Characterization of $\pi$ Let $x$ be the ratio of a circle's circumference to its diameter.
Let $y$ be twice the smallest positive number $t$ for which $\cos(t)$ equals $0$.
How to prove $x=y$?
Thanks.
 A: Here is a tedious proof the mainly relies on properties of ODEs. The power series are a little easier to work with (and easy to establish from the ODE), but I think using them loses some connection with the geometric aspects.
Amazingly (given the length of the proof) I have left out many details below.
The unit circle is $S^1 = \{ x \in \mathbb{R}^2 | \|x\| = 1 \}$.
A circle of radius $r>0$ corresponds to $r S^1$.
Length is a little complicated to define. Since I don't know how to directly ascribe a meaningful length to a shape such as $S^1$, instead I will define length by means of certain class of  parametrizations. (This is the connection between the intuitive geometric notion and the analytic perspective.)
Given a continuous function $\gamma:[a,b] \to \mathbb{R}^2$, we define the length $l(\gamma) = \sup_{{\cal P}} \sum_k | \gamma(t_{k+1}) -\gamma(t_k)|$ (which may be infinite in general), where ${\cal P}$ are the partitions of $[a,b]$. If $\gamma$ is $C^1$, we can show $l(\gamma) = \int_a^b \| \gamma'(t)\| dt$. For our purposes, there is no loss of generality in taking $a=0, b=1$.
Now let $\Gamma_r = \{ \gamma:[0,1] \to rS^1 | \gamma \text{ is continuous}, \gamma:[0,1) \to rS^1 \text{ is bijective} \}$. A little work shows that if $\gamma_1,\gamma_2 \in \Gamma_r$, then $l(\gamma_1) = l(\gamma_2)$, so we can assign a length to $rS^1$ (that is, the circumference) by choosing any $\gamma \in \Gamma_r$ and calling $l(\gamma)$ the length.
If $\gamma \in \Gamma_r$, then ${1 \over r} \gamma \in \Gamma_1$, and since $l({1 \over r} \gamma) = {1 \over r} l(\gamma)$, we see there is no loss of generality in just dealing with $r=1$. We will create a suitable $\gamma \in \Gamma_1$ below.
For the purposes of this question, cosine is defined as the solution of a differential equation. I will represent the relevant differential equation as a first order ODE as follows:
Let $A=\begin{bmatrix} 0 & -1 \\
1 & 0\end{bmatrix}$, and $\dot{x} = A x$, with $x(0) = e_1$. Note that   $A^4 = I$. The system is uniformly Lipschitz hence has a unique solution. Since it is linear and time invariant, the solution is smooth. By differentiating, it is straightforward to verify that $\|x(t)\|^2 = 1$ for all $t$, and consequently $\|\dot{x}(t)\|^2 = 1$.
Then $x_1$ corresponds to cosine and $x_2$ to sine. To establish the desired result, we show that there exists some smallest $T>0$ such that $x_1(T) = 0$ (then $T = { \pi \over 2}$, of course), and show that $t \mapsto x(t)$ traverses the unit circle for $t \in [0,4 T)$. Then we compute the length of the path $\gamma(x)=  x(4 T x)$ for $x \in [0,1]$ and show that this length is $4 T$.
First we show the existence of $T$ and show that $x$ traverses a quarter circle on $[0,T]$. 
Start with an eight of a circle: Let $B=[{1 \over \sqrt{2}},2] \times [0, {1 \over \sqrt{2}}]$. Note that $\partial B \cap S^1 = \{ e_1, {1 \over \sqrt{2}}(1,1)^T \}$. We have $x(0) \in B$, and $\dot{x_2}(0) = 1>0$, hence there is some $\delta>0$ such that $x(t) \in B^\circ$ for all $t \in (0,\delta)$.

Define $T' = \sup \{ t \ge 0 | x(\tau) \in B \ \text{for all } \tau \in [0,t] \} $. The previous remark shows that $T' >0$. Note that if $x(t) \in B$, then $\dot{x_2}(t) = x_1(t) \ge {1 \over \sqrt{2}}$. Hence we must have $T' < \infty$, and $x(T') \in \partial B$ from which it follows that $x(T') = {1 \over \sqrt{2}}(1,1)^T$.
Since $x_2$ is strictly increasing for $t \in [0,T']$, we see that the restricted map $x:[0,T'] \to B \cap S^1$ is a bijection.
Let $Q = \{ x  | x \in S^1, x_1\ge 0, x_2 \ge 0 \} $.
Let $y(t) = (x_2(T'-t), x_1(T'-t))^T$ and note that $y(0) = x(T')$ and $\dot{y} = A y$, from which we have $x(t+T') = y(t)$. Hence we have $x(2T') = e_2$. Note that $x_1(t) >0$ for $t \in [0,2T')$, hence $T=2T'$ is the first time that $x_1$ hits zero. It is straightforward to see that the restricted map $x:[0,T] \to Q$ is a bijection.
Now let $y(t) = A^{-1} x(t+T)$, and note that $y(0) = e_1$, and $y$ satisfies the above ODE. Hence $y = x$, and so we have $x(t+T) = A x(t)$ for all $t \ge 0$ (in fact it is true for all $t$, but that is not relevant here).
From this we have $x(n T) = A^n x(0)$, and so $x(4 T) = x(0)$. Also,
the maps $A^{-n} x : [n T , (n+1)T] \to Q$ are bijections. A small amount of work establishes that $x[0,4 T) \to S^1$ is a bijection, and since $x(4 T) = x(0)$, we see if we let  $\gamma(z) = x(4 T z)$, then $\gamma \in \Gamma_1$. Since ${\gamma}'(z) = 4 T $ for all $z$, we can quickly compute $l(\gamma) = 4 T$.
In terms of the original question, we have $x = {1 \over 2} 4 T$ (the ${1 \over 2}$ is because we used radius rather than diameter) and $y = 2 T$, from which we get $x=y$.
