Putnam 2004, Problem B-3

Determine all real numbers $a > 0$ for which there exists a nonnegative continuous function $f(x)$ defined on $[0,a]$ with the property that the region
\[
R = \{(x,y) : 0 \leq x \leq a, 0 \leq y \leq f(x)\}
\]
has perimeter $k$ units and area $k$ square units for some real number $k$.

Solution:
The answer is $\{a \mid a > 2\}$. If $a > 2$, then the function $f(x) = 2a/(a-2)$ has the desired property; both perimeter and area of $R$ in this case are $2a^2/(a-2)$. Now suppose that $a \leq 2$, and let $f(x)$ be a nonnegative continuous function on $[0,a]$. Let $P = (x_0,y_0)$ be a point on the graph of $f(x)$ with maximal $y$-coordinate; then the area of $R$ is at most $ay_0$ since it lies below the line $y = y_0$. On the other hand, the points $(0,0)$, $(a,0)$, and $P$ divide the boundary of $R$ into three sections. The length of the section between $(0,0)$ and $P$ is at least the distance between $(0,0)$ and $P$, which is at least $y_0$; the length of the section between $P$ and $(a,0)$ is similarly at least $y_0$; and the length of the section between $(0,0)$ and $(a,0)$ is $a$. Since $a \leq 2$, we have $2y_0 + a > ay_0$ and hence the perimeter of $R$ is strictly greater than the area of $R$.