h / t = sqrt(2 s / (k x) )

The minimum value of R for a given k occurs when s = x, and the fold goes through the corner.

4 – 2 k <= R <= 2 / k ]]>

From similar triangles s / y = t / (x / 2)

Eliminating s and y leads to t = x / sqrt(3)

And so alpha = 30 degrees and all of the triangles are similar.

From the bottom left triangle, t / (y – t) = 2 and so t = 2 y / 3

The cut = h = 2 t = 4 y / 3 = 32 cm

The diagram above shows the folded corner and illustrates that the fold produces two identical triangles and a set of angle relationships that allow the teaser to be solved. Using the symbols introduced in the diagram, we can see that:

\[y=s\sin{2\alpha}\]\[x=2t\sin{2\alpha}\]

Combining these two equations we obtain:

\[x\,y = 2\,s\,t\,\sin^2{2\alpha}\]

With \(A_r\) and \(A_t\) as the areas of the rectangle and the cut off triangle respectively, this equation becomes\[A_r = 4A_t\,\sin^2{2\alpha}\]showing that:

\[\alpha=\frac{1}{2}\sin^{-1}{ \left( \frac{\sqrt{A_r/A_t}}{2} \right)}\]

From the teaser text we know that \(A_r-A_t=2A_t\), giving \(A_r/A_t = 3\) and showing from this equation that \(\alpha=30^o\).

From the diagram we can see that\[\begin{array}{l}y&=&s\,\sin{2\,\alpha}\\&=&h\,\cos{\alpha}\sin{2\,\alpha}\\&=&2\,h\,\sin{\alpha}\cos^2{\alpha}\\ \end{array}\] With \(\alpha=30^0\) this shows that the length of the cut line \(h\) is equal to \(4\,y/3\), which gives the teaser solution as 32cm.

Other lengths are given by:\[t\,=\,2\,y/3\]\[s\,=\,x\,=\,2\,y/\sqrt{3}\] with lengths of 16cm and ~27.71cm respectively.

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