Drawing a Decagon

Geometry Level 2

It is not easy to draw a regular decagon without tools.

On a piece of writing paper (with equally spaced lines), I am trying to draw a regular decagon, as shown above. I started by drawing two sides so that their vertical extent is precisely 1 unit of the paper (black lines).

Now I want to draw the next side (red line), and I wonder how far it will extend vertically. To 3 decimal places, what is the distance marked with a question mark?


The answer is 0.61803399.

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3 solutions

Arjen Vreugdenhil
Feb 26, 2016

Method 1 : Apply the apothem formula, apothem, a = r × cos 18 0 n a = r \times \cos \dfrac{180^\circ}n , where r r denote the circumradius of this poylgon and n n denote the number of sides of this polygon, n = 10 n=10 .

By referring to the right triangle formed above, the angle formed between the hypothenuse ( r r ), and the horizontal line is 36 0 10 = 3 6 \dfrac{360^\circ}{10} = 36^\circ and the height of this right triangle is just 1 cm, then sin 3 6 = 1 r r = 1 sin 3 6 \sin36^\circ = \dfrac1r \Rightarrow r = \dfrac1{\sin36^\circ } .

The apothem of this regular decagon is just r cos ( 18 0 10 ) = cos 1 8 sin 3 6 = cos 1 8 2 cos 1 8 sin 1 8 = 1 2 sin 1 8 = 5 + 1 2 r \cdot \cos \left( \dfrac{180^\circ}{10} \right) = \dfrac{\cos18^\circ}{\sin36^\circ} = \dfrac{\cos18^\circ}{2\cos18^\circ\sin18^\circ} = \dfrac1{2\sin18^\circ} = \dfrac{\sqrt5 + 1}2 .

So the distance in question is just 5 + 1 2 1 = 5 1 2 0.618 \dfrac{\sqrt5 + 1}2 - 1 = \dfrac{\sqrt5 - 1}2 \approx \boxed{0.618} .

Method 2 : Without applying the apothem formula.

From one side to another in a regular decagon involves a turn of 360 / 10 = 3 6 360/10 = 36^\circ .

The two black lines make equal angles with the vertical; this must be 36 / 2 = 1 8 36/2=18^\circ . The red line will therefore make an angle of 18 + 36 = 5 4 18 + 36 = 54^\circ with the vertical.

If the side of the polygon is a a , the vertical extent of the black lines is y b = a cos 1 8 , y_b = a\cos 18^\circ, and that of the red line is y r = a cos 5 4 . y_r = a\cos 54^\circ. Therefore h = y r y b = a cos 5 4 a cos 1 8 = sin 3 6 cos 1 8 = 2 sin 1 8 = 2 ( 1 4 5 1 4 ) = 1 2 5 1 2 0.618 . h = \frac{y_r}{y_b} = \frac{a\cos 54^\circ}{a\cos 18^\circ} = \frac{\sin 36^\circ}{\cos 18^\circ} = 2\sin 18^\circ = 2(\tfrac14\sqrt 5-\tfrac14) = \tfrac12\sqrt 5 - \tfrac12 \approx \boxed{0.618}. Incidentally, this is the golden ratio.

Isn't golden ratio one more than the answer?

Kushagra Sahni - 5 years, 3 months ago

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It depends on your definition. If you write the golden ratio as larger : smaller, then its value is 1.618. If you write it as smaller : larger, then its value is 0.618.

In fact, the golden ratio can be defined as the ratio which differs by 1 from its inverse. :)

Arjen Vreugdenhil - 5 years, 3 months ago

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Golden ratio has 2 values, (1+√5)/2 and (1-√5)/2 from the equation which says x^2-1=x.

Kushagra Sahni - 5 years, 3 months ago

It not only differs by one it's actually one more than it's inverse.

Kushagra Sahni - 5 years, 3 months ago

The angle between the sides of a 10-gon = 180 - 360/10 =144.
Therefore between top black and horizontal the angle is 144/2=72 by symmetry of two black sides about the
horizontal line.
Vertical projection of the side is =1. So side=1/Sin72.
But then red side is turned by 360/10=36 degrees clock wise from the top black .
Implies red side is 72 - 36 = 36 with horizontal.
So vertical projection is
( s i d e s i n 36 ) = 1 s i n 72 S i n 36 = 1 2 c o s 36 = 0.6180339887 (side * sin 36)=\dfrac 1 {sin72}*Sin36=\dfrac 1 {2*cos36}=\Large ~~~~~~~~~~ \color{#D61F06}{ ~~\boxed{ ~~0.6180339887~~} } .



Using simple geometry, one may get the required distance as

4 sin 2 1 8 1 4 sin 2 1 8 0.618033988 c m \frac{4\sin^218^\circ}{1-4\sin^218^\circ}\approx 0.618033988\ cm

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