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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3How is pi defined? – miracle173 Nov 22 '15 at 12:09
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@miracle173 How does that help? – SchrodingersCat Nov 22 '15 at 12:13
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1How do you want to show something about pi if you have not definition of it? – miracle173 Nov 22 '15 at 12:15
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5Inscribe a regular hexagon! – Christian Blatter Nov 22 '15 at 12:16
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@miracle173 of course there is the conventional definition. – SchrodingersCat Nov 22 '15 at 12:17
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@ChristianBlatter Haven't tried that but know that works.Can you proceed with the triangle? – SchrodingersCat Nov 22 '15 at 12:18
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@DietrichBurde Think about the perimeter of the hexagon. – SchrodingersCat Nov 22 '15 at 12:29
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1An 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
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A related question. – Lucian Nov 22 '15 at 19:09
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A much more related question: https://puzzling.stackexchange.com/questions/111434/prove-that-%cf%80-3 – mathlander Feb 26 '24 at 18:38
4 Answers
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The inscribed hexagon in the unit circle has perimeter $6$. The perimeter of the circle is $2\pi$, hence $\pi > 3$.
SchrodingersCat
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user133281
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2How 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 08:25
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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 $\pi\ge 3$.
Dietrich Burde
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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
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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?
Yes
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Not really.. I just want the proof using a triangle.. without constructing a hexagon. – SchrodingersCat Nov 22 '15 at 12:53
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1@Aniket I would say the hint just provides a starting point; by inscribing suitable triangles twice we arrives at something useful, is not it? :) – Yes Nov 22 '15 at 12:54
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In that case.. I mean if you look at it in that way... It is fine. – 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 $\dfrac{\pi}3$.
Macavity
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