Congratulations to all the following who sent in very good solutions to this problem:
David Lowe, age 15, Trinity School, Carlisle (whose solution was the first to arrive); Babak Shirazi, age 17, Woodhouse SF College, London; Chen Yiwen, age 16, The Chinese High School, Singapore; Nathan Allpress, age 14, Riccarton High School, Churchtown, New Zealand; Lee Jia Hui, age 18, National Junior College, Singapore; Julian Steed (not a student); and finally Alexander Maryanovsky, age 18, 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.


Area (pentangle)
Area (pentagon)
=
Area (AOBC)
Area (AOB)
=
1- Area (ACB)
Area( AOB)
=
1- Area (DCB)
Area (DOB)
=
1- DC
DO
=
CO
DO
 < 1
2
 .

In order to calculate this ratio exactly we first find the angles and then use trigonometry.
ÐAOB = 72°

ÐDOB = 1
2
 ÐAOB = 36 °

As AC is parallel to PM and CB is parallel to MK,
ÐACB = ÐPMK = 108 ° .

ÐDCB = 1
2
 ÐACB = 54° .


Area (pentangle)
Area (pentagon)
=
1- DB tan36
DBtan54
=
1- tan36
tan54
=
1- tan2 36
=
1- (5 - 2Ö5)
=
2Ö5 - 4
=
0.47    approx.

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° , 36°, 54° and 72°. All the angles marked with a spot can be shown to be 36° using simple properties of triangles. Let CA=CB=x. The triangle PAC is an isosceles triangle with base angles of 72°. If PA=PC=1 then PB=1+x. Triangles PAB and ACB are similar, hence
x
1
 = 1
1+x

This gives a quadratic equation which can be solved to give
x = 1
2
 (Ö5 - 1).

Now we have AD=1/2 and so (using Pythagoras Theorem to find CD):
tan2 36° = 4CD2 = 4(x2 - 1
4
 ) = 4x2 - 1 = 5 - 2Ö5.