# Prove that $\pi>3$ using geometry

I was asked this question today in an interview.

Question: Prove that $\pi>3$ using geometry.

They gave me hints about drawing a unit circle and then inscribing an equilateral triangle and then proceeding. But I could not follow. Can anyone help?

• How is pi defined? – miracle173 Nov 22 '15 at 12:09
• @miracle173 How does that help? – SchrodingersCat Nov 22 '15 at 12:13
• How do you want to show something about pi if you have not definition of it? – miracle173 Nov 22 '15 at 12:15
• Inscribe a regular hexagon! – Christian Blatter Nov 22 '15 at 12:16
• An equilateral triangle inscribed in a unit circle has perimeter $3\sqrt{3}$, which is not of itself obviously helpful. Offhand I'd guess the interviewer was trying to guide you toward the idea of inscribing a suitable polygon, in this case a regular hexagon. – Andrew D. Hwang Nov 22 '15 at 13:33

The inscribed hexagon in the unit circle has perimeter $6$. The perimeter of the circle is $2\pi$, hence $\pi > 3$.

• Shouldn't it be $>$ and not $\ge$? – SchrodingersCat Nov 22 '15 at 12:30
• imgur.com/IbOElsj – Paulistic Nov 22 '15 at 12:48
• Arguably this involves six equilateral triangles – Henry Nov 22 '15 at 15:06
• How do you prove that the perimeter of the hexagon is smaller than the perimeter of the circle (really "prove", not say "it is clear that...")? – Sebastien Nov 23 '15 at 8:25

The inscribed $12$-gon in the unit circle has area $\frac{12}{2}\sin (2\pi/12)=3$. The area of the unit circle is $\pi$. Hence $\pi\ge 3$.

• Thanks, But can you use the triangle and do so? – SchrodingersCat Nov 22 '15 at 12:20
• Shouldn't it be $>$ and not $\ge$? – SchrodingersCat Nov 22 '15 at 12:30
• I like this argument based on area than another one on this thread based on perimeter. – Kim Jong Un Nov 22 '15 at 12:38

I am not sure if this is redundant, but:

If an equilateral triangle is inscribed in a unit circle, and if, on each side of the inscribed triangle, an isosceles triangle is further inscribed in the circle, then an equilateral hexagon with each side of length $=1$ results; but then $6 < 2\pi$ implies $3 < \pi$.

So is this something you are after?

• Not really.. I just want the proof using a triangle.. without constructing a hexagon. – SchrodingersCat Nov 22 '15 at 12:53
• @Aniket I would say the hint just provides a starting point; by inscribing suitable triangles twice we arrives at something useful, is not it? :) – Megadeth Nov 22 '15 at 12:54
• In that case.. I mean if you look at it in that way... It is fine. – SchrodingersCat Nov 22 '15 at 12:56

With a little bit of cheating, you don't need the whole hexagon...

Let $O$ be the centre of the unit circle with equilateral $\triangle ABC$ inscribed in it. Extend $\vec {AO}$ to meet the circle at D.

As $BC$ and $OD$ are perpendicular bisectors of each other, $\triangle OBD$ is isosceles, and hence $|BD|=1$. But this must be smaller than the minor arc subtended, which has length $\dfrac{\pi}3$.

• Could the downvoter comment why? – Macavity Dec 29 '15 at 14:50