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
$$y=(1-\sqrt{2x})^2$$
No comments:
Post a Comment