124. The ladder 
A ladder, leaning against a building, rests upon the ground and just touches a box, which is flush against the wall and the ground. The box has a height of 64 units and a width of 27 units.
Find the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall.
irst of all, we generalize the dimensions of the box to r × s; see below. Let the distance above the box at which the ladder touches the wall be x, and let the corresponding horizontal distance be y. Let the ladder have length d.
Applying Pythagoras' Theorem, (r + x)2 + (s + y)2 = d2.
By similar triangles, x/s = r/y, and so y = rs/x.
Hence (r + x)2 + (s + rs/x)2 = d2.
Expanding, and collecting terms, we obtain an equation in x:
f(x) = x2 + 2rx + (r2 + s2 - d2) + 2rs2/x + r2s2/x2 = 0. (1)
Multiplying by x2, we obtain (since x = 0 is not a root) an equivalent polynomial equation in x:
p(x) = x4 + 2rx3 + (r2 + s2 - d2)x2 + 2rs2x + r2s2 = 0. (2)
By similar triangles, x/s = r/y, and so y = rs/x.
Hence (r + x)2 + (s + rs/x)2 = d2.
Expanding, and collecting terms, we obtain an equation in x:
f(x) = x2 + 2rx + (r2 + s2 - d2) + 2rs2/x + r2s2/x2 = 0. (1)
Multiplying by x2, we obtain (since x = 0 is not a root) an equivalent polynomial equation in x:
p(x) = x4 + 2rx3 + (r2 + s2 - d2)x2 + 2rs2x + r2s2 = 0. (2)
Clearly d2 > r2 + s2, and so by Descartes' Sign Rule, this equation has 2 or 0 positive real roots (countingmultiplicity.)
We seek the value(s) of d for which the equation has two identical roots.
We seek the value(s) of d for which the equation has two identical roots.
p has repeated root a if, and only if, p(x) = (x - a)2(x2 + bx + r2s2/a2), for some a and b, to be found. (The constant term in the second quadratic factor is determined by that in the first, given that their product must equal r2s2.)
Expanding, p(x) = x4 + (b - 2a)x3 + (a2 - 2ab + r2s2/a2)x2 + (a2b - 2r2s2/a)x + r2s2 = 0.
Equating coefficients of x3 and x with (2), we obtain:
Expanding, p(x) = x4 + (b - 2a)x3 + (a2 - 2ab + r2s2/a2)x2 + (a2b - 2r2s2/a)x + r2s2 = 0.
Equating coefficients of x3 and x with (2), we obtain:
b - 2a = 2r,
a2b - 2r2s2/a = 2rs2. (3)
a2b - 2r2s2/a = 2rs2. (3)
Substituting b = 2(a + r) into (3), multiplying by a, and rearranging, we get
2a3(a + r) = 2(a + r)rs2.
Since a + r
0, we deduce that a3 = rs2.
Hence a = r1/3s2/3, and b = 2(a + r) = 2r1/3s2/3 + 2r.
2a3(a + r) = 2(a + r)rs2.
Since a + r
0, we deduce that a3 = rs2.Hence a = r1/3s2/3, and b = 2(a + r) = 2r1/3s2/3 + 2r.
Equating coefficients of x2, we obtain:
| r2 + s2 - d2 | = a2 - 2ab + r2s2/a2. |
| = r2/3s4/3 - 4r2/3s4/3 - 4r4/3s2/3 + r4/3s2/3. | |
| = - 3r2/3s4/3 - 3r4/3s2/3. |
Hence d2 = r2 + s2 + 3r2/3s4/3 + 3r4/3s2/3 = (r2/3 + s2/3)3.
(Note the pleasingly symmetrical form: d2/3 = r2/3 + s2/3.)
Finally, substituting r = 64 = 43 and s = 27 = 33, we get d2 = (42 + 32)3 = (52)3, so that d = 53 = 125.
Therefore, the length of the ladder so that there is only one position in which it can touch the ground, the box, and the wall, is 125 units.
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