Without calculator, how to prove that $2 \sqrt{3} > \pi$?
The level is baccalauréat grade. I confirm it's not a school exercise at all, as I left school like 35 years ago.
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4$\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$– José Carlos SantosNov 26, 2019 at 6:55
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2$\begingroup$ If you know that $\sqrt 3 \approx 1.732$ and $\pi \approx 3.1416$ you don't need a calculator $\endgroup$– Ross MillikanNov 26, 2019 at 6:57
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10$\begingroup$ The perimeter of a circumscribed hexagon of the unit circle is $4\sqrt3$, while the perimeter of the circle, which is clearly smaller, is $2\pi$. $\endgroup$– Rushabh MehtaNov 26, 2019 at 7:00
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4$\begingroup$ @DonThousand "clearly"? Clearly a word to avoid in maths. $\endgroup$– Jean-Claude ArbautNov 26, 2019 at 7:01
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4$\begingroup$ @Jean-ClaudeArbaut I'm offering a sketch. I'm not submitting an answer ... Clearly is perfectly appropriate in sketching proofs. $\endgroup$– Rushabh MehtaNov 26, 2019 at 7:02
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6 Answers
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(Credit to David G. Stork for the image).
I am showing the area-based argument explicitly because it seems that the other answers rely on a perimeter-based argument (which I find unconvincing without a rigorous proof). In contrast, it is quite easy to conclude by simple inspection that the circumscribed hexagon has a larger area than the inscribed circle.
The area of the circle is clearly simply $\pi$.
The hexagon can be decomposed into six congruent equilateral triangles. The height of each is $1$. The base can be computed with trigonometry as $(2)\tan\frac{\pi}{6} = \frac 2{\sqrt 3}$. Hence the area of a single triangle is $\frac 12 (1)(\frac 2{\sqrt 3})= \frac 1{\sqrt 3}$. The area of the hexagon is therefore $\frac 6{\sqrt 3} = 2\sqrt 3$.
This allows us to immediately conclude $2\sqrt 3> \pi$ as required.
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$\begingroup$ Sorry, I was not clear. I just meant, for geometric proofs what one considers "unconvincing without a rigorous proof" vs "quite easy to conclude by simple inspection" is rather subjective. $\endgroup$– philipxyDec 8, 2019 at 0:00
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1$\begingroup$ @philipxy Fair enough. It is subjective. But if you consider that even a child can draw a very squiggly curve inside a circle, with the squiggly curve having a greater perimeter than the enclosing circle, the issue becomes clearer. You have to bring in the notion of convexity to make the argument convincing in the case of the perimeter, and showing it properly relies on things like Crofton's formula (which are in the realm of analysis). Whereas even with the squiggly curve, the area argument remains visually obvious. I hope you can see this too. It's not all that subjective after all. $\endgroup$– DeepakDec 8, 2019 at 1:30
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$\begingroup$ It is extremely difficult to prove anything geometrically non-analytically. (And when the question is analytic in the first place it adds a pile of unnecessary weight.) It's very easy to suggest with a diagram; but a suggestion is not a proof. Little is clear or obvious; much is merely plausible, and when so, not the general case. PS Anyway, when people ask for proofs they need to say what they accept as given, or the question is ill-defined and/or trivial--take your pick(s)--and indeed what they even mean by "prove". $\endgroup$– philipxyDec 15, 2019 at 4:57
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By a geometric argument just consider a circle inscribed in an hexagon and compare the two perimeters to obtain
$$3 \cdot \frac23 \sqrt 3 > \pi \cdot 1 \iff 2\sqrt 3>\pi$$
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We know that $\tan x \gt x$ for $x\in (0,\frac{\pi}{2})$. So, when $x = \frac{\pi}{6}$
$\tan \frac{\pi}{6} \gt \frac{\pi}{6} \implies \frac{1}{\sqrt3} \gt \frac{\pi}{6} \implies 2\sqrt3 \gt \pi$
answered Nov 26, 2019 at 7:39
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Consider the perimeters in this figure:
$$2 \pi < 4\sqrt{3}$$
answered Nov 26, 2019 at 7:11
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2$\begingroup$ Nice diagram. As I commented, I think the perimeter argument is a bit cleaner since the numbers work out exactly. But this is, in my opinion, the best approach. $\endgroup$ Nov 26, 2019 at 7:12
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1$\begingroup$ Now you have to prove that the circumscribed polygon has a larger perimeter, which, as "obvious" as it may seem, is not that trivial. See math.stackexchange.com/questions/2581157/… $\endgroup$ Nov 26, 2019 at 7:14
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3$\begingroup$ The area argument is neater and more convincing for elementary level students (all that's needed is a bit of trigonometry). But this answer appears to use a perimeter argument (the area argument immediately yields the required inequality without any cancellation needed). $\endgroup$– DeepakNov 26, 2019 at 7:28
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1$\begingroup$ @Deepak: Oh... it is fine. Thanks for the credit. $\endgroup$ Nov 26, 2019 at 18:29
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Let us consider an equilateral triangle of side length $a$. The area of equilateral triangle is $$A_e=\frac{\sqrt3a^2}{4}$$ Let there be incentre in the triangle. Incentre Radius $$r=\frac{a}{2\sqrt3}$$ Area of circle $$A_c=\pi r^2 = \frac{\pi a^2}{12}$$ Since $$A_e>A_c$$ $$\frac{\sqrt3a^2}{4} > \frac{\pi a^2}{12}$$ $$3\sqrt3>\pi$$
If you consider a circle in a square, then $$a^2>\pi a^2/4$$ $$4>\pi$$
If you consider a circle in hexagon then $$\frac{3\sqrt3 \cdot a^2} 2 > \pi\left(\frac{\sqrt3}2\cdot a \right)^2$$ $$2\sqrt3>\pi.$$
In general, incentre Radius of a circle inscribed in a polygon of side $n$ and length $a$ is $$\frac{a}{2\tan{180^\circ/n}}$$
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Noting the fact that $$\tan\theta \geq \theta~~\forall x\in\left[0,\frac\pi2\right)$$ with equality for $x=0$
we have $$\tan\frac\pi6>\frac\pi6$$ $$\implies \frac{\sqrt 3}3 >\frac\pi6$$ $$\implies \sqrt3 >\frac\pi2$$ $$\implies\boxed{2\sqrt3>\pi}$$
answered Dec 10, 2019 at 7:23
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