Source: Internet newsgroup rec.puzzles. Used in Ken's POTD 8/16/94.
1) 1/4=1/sqrt(13^2-x^2) +1/sqrt(11^2-x^2) ===> x=8.64450372894493meters
width of road = 56; length of shorter ladder = 70; length of longer ladder = 119; giving height of crossing= 30.
width of road = 63; length of shorter ladder = 87; length of longer ladder = 105; giving height of crossing= 35.
If you look for a solution in integers in which the
longer ladder is as short as possible, I can improve
on the solution given in your answers:
width of road = 80;
length of shorter ladder = 100;
length of longer ladder = 116;
giving height of crossing= 35.
It was a nice problem; particularly the relationship
1 1 1
--- + --- = ---
h1 h2 h
between the heights of points where ladders rest on the
walls and height of crossing.
Obviously the easist place to look for integer solutions
is with integer values for h1 and h2; but it took
quite a bit more effort to prove that h1 and h2 must
be integers.
The following annotations are defined on the graph at the end of the text.
Let AM = a, BN = b, AN = c, BM = d, AB = z, IJ = h.
We have the assumption that a = 13, b = 11 and h = 4.
We have the 3 following relations:
1) z^2 + c^2 = b^2 = 121
2) z^2 + d^2 = a^2 = 169
3) AJ+JB = z = zh/d + zh/c so 1/c +1/d = 1/h = 1/4 and d = 4c/(c-4)
Therefore, we obtain the following 4th degree equation:
c^4 - 8c^3 + 48c^2 - 384c +768 = 0
The corresponding solution is c = 6,80239.
Then z = sqr(b^2 - c^2) = 8,6445
We look for solutions in which a, b, h and z are integers.
ABM and ABN are rectangular triangles which are similar to the
triangles 3,4,5 or 5,12,13 or 8,15,17... where the hypotenuses
AM and BN take the highest values and AB the side of the right
angle common to the 2 triangles take one of the other values.
So we can consider the 2 triangles 9,12,15 and 5,12,13 where z = 12, a = 15 and b = 13.
Therefore 1/h = 1/9 + 1/5 = 14/45 and h = 45/14.
If we multiply by the coefficient 14 all the figures above mentioned,
we obtain the set of required integers:
z =168 a = 210 b = 182 c = 126 d = 60 and h = 45.
Another solution among an infinite set of solutions:
z = 56 a = 119 b = 70 c = 42 d = 105 and h = 30.
Best regards.
[Philippe originally had a nice graphic for the drawing below. I copied my text version into
his text here to be sure everyone could read it - my apologies for its crude form.-KD.]
N M
|\ /|
| \ a / |
| \ / |
| \ I / |
c| \ / |d
| /|\ |
| / | \ b |
| / |h \ |
| / | \ |
|/________|________\|
A J z B
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.
(The first set of integral values with an integral solution I know of
is: s=144, L1=375, L2=500, w=300. Or s=180, L1=350, L2=546, w=210.)