Congratulations to all the following who sent in very good solutions to this problem: David from Trinity School, Carlisle (whose solution was the first to arrive); Babak Shirazi from Woodhouse SF College, London; Chen Yiwen from The Chinese High School, Singapore; Nathan from Riccarton High School, Churchtown, New Zealand; Lee Jia Hui from National Junior College, Singapore and finally Alexander from Shevah-Mofet School, Israel. The following solution is made up of bits supplied by several of these contributors.

The pentagon is made of 5 triangles exactly the same as $AOB$, and the pentangle is made of 5 shapes exactly the same as $AOBC$. In order to calculate this ratio exactly we first find the angles and then use trigonometry. $$ \angle AOB = 72^\circ $$ $$\angle DOB = \frac{1}{2}\angle AOB = 36 ^\circ $$As $AC$ is parallel to $PM$ and $CB$ is parallel to $MK$, $$\angle ACB = \angle PMK = 108 ^\circ . $$ $$\angle DCB = {1\over 2}\angle ACB = 54^\circ .$$ This ratio is just less than 0.5 meaning the pentangle is a bit smaller than half the pentagon.

Footnote: You don't need a calculator, from the diagram it is possible to calculate exact values for the trig. ratios for $18^\circ $, $36^\circ$, $54^\circ$ and $72^\circ$. All the angles marked with a spot can be shown to be $36^\circ$ using simple properties of triangles. Let $CA=CB=x$. The triangle $PAC$ is an isosceles triangle with base angles of $72^\circ$. If $PA=PC=1$ then $PB=1+x$. Triangles $PAB$ and $ACB$ are similar, hence $$\frac{x}{1} = \frac{1}{1+x}$$

This gives a quadratic equation which can be solved to give $x = \frac{1}{2}({\sqrt 5} - 1).$

Now we have $AD={1\over 2}$ and so (using Pythagoras Theorem to find $CD$): $$\tan^2 36^\circ = 4CD^2 = 4(x^2 - \frac{1}{4}) = 4x^2 - 1 = 5 - 2{\sqrt 5}.$$