This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Folding A4 Paper

The sides of A4 paper are in the ratio \(1:\sqrt{2}\). Let the width of the paper be 1 unit. This means that the height of the paper is \(\sqrt{2}\) units.

Therefore on the diagram, \(AE=1, BE=1, AC=\sqrt{2}\). By Pythagoras' Theorem, \(AB=\sqrt{2}\), so \(AB=AC\).

\(BF=DF=\sqrt{2}-1\) so by Pythagoras' Theorem again, \(BD=2-\sqrt{2}\). \(CD=1-(\sqrt{2}-1)=2-\sqrt{2}\). Hence, \(CD=BD\) and so the shape is a kite.

##### Extension

Prove that for a starting rectangle with the sides in any ratio, the resulting shape is a cyclic quadrilateral.

#### Bézier Curves

If

$$P_0=\left(0,1\right)$$
$$P_1=\left(0,0\right)$$
$$P_2=\left(1,0\right)$$
then

$$Q_0=t\left(0,0\right)+(1-t)\left(0,1\right)=\left(0,1-t\right)$$
$$Q_1=t\left(1,0\right)+(1-t)\left(0,0\right)=\left(t,0\right)$$
and so

$$R_0=t\left(t,0\right)+(1-t)\left(0,1-t\right)=\left(t^2,(1-t)^2\right)$$
which means that

$$x=t^2$$$$y=(1-t)^2$$
or

$$y=(1-\sqrt{x})^2$$
##### Extension

What should \(P_0\), \(P_1\) and \(P_2\) be to get a curve with Cartesian equation

$$y=(1-\sqrt{2x})^2$$
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