MEI Maths Item of the Month
December 2014
A Christmas Star
An eight pointed Christmas star is made with a gold layer and a silver layer.
The star for the gold layer is formed from a regular octagon ABCDEFGH by finding the
points of intersection of the squares ACEG and BDFH.
The silver layer is formed from the same regular octagon ABCDEFGH by creating the star
ADGBEHCF.
What fraction of the gold layer is covered by the silver layer?
1 of 3
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MEI Maths Item of the Month
Solution
Adding a central point O shows that the start can be split up into 8 equal sections.
Therefore the fraction covered by silver in the whole star is the same as the fraction of the
shape APBO covered by the shape AQBO.
Angle AOB is 45°, therefore, by calling the length AB as 2a, the triangle AOB can be drawn
as:
2 tan x
and the value tan 45  1 gives tan 22.5  2  1 .
1  tan 2 x
a2
a
Therefore the height of the triangle is
and its area is
.
2 1
2 1
Using the identity tan 2 x 
2 of 3
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MEI Maths Item of the Month
Angle AQB is 90°, therefore the triangle AQB can be drawn as:
Therefore the height of the triangle is a and its area is a 2 .
Area of Kite AQBO
 Area of triangle AOB – Area of triangle AQB
a2

 a2
2 1

a 2  2a 2  a 2
2 1

2( 2  1)a 2
2 1
 2a 2
Angle APB is 135°, therefore the triangle APB can be drawn as:
Therefore the height of the triangle is
Area of Kite AQBO


2  1 a and its area is


2 1 a2 .
 Area of triangle AOB – Area of triangle APB
a2

 2 1 a2
2 1


a 2  ( 2  1) 2 a 2

2 1
2( 2  1) a 2
2 1
2
 2a

Therefore the fraction of the gold layer covered by the silver layer is:
2a 2
1
.

2
2a
2
3 of 3
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27/02/2015