# Looking for a better (geometric) approximation of $\pi$...

I was hanging around with circles when I did the following "workout" :

(Sorry for not including any diagram of my construction. I could not find any suitable software for drawing. I have tried my best to describe each and every step so that you can follow along.)

Construction : I took a circle with $O$ as center and $\overline{AB}$ as diameter. The perpendicular bisector of $AB$ meets the "upper" arc $\overline{AB}$ at $C$. Hence, $\overline{OC}$ divides the upper hemisphere into two quadrants. Next I constructed $\angle OCD = 60^o$, $D$ lying in the quadrant $BOCB$ and on the circumference. This gives me an equilateral triangle $\triangle OCD$. I joined $\overline{AD}$ and $\overline{BD}$.

Method :

Let $r$ be the radius of the circle.

As $\overline {OD}=\overline {OB}=r$, so $\angle DBO=\angle ODB$. But $\angle DOB = 90^o-60^o = 30^o$. So $\angle DBO = \angle ODB=75^o$.

Using Sine Rule in $\triangle BDO$ :

$$\frac{\sin 30^0}{\overline {BD}} = \frac {\sin 75^o}{r}$$

$$\implies \overline{BD}=\frac {r}{2} \csc 75^o$$

Clearly,

$$\overline {AC}+\overline{CD}+\overline{BD} < \frac {2\pi r}{2}$$

By Pythagorean Theorem, $\overline{AC}=r\sqrt{2}$.

$$\therefore r\sqrt{2}+r+\frac{r}{2}\csc 75^o < \pi r$$

Cancelling out the common factor $r$ :

$$\sqrt{2}+1+\frac{\csc 75^o}{2} < \pi$$

$$OR$$

$$\pi > 2.9318516525781364$$

My thoughts : This "workout" gives a very "broad" approximation for $\pi$. I was overjoyed when I derived this result myself, but I am looking for a better approximation. So here is/are my question(s) :

Is there a way to refine/fine-tune/improve my approximation by using the same geometric setup ? Can $\pi$ be approximated ("in the same diagram") without using trigonometry? Is my "method" something which already has been "done" or tested ?

Any help will be gratefully appreciated.

• You could simply use $\pi>6\overline{BD}$. Feb 9 '17 at 20:38
• As $\angle DOB=30°$, $BD$ is the side of a regular dodecagon inscribed in the circle, so $12 BD<2\pi r$. This gives a better approximation. Feb 10 '17 at 7:52
• And of course, you can go up to a $24$-gon, and more. Using trig bisection formulas to have an exact result. Feb 10 '17 at 8:00
A first step in refinement is to construct point $E$ on the circumference in quadrant $AOCA$ such that angle $COE$ is $60^\circ$. And draw line segments $\overline{OE}$, $\overline{AE}$ and $\overline{CE}$.
Then, by the same trig reasoning you applied above, $$\overline{AE} = \frac r2 \csc(75^\circ) = \frac{\sqrt{2}(\sqrt{3}-1)}{2}\\ \overline{EC} = r\\ \overline{CD} = r\\ \overline{BD} = \frac r2 \csc(75^\circ) = \frac{\sqrt{2}(\sqrt{3}-1)}{2}$$ And again the same form of reasoning you used above gives $$\pi > \frac12 \csc(75^\circ) = \frac{\sqrt{2}(\sqrt{3}-1)}{2} + 1 + 1 + \frac12 \csc(75^\circ) + \frac{\sqrt{2}(\sqrt{3}-1)}{2}$$ or $$\pi > 2 + \sqrt{2}(\sqrt{3}-1) \approx 3.03528$$ Next, you can place points $F$ and $G$ on the midpoints of arcs $DC$ and $EC$ respectively, and draw line segments $\overline{EG}$, $\overline{GC}$, $\overline{CF}$, and $\overline{FB}$. The same trig reasoning then says $$\pi > 6 \frac{\sqrt{2}(\sqrt{3}-1)}{2}\approx 3.10583$$ This is the method Archimedes used; it inscribes a regular $12$-sided figure in the circle, and observes that the circumference of the circle exceeds the perimeter of the dodecagon.