2012 Putnam Exam A3 This question mostly eluded me during the exam itself.
Problem: Suppose that $f: [-1, 1] \rightarrow \mathbb{R}$ continuously, and that 
$$\begin{align}
\text{(i)}\qquad   &f(x) = \frac{2 - x^2}{2} f\left(\frac{x^2}{2-x^2}\right)\\
\text{(ii)}\qquad  &f(0) = 1\\[6pt]
\text{(iii)}\qquad &\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}\ \text{exists}
\end{align}$$
Determine the closed form of $f$, and prove that it is unique. ---
All that I managed to do was - using the eventual heuristic of "powers of $1 - x^2$ will cancel neatly in (iii) and work in (ii), so let's see if they work in (i)" - discover that 
$$ f(x) = \sqrt{1 - x^2}$$
Now, this $f$ is an involution, so it suffices to prove that if $g$ is a function satisfying (i) to (iii), then $g(f(x)) = x$. This, however, was more than I could do.
How do you prove uniqueness? (I ran across one solution I did not understand; please, be gentle.)
 A: Disclaimer: I am borrowing much from the Art of Problem Solving link alluded to in the comments.
Since you have guessed a solution, namely $x \mapsto \sqrt{1-x^2}$, so let us exploit it. Denote $s(x) := \sqrt{1-x^2}$, and note, as you did, that $s(s(x)) = x$. It is a good idea to express $f$ in an alternative way so that for $f(x) = s(x)$ things become very simple. This is vague, but a way to proceed is to write $f(x) = h(s(x))$ for some continuous $h:[0,1] \to \mathbb{R}$; since this is equivalent to saying that $h(y) = f(s(y))$, there is a 1 to 1 correspondence between possible $f$'s and $h$'s.
Let us see what the conditions say about $h$:
(i) $$h(y) = f(s(y)) = \frac{1 + y^2}{2} f\left(\frac{1-y^2}{1+y^2}\right) = 
\frac{1 + y^2}{2}h\left(s\left(\frac{1-y^2}{1+y^2}\right)\right) =
\frac{1 + y^2}{2} h\left( \frac{2y}{1+y^2} \right)$$
(ii) $h(1) = 1$
(iii) $\lim_{y \to 0+} \frac{h(y)}{y}$ exists.
There are several ways to proceed at this stage. To keep things elegant, let us notice the striking resemblance between the expression $\frac{2y}{1+y^2}$ and the formula for $\tanh$ of doubled angle. In fact, if $y = \tanh \alpha$, then $\frac{2y}{1+y^2} = \tanh 2 \alpha$. Writing additionally $\frac{1+y^2}{2}$ as $\frac{y}{ \frac{2y}{1+y^2}} $ we conclude that (i) can be re-expressed in a nicer form:
$$ h(\tanh \alpha) = \frac{\tanh \alpha}{\tanh 2 \alpha} h(\tanh 2\alpha)$$
Iterating this as many times as we like, we conclude:
$$ h(\tanh \alpha) 
= \frac{\tanh \alpha}{\tanh 2 \alpha}  \frac{\tanh 2 \alpha}{\tanh 4 \alpha}
\dots \frac{\tanh 2^{k-1} \alpha}{\tanh 2^{k} \alpha}
h(\tanh 2^k\alpha)
= \tanh \alpha  \frac{h(\tanh 2^k\alpha)}{\tanh 2^k \alpha}$$
Passing to the limit $k \to \infty$ (and remembering that $\tanh \beta \to 1 $ as $\beta \to \infty$ we conclude that:
$$ h(\tanh \alpha) = \tanh \alpha$$
This means that $h(y) = y$. Translating this back to $f$ we conclude that:
$$ f(x) = h(s(x)) = s(x) = \sqrt{1-x^2}$$
