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 $AOBC$.
\begin{eqnarray} \\ \frac{{\rm Area (pentangle)}}{{\rm Area
(pentagon)}} &=& \frac{{\rm Area (AOBC)}}{{\rm Area
(AOB)}}\\ &=& 1-\frac{{\rm Area (ACB)}}{{\rm Area(
AOB)}} \\ &=& 1-\frac{{\rm Area (DCB)}}{{\rm Area
(DOB)}} \\ &=& 1-\frac{DC}{DO}\\ &=&
\frac{CO}{DO} < \frac{1}{2}. \end{eqnarray}
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 $MK$, $$\angle ACB =
\angle PMK = 108 ^\circ . $$ $$\angle DCB = {1\over 2}\angle ACB
= 54^\circ .$$
\begin{eqnarray} \\ \frac{{\rm Area (pentangle)}}{{\rm Area
(pentagon)}} &=& 1-\frac{DB \tan 36}{DB\tan 54}\\
&=& 1-{\tan 36 \over \tan 54} \\ &=& 1- \tan^2
36 \\ &=& 1- (5 - 2{\sqrt 5}) \\ &=& 2{\sqrt 5}
- 4 \\ &=& 0.47\quad {\rm approx.} \end{eqnarray}
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 $CD$): $$\tan^2 36^\circ = 4CD^2 = 4(x^2 - \frac{1}{4}) =
4x^2 - 1 = 5 - 2{\sqrt 5}.$$