Question

A fold is formed on a `20 cm × 30 cm` rectangular sheet of paper running from the short side to the long side by placing a corner over the long side. Find the minimum possible length of the fold?

Answer 1

`L = 15sqrt(3) ~~ 25.98 \ cm`

Explanation:

Let us set up the following variables:

# { (L, "Length of the fold", cm), (x, "DE", cm), (theta, "Angle " hat(DGE), "radians") :} #

We are given that `AD=30 cm" and "AB=20 cm`. Our aim is to find the minimum length `L` of the fold.

By trigonometry for `triangle DEG` we have:

`sin hat(DGE) = (DE)/(EG) => sin theta = x/L` ..... [A]

When folded the point `D` must touch the side `BC`

`hat(HFG) = hat (FGD) = 2theta`

Now:

`hat(HFG) + hat(GFE) + hat(EFC) = pi`
`:. 2theta + pi/2 + hat(EFC) = pi`
`:. hat(EFC) = pi/2 - 2theta`

By trigonometry for `triangle CEF` we have:

`sin hat(EFC) = (EC)/(EF )`
`:. sin (pi/2 - 2theta) = (CD-DE)/(EF )`
`:. sin (pi/2 - 2theta) = (20-x)/(x)`

Using the sum of angle formula:

`sin(A-B) -= sinAcosB - cosAsinB`

we have:

`sin(pi/2)cos(2theta) - cos(pi/2)sin(2theta) = (20-x)/(x)`
`:. cos(2theta) = 20/x - 1`

Using the identity:

`cos 2A -= 1-2sin^2A`

we have:

`1-2sin^2theta = 20/x - 1`

Using `Eq [A]` this becomes:

`\ \ \ \ \ 1-2(x/L)^2 = 20/x - 1`
`:. 2 -2 x^2/L^2 = 20/x`
`:. x^2/L^2 = 1-10/x`
`:. x^2/L^2 = (x-10)/x`
`:. L^2/x^2 = x/(x-10)`
`:. L^2 = x^3/(x-10)`

As `L` changes then `theta` and `x` change accordingly, we know have `L^2` as a function of `x` alone, and minimizing `L` is the same as minimizing `L^2`, so we seek a critical point of `L^2`

Differentiating wrt `x` by applying the quotient rule we have:

`d/dx L^2 = { (x-10)(d/dx(x^3)) - (d/dx(x-10))(x^3) } / (x-10)^2`
`\ \ \ \ \ \ \ \ \ \ = { (x-10)(3x^2) - (1)(x^3) } / (x-10)^2`
`\ \ \ \ \ \ \ \ \ \ = x^2{ 3(x-10) - x } / (x-10)^2`
`\ \ \ \ \ \ \ \ \ \ = x^2{ 3x-30 - x } / (x-10)^2`
`\ \ \ \ \ \ \ \ \ \ = x^2{ 2x-30 } / (x-10)^2`

At a critical point this derivative vanishes, and so:

`x^2{ 2x-30 } / (x-10)^2 = 0`
`x^2(2x-30 ) = 0`
`x=0,15`

`x =0` result in no fold and so we required `x=15`, with this value of `x` we have:

`L^2 = 15^3/(15-10)`
`\ \ \ \ = 3375/5`
`\ \ \ \ = 675`

`:. L = 15sqrt(3) ~~ 25.98 \ cm`