At a glance

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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. $$\mathrm{\angle }AOB={72}^{\circ }$$ $$\mathrm{\angle }DOB=\frac{1}{2}\mathrm{\angle }AOB={36}^{\circ }$$As $AC$ is parallel to $PM$ and $CB$ is parallel to $MK$, $$\mathrm{\angle }ACB=\mathrm{\angle }PMK={108}^{\circ }.$$ $$\mathrm{\angle }DCB=\frac{1}{2}\mathrm{\angle }ACB={54}^{\circ }.$$ 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 $CD$): $${\tan }^2{36}^{\circ }=4C{D}^2=4(x^2-\frac{1}{4})=4x^2-1=5-2\sqrt{5}.$$