What is the length of the longest 1-unit wide rectangle which can be placed inside a 3x4 rectangle?
Source: Past puzzle experience.
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)
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...
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