Happy $\pi$ Day to one and all. The day is special because of the epic time : March 14, 2015 - 9:26:53 which is **3.141592653**

On this special occasion, I've got an amazing problem for you all.

**Prove that $\pi$ is a constant**

This claim is equivalent to saying "all circles are similar."

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## Comments

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TopNewestHappy $\pi$ day to you too ! Frankly I didn't know that such a question could be asked !

Here's a proof that everyone here on Brilliant knows :

The two triangles are similar , so the ratio of their radii is constant and Hence the two circles are similar .

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That's also a nice proof!

You maybe interested in trying this one also. Prove that the ratio of circumference to diameter of a circle equals the constant $\pi = 3.141592653 \ldots$

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@Kishlaya Jaiswal - please provide a proof for this one, without using calculus because i think using calculus is a round about method because then we will use radians and the system of radians is now based indirectly upon the fact that $2 \pi r = c$ , so it is not correct.

I have tried and thought about it but it seems that all methods of approach indirectly need to include $\pi$ from some other source (like trignometry, or radians) which are again derived by using a unit circle and the result that we have to prove,

I mean , how do you find the value without any other thing,

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this ?

Hi, can you pls seeLog in to reply

Two mathematicians of international repute meet to share notes about the circle. Both each have brought to the meeting a picture of a circle, and both have determined the numerical value of $\pi$. Their figures agree. Then one of them says, "But wait a minute, your figure is based on the metric system, I'm using the English! How that can be?"

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Well, "pi" is the ratio of two similar quantities(length in this case), hence whichever units we use, they'll get cancelled out when we divide the circumference by the diameter(so it becomes unitless) and the value of pi will remain the same no matter what.

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Pi is a dimensionless quantity (because it's a ratio), which means it's scale invariant. Invariants are extremely important in theoretical physics.

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To prove that all circles are similar, first draw 2 concentric circles with radius ${r}$ and ${R}$ respectively (where R > r ). Now draw two lines from the centre of both circles and make a triangle in each individual circle and call the third side of each one ${x}$ and ${y}$ (${x}$ < ${y}$). Also, suppose that the angle at the centre is given by $\frac{360}{n}$. By similar triangles: $\frac{y}{R} = \frac{x}{r}$ It is clear from the diagram that $as \ n \rightarrow\infty \sum_{}^{} y = \ circumference \ = C_{1}$ $and \ as \ n \rightarrow\infty \sum_{}^{} x = \ circumference \ = C_{2}$ $\therefore\frac{C_1}{R} = \frac{C_{2}}{r}$ Which shows that all circles are similar because the ratio of the circumference to the radius is constant (namely, 2$\pi$ which I may prove later). Q.E.D.

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Here's the diagram :

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That's interesting!

Thanks.

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Yes!! Yesterday it was sir Einstein's B'day !!

I waited for one whole day to see if anyone mentions his name , but no ! No one did so .

Poor guy $\ddot\frown$

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Yeah, I came to know about it a day later. $\ddot \frown$

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haha me too!

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Prove how pie is considered as 180 degree.

All circles are similar because they always have the same round shape...

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That's too vague , don't you think ?

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pi radians = 180 degrees

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Frankly pi is not a constant!! It depends on the geometry of the space. It is for a flat space pi=3.141592653......

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The amazing thing about $\pi$ is that it STILL crops up in constant-curvature Riemannian or hyperbolic geometries! Even though not necessarily as the ratio between the radii and the circumference of circles. It just keeps popping up everywhere, not just in "flat space".

To drive this point home, let's not forget Einstein's famous General Relativity equation for curved spacetime

$G=8\pi T$

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Let Circle1 represent a circle of radius r1, and construct it. Let Circle2 represent a circle of radius r2, such that r2 > r1 and Circle2 is concentric with Circle1, and construct it. Choose an integer n \ge 3 and let \theta = 360^\circ/n. Construct an isosceles triangle, \triangle T

2, such that each equal leg has length r2, the vertex adjacent to both equal legs (v2) lies at the center of Circle2, and the measure of the angle at v2 is equal to \theta. Let s2 represent the length of the side opposite \theta. By this, it can be seen that filling Circle2's interior with n triangles congruent to \triangle T2 such that all share a vertex at the center of Circle2, the equal legs extend outward from that vertex, and none overlap, a regular polygon is created whose perimeter approximates the circumference of Circle2.By the above constructions, a similar isosceles triangle, \triangle T

1, is made such that each equal leg has length r1, the vertex adjacent to both equal legs (v1) lies at the center of Circle1, and the measure of the angle at v1 is equal to \theta. Let s1 represent the length of the side opposite \theta in \triangle T1. Let c2 represent the circumference of Circle2, and c1 represent the circumference of Circle1. Then:c

2\approx ns2 c1\approx ns1 Further, let n increase without bound. Then:c

2 = \lim{n \to \infty}ns2 c1 = \lim{n \to \infty}ns1 and{c

2\over 2r2}={\lim{n \to \infty}ns2\over 2r2}=\lim{n \to \infty}{ns2\over 2r2} (the ratio of Circle2's circumference to its diameter){c

1\over 2r1}={\lim{n \to \infty}ns1\over 2r1}=\lim{n \to \infty}{ns1\over 2r1} (the ratio of Circle1's circumference to its diameter)To prove the proposition, it suffices to show that:

{\lim

{n \to \infty}}{ns2\over 2r2}=\lim{n \to \infty}{ns1\over 2r1} By the laws of similar triangles:{s

1\over r1}={s2\over r2} Substituting:\lim

{n \to \infty} \frac{ns2}{2r2}=\lim{n \to \infty}\frac{ns2}{2r2}Log in to reply

Click here for the relevant diagrams. Firstly, I shall prove that C = 2$\pi$r. To start we shall consider a circle that is being excribed and inscribed by two squares, which creates the following inequality: $8rsin45 < C < 8rtan45$ because the square inside the circle has a smaller perimeter and the outside circle has a larger perimeter. Using this principle with hexagons gives: $12rsin30 < C < 12r tan30$. Now it clear that if ${n}$ = number of sides of a regular polygon, then generally: $2nr \ sin(\frac{180}{n}) < C < 2nr \ tan(\frac{180}{n})$ $n \ sin(\frac{180}{n}) <\frac{C}{2r} < n \ tan(\frac{180}{n})$ $\therefore\lim_{n \rightarrow\infty} nsin(\frac{180}{n}) = \lim_{n \rightarrow\infty} ntan(\frac{180}{n}) = \pi \Rightarrow\ C =2\pi r$ Now the area of a triangle can be given by $\frac{1}{2}absin(c)$, so by splitting up a n-sided polygon into 2n congruent triangles : $A = \lim_{n \rightarrow\infty} r^2 nsin(\frac{180}{n}) = \pi r^2$

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However, we can also prove that C = 2$\pi$r by using radians, so we are making the assumption that the equation is true (a bit like mathematical induction, but not quite). $C = \lim_{n \rightarrow\infty} 2rn.sin(\frac{\pi}{n}) = 2\pi r \lim_{n \rightarrow\infty} \frac{sin(\frac{\pi}{n})}{\frac{\pi}{n}} \ let \ \psi = \frac{\pi}{n}$ $\Rightarrow\ = 2\pi r \lim_{\psi \rightarrow\ 0} \frac{sin(\psi)}{\psi} = 2 \pi r$ (as $\lim_{\psi \rightarrow\ 0} \frac{sin(\psi)}{\psi}$ = 1 from the taylor series of sin(x) )

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Wow! that was an amazing proof.

Just a little typo : $\lim_{\psi \rightarrow \infty}\frac{\sin \psi}{\psi} = 0$ and $\lim_{\psi \rightarrow 0}\frac{\sin \psi}{\psi} = 1$

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$\frac{\pi}{4} = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} = (1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} +...) \ and \ Ln(2) = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} -...)$ $\Rightarrow\ 4Ln 2 < \pi$ If you really want to challenge yourself :)

Thanks - I was bound to make a typo somewhere. Also, it is possible to prove that:Log in to reply

$\ddot \smile$

Yea! that inequality seems so nice. I'm working on the proof.Btw I posted something similar here

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$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \int_0^1 \sum_{k=0}^\infty (-x)^{2k} \mathrm{d}x = \int_0^1 \frac{1}{1+x^2} \mathrm{d}x = \left(\tan^{-1}x\right|_0^1 =\frac{\pi}{4}$ and $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} = \ln(2)$ Subtracting them up, we get $\ln(2)-\frac{\pi}{4} = \left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} + \ldots\right)-\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7} + \ldots\right)$ $\quad = -\frac{1}{2}+\frac{2}{3}-\frac{1}{4}+\frac{2}{7}-\frac{1}{6}+\ldots$ Rearranging gives $\quad = \frac{2}{3}+\left(\frac{2}{7}-\frac{1}{2}\right)+\left(\frac{2}{11}-\frac{1}{4}\right)+\left(\frac{2}{15}-\frac{1}{6}\right)+\ldots$ $\quad = \frac{2}{3}+\left(\frac{-3}{14}\right)+\left(\frac{-3}{44}\right)+\left(\frac{-3}{90}\right)+\ldots < 0$

Thus $\ln(2)-\frac{\pi}{4} < 0 \Rightarrow 4\ln(2) < \pi$

But I know that this is obviously wrong proof because rearranging it the other way makes the sum positive.

I am interested in seeing your solution.

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$Ln(2) = (1-\frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) +(\frac{1}{5} - \frac{1}{6})+...= \displaystyle\sum_{n=1}^{\infty} (\frac{1}{2n-1} - \frac{1}{2n}) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n(2n-1)}$ $\frac{\pi}{4} = (1-\frac{1}{3}) + (\frac{1}{5} - \frac{1}{7}) +(\frac{1}{9} - \frac{1}{11}) +... = \displaystyle\sum_{n=1}^{\infty} (\frac{1}{4n-3} - \frac{1}{4n-1}) = \displaystyle\sum_{n=1}^{\infty} \frac{2}{(4n-3)(4n-1)}$ Now if the individual terms of the expansion of $\frac{\pi}{4}$ are larger than that of Ln(2), then the whole expression will be larger..... $\frac{2}{(4n-1)(4n-3)} > \frac{1}{2n(2n-1)} \Rightarrow\ 4n(2n-1) > (4n-1)(4n-3)$ $\Rightarrow\ 8n^2 -12n - 3 = 0 \Rightarrow\ n^2 - \frac{3}{2}n -\frac{3}{8} = 0 \Rightarrow\ (n-\frac{3}{4})^2 > \frac{15}{16} \therefore\ n > \frac{3+\sqrt{15}}{4} \approx\ 1.75$ Now the first term of $\frac{\pi}{4}$ can be seen to be larger, so all the pairs of terms are larger. $\ Hence \ Ln(2) < \frac{\pi}{4} \therefore\ 4Ln(2) < \pi \ \ \ \ \ \ \ Q.E.D.$

I factored both expressions into brackets with pairs of terms, which are in their original order. Each bracket can be considered as a single term to produce a different way of writing the sum, as follows:Log in to reply

$\frac{\pi}{4}<\ln(2)$. For example

But there's just a little problem. How can you be sure about grouping the terms like the way you have done because you can combine different terms to show that$\frac{\pi}{4} = 1 + \left(-\frac{1}{3}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{9}\right)+\ldots$ $\quad = 1 - \frac{2}{15} - \frac{2}{63}-\ldots$

Now with the above terms, you can show that the individual terms in expansion of $\dfrac{\pi}{4}$ are less than those of $\ln(2)$

Sir @Calvin Lin , can you help us with this inequality.

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Pi is a constant because it is ratio of the circumference to the radius (or the diameter which is twice the radii).

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But actually you need to prove it - Prove that the ratio circumference to diameter equals the constant $\pi=3.141592653 \ldots$.

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its not a day its only a second hahaha

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For 2 circles haves legth each one c1 and c2 each diameter d1and d2 we know that circles are directly porpotional for c1/d1=c2/d2 ...therfor...aslo we can to write K=c1/d1=c2/d2 ... Were k is pi .... Then for every legth C of circles with diameter d... We have.. Pi=C/d then C=Pi * d or C = 2

Pid !!!! CVD[it] / QED[en] sorry for my english! :DLog in to reply

think of the ratio of the distance from the foci to the curve and length of the curve. {directrix==diameter?? } since both foci meet at one point, e=0, the curve becomes a circle.. radius of curvature and length of curve here depends only on radius of the curve., which is fixed for a single circle.. So, their ratio becomes fixed.,.. for all circles. that is the definition of pi. more formal proof is out there.. :D

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pi is not constant ,,, its an add numbers,,\

hehhe

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I memorized 94 more digits of pi this week so that I could have 314 on pi day. At 9:26 I tried to recite all the digits before the minute ended (although before the 53rd second would have been best). I failed... It took me nearly 1.5 minutes to finish.

The technique of approximating the circle with polygons is well known and allowed ancient peoples to approximate pi as well. The Rhind Mathematical Papyrus describes how the Egyptians used an octagon to determine a proportion between the radius^2 and the area. At best, their approximation would only reveal pi to be 3.11 (The figure is 63/20.25 but they rounded to 64/20.25 which is 3.16). More accurate representations are achieved by using polygons with more sides and by using a polygon that "surrounds" the circle and another that lies in the circle.

(sorry, I know this is not a proof :) )

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pi doesn't change, no matter how many times you divide the circumference of a circle by its diameter. Done.

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But you need to prove it somehow, no matter you prove it mathematically or by giving arguments.

Just saying that the ratio of circumference to diameter is a constant and equals $\pi$ doesn't works. Try proving it. (Hint - look at the solution(s) above). Best of luck.

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For any circle with radius r ( irrespective of its value or unit), circumference = 2cr where c is a constant diameter=2r pi=circumference/diameter=2cr/2r=c=constant

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