Putnam 2001, Problem A-6

Can an arc of a parabola inside a circle of radius $1$ have a length greater than $4$?

Solution:
The answer is yes. Consider the arc of the parabola $y = Ax^2$ inside the circle $x^2 + (y-1)^2 = 1$, where we initially assume that $A > 1/2$. This intersects the circle in three points, $(0,0)$ and $(\pm\sqrt{2A-1}/A, (2A-1)/A)$. We claim that for $A$ sufficiently large, the length $L$ of the parabolic arc between $(0,0)$ and $(\sqrt{2A-1}/A, (2A-1)/A)$ is greater than 2, which implies the desired result by symmetry. We express $L$ using the usual formula for arclength:
$$L = \int_0^{\sqrt{2A-1}/A} \sqrt{1 + (2Ax)^2} dx$$
$$= \frac{1}{2A} \int_0^{2\sqrt{2A-1}} \sqrt{1+x^2} dx$$
$$= 2 + \frac{1}{2A} \left(\int_0^{2\sqrt{2A-1}} (\sqrt{1+x^2} - x) dx - 2\right),$$

where we have artificially introduced $-x$ into the integrand in the last step. Now, for $x \geq 0$,
$$\sqrt{1+x^2} - x = \frac{1}{\sqrt{1+x^2} + x} > \frac{1}{2\sqrt{1+x^2}} \geq \frac{1}{2(x+1)};$$

since $\int_0^\infty dx/(2(x+1))$ diverges, so does $\int_0^\infty (\sqrt{1+x^2} - x) dx$. Hence, for sufficiently large $A$, we have $\int_0^{2\sqrt{2A-1}} (\sqrt{1+x^2} - x) dx > 2$, and hence $L > 2$.

Note: a numerical computation shows that one must take $A > 34.7$ to obtain $L > 2$, and that the maximum value of $L$ is about 4.0027, achieved for $A \approx 94.1$.