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188 geometry/ladders.p




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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

188 geometry/ladders.p


Two ladders form a rough X in an alley. The ladders are 11 and 13 meters
long and they cross 4 meters off the ground. How wide is the alley?

geometry/ladders.s

Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two
walls (taken to be perpendicular to the ground), and they will
intersect at a point O = (a,s), a height s from the ground. Find the
largest s such that this is possible. Then find the width of the
alley, w = a+b, in terms of L1, L2, and s. This diagram is not to
scale.

                 B                     D
                  |\ L1           L2 /|
                  |  \             /  |           BC = length of L1  
                  |    \         /    |           AD = length of L2
                  |      \  O  /      |            s = height of intersection
                 x|        \ /        |y           A = (0,0)
                  |        /|\        |           AE = a 
                  |    m /  |  \ n    |           EC = b
                  |    /    |s   \    |           AO = m
                  |  /      |      \  |           CO = n
                  |/________|________\|     
         (0,0) = A    a     E    b     C

-----------------------------------------------------------------------------
Without loss of generality, let L2 >= L1.

Observe that triangles AOB and DOC are similar. Let r be the ratio of
similitude, so that x=ry. Consider right triangles CAB and ACD. By
the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry,
this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0),
and factoring, this becomes

(*) y^2 (1+r)(1-r) = L

Now, because parallel lines cut L1 (a transversal) in proportion, r =
x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x =
s(r+1). Solving for r, one obtains the formula r = s/(y-s).
Substitute this into (*) to get

(**) y^2 (y) (y-2s) = L (y-s)^2

NOTE: Observe that, since L>=0, it must be true that y-2s>=0.

Now, (**) defines a fourth degree polynomial in y. It can be written in the
form (by simply expanding (**))

(***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0

L1 and L2 are given, and so L is a constant. How large can s be? Given L,
the value s=k is possible if and only if there exists a real solution, y',
to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are
constants, and (***) gives the desired value of y. (Make sure to choose the
value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e.,
feasible), then there will exist exactly one such solution.)
Now, w = sqrt(L2^2 - y^2), so this concludes the solution.

L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes

y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0

Numerically find root y ~= 9.70940555, which yields w ~= 8.644504.

 

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