triangle formed by the upper edge of the box the upper part of the wall and the upper part of the ladder is the same shape as the triangle formed by the floor the wall and the entire ladder.
Three triangles - and also of course the triangle formed by the part of the floor under the ladder that is not occupied by the box, the side of the box and the lower section of the ladder up to the top of the box.
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>>Three triangles
Yes, ...., and no.
Yes, in that there are clearly more than two similar triangles there.
No in that the equation you would get by considering another similar triangle would give you no new information, and doesn't mathematically help in the solution of the problem. When solving simultaneous equations, it's vitally important that the equations used are independent.
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This link gives you 2 methods to do the calculation:
other links:
www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/la...x
mathforum.org/library/drmath/view/55250.html
mathcentral.uregina.ca/QQ/database/QQ.09.98/blade1...l
I think another way of describing this puzzle is that in any given right-angle triangle, there can only be one size of a square that can be placed on the base such that it touches all three sides of the triangle.
Edited by jbif on 18/09/2008 at 16:11
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Thanks for the replies and links. I had a page of equations but got somewhat stuck sorting out my x squareds from my y's to the fourths!
I had the basic pythagorous but missed out on the similar triangle bit and got sidetracked trying to do some trig. Sanity is now approaching normal.
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The McKinsey answer:
Get a cherrypicker and a tape measure and measure it.
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Or Malibu Stacy's answer.
Maths is hard. Let's go shopping.
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