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## Monday, 14 July 2014

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