A popular math problem is proving that \(22/7\) exceeds \(\pi\), without resorting to known values of \(\pi\) or approximations by calculator. One method uses certain definite integrals which work out to \(22/7\) \( -\pi \) which is known to be greater than \(0\). But back around 250 BC, Archimedes was the first to prove that \(22/7\) exceeds \(\pi\), by successive use of the trigonometric identity

\[Tan(2x)=\dfrac { 2Tan(x) }{ 1-{ (Tan(x)) }^{ 2 } } \]

Starting with the regular hexagon, and doubling the number of sides until he reached a regular polygon of \(96\) sides, he was able to show that \(22/7\) does indeed exceed \(\pi\). Unfortunately, he relied on a series of inequalities in his computations, the notes of which are lost, so we have no complete proof left to us from his works. So, here’s an updated version of his approach that will be an exact proof, beginning with the following diagram

where\(\angle OPB=90°\)

\(OP=OA=1\)

\(OB=Sec(\dfrac { 45° }{ n } )\)

\(OC=(\dfrac { a }{ b } )(\dfrac { 1 }{ 4n } )Csc(\dfrac{45°}{n})\)

In this diagram shown, \(n=2\) and \(\dfrac { a }{ b } =\dfrac { 25 }{ 7 } \), a very crude apprximation of \(\pi\). \(\dfrac { a }{ b } \) is selected so that \(OC>OB\), so that it is clear that the areas are

\(\left[ OPC \right] >\left[ OPB \right] >\left[ OPC \right] \)

That is

\((\dfrac { 1 }{ 8n } )(\dfrac { a }{ b } )>(\dfrac { 1 }{ 2 } )Tan(\dfrac { 45° }{ n } )>(\dfrac { 1 }{ 8n } )\pi \)

so that if we can prove that for some sufficiently large \(n\)

\((\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } )>Tan(\dfrac { 45° }{ n } )\)

then we've proven that

\(\dfrac { a }{ b } >\pi \)

Now but there is a delightfully useful trigonometric identity to use for this purpose, which Archimedes didn't have

\(Tan(nx)=F(n,Tan(x))=i\dfrac { { (1-iTan(x)) }^{ n }-{ (1+iTan(x)) }^{ n } }{ { (1-iTan(x)) }^{ n }+{ (1+iTan(x)) }^{ n } } \)

If \((\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } )>Tan(\dfrac { 45° }{ n } )\), then

\(F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))>F(n,Tan(\dfrac { 45° }{ n } )) \), (a little handwaving here), but since

\(F(n,Tan(\dfrac { 45° }{ n } ))=1\), we have

\(F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))>1\)

So, for \(\dfrac { a }{ b } =\dfrac { 22 }{ 7 } \), all we have to do is to show that for some sufficiently large \(n\), let's say \(n=24\), which corresponds to the regular \(96\) sided polygon Archimedes used

\(F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))=F(24,(\dfrac { 1 }{ 96 } )(\dfrac { 22 }{ 7 } ))>1\)

Using that trigonometric identity above, the exact final rational fraction is

\(\dfrac { 3070617780371250623172531488874792422422762969087263790473600 }{3070399174588386853835740065608415583799863505007168821897761 } \)

Note that the first \(6\) digits of the numerator and denominator are

\(307061>307039\)

which conclusively and exactly proves that \(22/7\)\(>\pi\), without resorting to approximations, without using known computed values of \(\pi\), nor infinite series, nor infinite products, nor infinite continued fractions.

Going further, we can let \(n=1557\), corresponding to a regular polygon of \(6228\) sides, and let \(a/b=355/113\), another famous and very accurate approximation. We end up with a rational fraction with a numerator and a denominator both \(9105\) digits long, the first \(12\) digits of each being

\(196554866601 > 196554866571\)

thus proving that \(355/113>\pi\) as well.

Addendum: Note that as \(n\rightarrow \infty \)

\(Tan(x)=Tan(n(\dfrac { x }{ n } ))=F(n,Tan(\dfrac { x }{ n } ))\approx F(n,(\dfrac { x }{ n } ))\)

\(F(n,(\dfrac { x }{ n } ))=i\dfrac { { (1-\dfrac { ix }{ n } ) }^{ n }-{ (1+\dfrac { ix }{ n } ) }^{ n } }{ { (1-\dfrac { ix }{ n } ) }^{ n }+{ (1+\dfrac { ix }{ n } ) }^{ n } } =i\dfrac { { e }^{ -ix }-{ e }^{ ix } }{ { e }^{ -ix }+{ e }^{ ix } } =Tan(x)\)

which shows the relationship between the exponential form of the \(Tan(x)\) function and the trigonometric identity given above for \(Tan(nx)\)

## Comments

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TopNewestMy favorite proof is this funny integral. \[0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \,\mathrm dx = \frac{22}7 - \pi\] – Ali Caglayan · 3 years ago

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– Michael Mendrin · 3 years ago

Well, it's true that it's hard to beat that one as "something that can be done on the back of an envelope". A large envelope anyway.Log in to reply

Eh, but you know, at the point where I said, "a little handwaving here", there's actually more that needs to be said before this really becomes a complete proof. But this tiny detail will take up too much space to resolve. – Michael Mendrin · 3 years, 1 month ago

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That's really something, u know!! Thankx for posting it.. Quite enlightening. :-) – Chinmay Raut · 3 years, 1 month ago

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@Michael Mendrin , Your note inspired my next sum!

Integrate It! Part-V

https://brilliant.org/community-problem/integrate-it-part-v/?group=DvYD57CYrdYl&just_created=true – Avineil Jain · 3 years, 1 month ago

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Wow. – Joshua Ong · 3 years, 1 month ago

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– Yuxuan Seah · 3 years, 1 month ago

I know right!? :DLog in to reply

Another proof that \(\large\frac{22}{7}>\pi\). – Anastasiya Romanova · 3 years, 1 month ago

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– Avineil Jain · 3 years, 1 month ago

Dude, you just published the solution of my problem! Why?!Log in to reply

– Michael Mendrin · 3 years, 1 month ago

Probably because she didn't know you posted this as a problem?Log in to reply

– Anastasiya Romanova · 3 years, 1 month ago

Like you said Sir. I did not know that. Besides, I already knew this problem long time ago. This problem is a problem in Putnam competition.Log in to reply

– Avineil Jain · 3 years, 1 month ago

All right, but now she knows.Log in to reply

– Anastasiya Romanova · 3 years, 1 month ago

Gimme a break!? How do I know that?? As Mr. Mendrin said, I did NOT know that. Why didn't you comment by inserting your problem + link direction instead of link only?? (¬‿¬)Log in to reply

– Avineil Jain · 3 years, 1 month ago

Also, can you tell me how to use + link that you are telling?Log in to reply

`[your problem](link direction)`

– Anastasiya Romanova · 3 years, 1 month agoLog in to reply

– Avineil Jain · 3 years, 1 month ago

All right, since you now know, can you please delete your comment?Log in to reply

– Michael Mendrin · 3 years, 1 month ago

Valentina, actually, I didn't mind you posting the solution to that classic definite integral. That was the one I was thinking about when I referred to them at the beginning of this short paper. But it's not the only one possible. Maybe you can post another that isn't the same as Avineil's posted problem?Log in to reply

– Anastasiya Romanova · 3 years, 1 month ago

Right now, I have no idea. If something crosses to mind, I will post it here.Log in to reply