Putnam Challenge Question 
Determine all real numbers $a$, where $a>0$, for which there exists a nonnegative continuous function $f(x)$, defined on $[0,a]$ with the property that the region $$R=\big\{(x,y)\,\big|\, 0\le x\le a\text{ and } 0\le y\le f(x)\big\}$$ has perimeter $k$ units and area $k$ units for some real number of $k$.

I apologize for not yet knowing how to type equations as they are seen in most questions on this forum. I joined about 15 minutes ago.
 A: We first claim that $a>2$ must be satisfied for such a function $f$ to exist.  Suppose that $f:[0,a]\to\mathbb{R}_{\geq 0}$ satisfies the condition that the perimeter $\sigma$ of $R=\big\{(x,y)\in [0,a]\times \mathbb{R}_{\geq 0}\,\big|\,y\leq f(x)\big\}$ equals its area $$\alpha:=\displaystyle \int_0^a\,f(x)\,\text{d}x\,.$$ 
Write $m$ for the maximum value of $f$ in the interval $[0,a]$ (which exists as $f$ is continuous and $[0,a]$ is a compact space).  Let $c\in[0,a]$ be such that $f(c)=m$.  Then,
$$\sigma \geq a+f(0)+f(a)+\sqrt{c^2+\big(m-f(0)\big)^2}+\sqrt{(a-c)^2+\big(m-f(a)\big)^2}\,.$$
Using the Triangle Inequality, we get that
$$\begin{align}\sigma &\geq a+f(0)+f(a) +\sqrt{\big(c+(a-c)\big)^2+\Big(\big(m-f(0)\big)+\big(m-f(a)\big)\Big)^2}\\
&\geq a+f(0)+f(a)+\sqrt{a^2+\big(2m-f(0)-f(a)\big)^2}\\
&>a+f(0)+f(a)+\big(2m-f(0)-f(a)\big)=a+2m>2m\,.
\end{align}$$
On the other hand, we see that
$$\alpha \leq \int_0^a\,m\,\text{d}x=am\,.$$
Because $\alpha=\sigma$, we conclude that
$$am\geq \alpha = \sigma>2m\text{ or }a>2\,,$$
since $m$ is clearly a positive real number.
To finish the proof, we shall now verify that, when $\alpha>2$, such a function $f$ exists.  As discovered in the comment section, the constant function $f$ defined via $$f(x)=\frac{2a}{a-2}\text{ for all }x\in[0,a]$$ has the required property (with $k=\dfrac{2a^2}{a-2}$).
