Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers An exam for high school students had the following problem:
Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle?

Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square. 
The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?

This isn't a full answer, just an observation to supplement the other answers (and celebrate Pi Day).
If you construct tangents at the 20 points $(0, \pm25)$, $(\pm25, 0)$, $(\pm15, \pm20)$, $(\pm20, \pm15)$, $(\pm7, \pm24)$, $(\pm24, \pm7)$ on the circle of radius 25 centred on the origin, they intersect at the vertices $(\pm\frac{25}{7}, \pm25)$, $(\pm25, \pm\frac{25}{7})$, $(\pm\frac{125}{11}, \pm\frac{250}{11})$, $(\pm\frac{250}{11}, \pm\frac{125}{11})$, $(\pm\frac{125}{7}, \pm\frac{125}{7})$ of an irregular circumscribing icosagon, whose perimeter is $8 \times \left(\frac{25}{7} + 2 \times \frac{625}{77}\right) = \frac{12200}{77} = 158\frac{34}{77} < 160,$ as required.
Comparing this with the circumference, $50\pi$, of the inscribed circle, we get the upper bound for $\pi$:
$$
\pi < \frac{244}{77} = 3\frac{13}{77} < 3.17.
$$
Here, the circumscribing icosagon is shown in red, with the circle (just) visible behind it in black.

The points of tangency are marked with smaller blue circles. The 16 right-angled triangles used in construction, shown in green, are all Pythagorean, in ratio 3:4:5 or 7:24:25, with hypotenuse $\frac{625}{77}$. The other 4 edges of the icosagon are of length $\frac{50}{7}$.
There is no need to calculate any angles, but the symmetry of the figure is illuminated by knowing the sequence of angles subtended at the centre of the circle. Running counter-clockwise from the positive $x$-axis to the positive $y$-axis, it is: $\underline{\alpha\alpha\beta\beta\alpha\alpha\beta\beta\alpha\alpha}$, where $\alpha = \tan^{-1}\frac{1}{7}$, $\beta = \tan^{-1}\frac{2}{11}$, and:
\begin{gather*}
2\alpha = \tan^{-1}\frac{7}{24}, \
\alpha + \beta = \tan^{-1}\frac{1}{3}, \
2\alpha + \beta = \tan^{-1}\frac{1}{2}, \\
2\alpha + 2\beta = \tan^{-1}\frac{3}{4}, \
3\alpha + 2\beta = \frac{\pi}{4}, \
4\alpha + 2\beta = \tan^{-1}\frac{4}{3}, \\
4\alpha + 3\beta = \tan^{-1}2, \
5\alpha + 3\beta = \tan^{-1}3, \
4\alpha + 4\beta = \tan^{-1}\frac{24}{7}.
\end{gather*}
