A golden rectangle
Part 1 :
Here is a diagram which has been drawn from the line segment [ AB] :
Describe this diagram to the other group so that they could draw it using GeoGebra.
Part 2 :
AP
BC
AP
BC
and
using GeoGebra :
� ..................... and
� ......................
AD
BP
AD
BP
What do you notice ?
Work out the ratios
√
1+ 5
The Golden Number is equal to
.
2
Its approximate value to 3 decimal places is : ........................
Part 3 :
A rectangle is a golden rectangle when the ratio of the larger side to the smaller side is equal to the
golden number.
1. Let us prove that the previous construction leads to APQD be a golden rectangle.
Take 1 as the value of the side of the square ABCD.
√
5
.
2
(b) Deduce from this result that APQD is a golden rectangle.
(a) Using Pythagora’s theorem, prove that IC =
2. Prove that the rectangle BPQC is also a golden rectangle.
√
2
Tip : change the ratio √
by multiplying the numerator and the denominator by 5 + 1.
5−1
Part 3 :
1.
(a) Using Pythagora’s theorem in the right-angled triangle IBC :
BP = AP − .........
= ......... − .........
= ......... − .........
√
5−1
=
2
IC2 = ......... + ......... ⇔ IC2 = ......... + .........
⇔ IC2 = ......... + .........
⇔ IC2 = .........
�
⇔ IC =
Thus
( because ...................................... )
�
⇔ IC = �
=
=
=
⇔ IC =
(b) APQD is a golden rectangle if and only if
AP
= .....................
AD
AP = AI + IP
= AI + ......... ( because IP and ......... are radiuses of the same circle )
= ......... + .........
√
........... × ( 5 + 1)
√
(..............) × ( 5 + 1)
=
=
=
Therefore
AP
AD
BC
BP
So BPQC is a .....................................................................
=
So APQD is a .....................................................................
2. BPQC is a golden rectangle if and only if
BC
= .....................
BP