# Happy Pi Day!

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." Note by Kishlaya Jaiswal
6 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Happy $\pi$ 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 .

- 6 years, 4 months ago

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 $\pi = 3.141592653 \ldots$

- 6 years, 4 months ago

@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,

- 6 years, 4 months ago

Hi, can you pls see this ?

- 6 years, 4 months ago

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?"

- 6 years, 4 months ago

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.

- 6 years, 4 months ago

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

- 6 years, 4 months ago

So, isn't dimensionless the same as unitless? As far as we don't consider what units we're using for measuring a particular quantity. If a quantity is unitless, then it is dimensionless as well, I suppose. So according to you, what I should've said is, circumference and diameter belong to the same dimension; length- "[L]". So, when we take their ratio, the dimensions get cancelled and thus pi becomes dimensionless and ultimately unitless. is that right?

- 6 years, 4 months ago

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.

- 6 years, 4 months ago

Here's the diagram : - 6 years, 4 months ago

That's interesting!

Thanks.

- 6 years, 4 months ago 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 $\ddot\frown$

- 6 years, 4 months ago

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

- 6 years, 4 months ago

haha me too!

- 6 years, 4 months ago

Prove how pie is considered as 180 degree.

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

$G=8\pi T$

- 6 years, 4 months ago

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 T2, 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 T1, 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:

c2\approx ns2 c1\approx ns1 Further, let n increase without bound. Then:

c2 = \lim{n \to \infty}ns2 c1 = \lim{n \to \infty}ns1 and

{c2\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)

{c1\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:

{s1\over r1}={s2\over r2} Substituting:

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

- 6 years, 4 months ago

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 : $A = \lim_{n \rightarrow\infty} r^2 nsin(\frac{180}{n}) = \pi r^2$

- 6 years, 4 months ago

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) )

- 6 years, 4 months ago

Wow! that was an amazing proof.

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

- 6 years, 4 months ago

Thanks - I was bound to make a typo somewhere. Also, it is possible to prove that: $\frac{\pi}{4} = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k-1} = (1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} +...) \ and \ Ln(2) = \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} -...)$ $\Rightarrow\ 4Ln 2 < \pi$ If you really want to challenge yourself :)

- 6 years, 4 months ago

Yea! that inequality seems so nice. I'm working on the proof. $\ddot \smile$

Btw I posted something similar here

- 6 years, 4 months ago

This is the best I could do.

$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = \int_0^1 \sum_{k=0}^\infty (-x)^{2k} \mathrm{d}x = \int_0^1 \frac{1}{1+x^2} \mathrm{d}x = \left(\tan^{-1}x\right|_0^1 =\frac{\pi}{4}$ 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.

- 6 years, 4 months ago

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: $Ln(2) = (1-\frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) +(\frac{1}{5} - \frac{1}{6})+...= \displaystyle\sum_{n=1}^{\infty} (\frac{1}{2n-1} - \frac{1}{2n}) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{2n(2n-1)}$ $\frac{\pi}{4} = (1-\frac{1}{3}) + (\frac{1}{5} - \frac{1}{7}) +(\frac{1}{9} - \frac{1}{11}) +... = \displaystyle\sum_{n=1}^{\infty} (\frac{1}{4n-3} - \frac{1}{4n-1}) = \displaystyle\sum_{n=1}^{\infty} \frac{2}{(4n-3)(4n-1)}$ 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. $\ Hence \ Ln(2) < \frac{\pi}{4} \therefore\ 4Ln(2) < \pi \ \ \ \ \ \ \ Q.E.D.$

- 6 years, 4 months ago

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

$\frac{\pi}{4} = 1 + \left(-\frac{1}{3}+\frac{1}{5}\right)+\left(-\frac{1}{7}+\frac{1}{9}\right)+\ldots$ $\quad = 1 - \frac{2}{15} - \frac{2}{63}-\ldots$

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.

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

- 6 years, 4 months ago

its not a day its only a second hahaha

- 6 years, 4 months ago

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 = 2Pid !!!! CVD[it] / QED[en] sorry for my english! :D

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

hehhe

- 6 years, 4 months ago

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 :) )

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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.

- 6 years, 4 months ago

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

- 6 years, 4 months ago