Archive for March, 2013

happy “Pi” day

Posted: March 14, 2013 in Mathematics

dodecagonToday I am going to show you how to find the exact area of a regular dodecagon with side length of 1 without using a calculator.  First of all, let me tell you what a regular dodecagon is.  A regular dodecagon is a regular 12-gon, a 2D shape with 12 equal sides.

Let’s begin!

See the diagram, a regular dodecagon consists of 12 isosceles triangles with central angle of 30 degrees.  Let the base of that triangle be 1, then

15degree

\dfrac{1/2}{h}=\tan{15^\circ}

h=\dfrac{1}{2\tan{15^\circ}}

then the area of one triangle is h/2 and the exact area of the regular dodecagon is 6h.  Now, the difficult part is to find the exact value of h by hand.  (Remember this is a math contest question, no calculator is allowed.  Just your brain, a pencil, and papers. )

Now, if it is tan(30), it will be easy because \tan{30^\circ}=\dfrac{1}{\sqrt{3}}

but notice that 15=30/2, so we can use the half angle trig identity.

Since

\cos{2x}=\cos^2{x}-\sin^2{x}

\cos{2x}=1-2\sin^2{x}

\cos{2x}=2\cos^2{x}-1

then we have

\sin^2{x}=\dfrac{1-\cos{2x}}{2}

and

\cos^2{x}=\dfrac{1+\cos{2x}}{2}

And since our angle here is less than 90 degrees, we can drop the negative sign when taking square roots.

\sin{x}=\sqrt{\dfrac{1-\cos{2x}}{2}}

\cos{x}=\sqrt{\dfrac{1+\cos{2x}}{2}}

then replace x with x/2 yields,

\sin{\dfrac{x}{2}}=\sqrt{\dfrac{1-\cos{x}}{2}}

\cos{\dfrac{x}{2}}=\sqrt{\dfrac{1+\cos{x}}{2}}

Then tangent is sine over cosine,

\tan\dfrac{x}{2}=\sqrt{\dfrac{1-\cos{x}}{1+\cos{x}}}

Then substitute in x=30, we have

\tan15^\circ=\tan\dfrac{30^\circ}{2}=\sqrt{\dfrac{1-\cos30^\circ}{1+\cos30^\circ}}

=\sqrt{\dfrac{1-\sqrt{3}/2}{1+\sqrt{3}/2}}

=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}

=\sqrt{\dfrac{(2-\sqrt{3})^2}{(2+\sqrt{3})(2-\sqrt{3})}}

=2-\sqrt{3}

Therefore, the area of the regular dodecagon with side length of 1 is

6h=\dfrac{6}{2\tan15^\circ}=\dfrac{3}{2-\sqrt{3}}

=3(2+\sqrt{3})=6+3\sqrt{3}

DONE.