Sunday Afternoon Maths XX Answers & Extensions

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

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