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The Length of a Rectangle

What is the length of the longest 1-unit wide rectangle which can be placed
inside a 3x4 rectangle?

Source: Past puzzle experience.

Solutions were received from Alan O'Donnell, David Bachtel, Al
Zimmermann, Moandi Maurizio, Philippe Fondanaiche, Erika Brandner, and Denis Borris. David
Bachtel decided to turn the rectangle into the third dimension and slide a
length-5 rectangle through the diagonal. (Of course this could be
extended to an infinite rectangle, too.) Some 2-D solutions follow below.
From Alan O'Donnell:

The horizontal length of the 1-width rectangle length r, at angle Q to the
horizontal is sinQ + r cosQ
vertical height at angle Q is cosQ + r sinQ.
Thus we want to solve for r:
sinQ + r cosQ = 4
r sinQ + cosQ = 3
Multiply each in turn, subtracting from the other, we get:
(r^2 - 1) cosQ = 4r - 3
(r^2 - 1) sinQ = 3r - 4
->
cosQ = (4r-3)/(r^2 -1)
sinQ = (3r-4)/(r^2 -1)
Now, cos^2(x) + sin^2(x) = 1, so
(4r-3)^2 + (3r-4)^2 = (r^2-1)^2
expands to:
-r^4 + 27r^2 -48r + 24 = 0.
Solving for r, we find this is zero when r=4.085114867
Internal rectangle inclined at 0.554143 radians (31.75 deg)

From Morandi Maurizio:
When the four vertices of the internal rectangle are on the four sides of the
external rectangle the length is maxima and it is 4.085115866...units.
__________________________________C
| \ |
| \ |
| ____\| F
| ____/ |
| ____/ |
G| ____/ |
|\ ____/ |
| \ ___/ |
|_ \/_____________________________|
A x E B
If AE =x we have AG = CF = sqrt(1-x^2).
The triangles GAE and EBF are similar threfore we can write the proportion:
AE : AG = BF : BE.
But EB = 4 - x and BF = 3 - CF = 3 - sqrt(1-x^2), then it becomes:
x : sqrt(1 - x^2) = (3 - sqrt(1 - x^2)) : (4 - x).
This proportion is equivalent to the equation:
4x^4 - 16x^3 + 21x^2 + 8x - 8 = 0
Its solution is x = 0.5262.
The length of the internal rectangle is:
EF = sqrt(EB^2 + BF^2)
= sqrt(26 - 8x - 6sqrt(1 - x^2))
= 4.085115866...

Mail to Ken