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) rmArea(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.