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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?

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    $\begingroup$ How is pi defined? $\endgroup$ – miracle173 Nov 22 '15 at 12:09
  • $\begingroup$ @miracle173 How does that help? $\endgroup$ – SchrodingersCat Nov 22 '15 at 12:13
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    $\begingroup$ How do you want to show something about pi if you have not definition of it? $\endgroup$ – miracle173 Nov 22 '15 at 12:15
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    $\begingroup$ Inscribe a regular hexagon! $\endgroup$ – Christian Blatter Nov 22 '15 at 12:16
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    $\begingroup$ 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. $\endgroup$ – Andrew D. Hwang Nov 22 '15 at 13:33
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The inscribed hexagon in the unit circle has perimeter $6$. The perimeter of the circle is $2\pi$, hence $\pi > 3$.

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  • $\begingroup$ Shouldn't it be $>$ and not $ \ge $? $\endgroup$ – SchrodingersCat Nov 22 '15 at 12:30
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    $\begingroup$ imgur.com/IbOElsj $\endgroup$ – Paulistic Nov 22 '15 at 12:48
  • $\begingroup$ Arguably this involves six equilateral triangles $\endgroup$ – Henry Nov 22 '15 at 15:06
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    $\begingroup$ 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...")? $\endgroup$ – Sebastien Nov 23 '15 at 8:25
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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 $\pi\ge 3$.

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

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  • $\begingroup$ Not really.. I just want the proof using a triangle.. without constructing a hexagon. $\endgroup$ – SchrodingersCat Nov 22 '15 at 12:53
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    $\begingroup$ @Aniket I would say the hint just provides a starting point; by inscribing suitable triangles twice we arrives at something useful, is not it? :) $\endgroup$ – Megadeth Nov 22 '15 at 12:54
  • $\begingroup$ In that case.. I mean if you look at it in that way... It is fine. $\endgroup$ – SchrodingersCat Nov 22 '15 at 12:56
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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 $\dfrac{\pi}3$.

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

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