Curved Square
There were several interesting solutions to
this problem, using a variety of different approaches, along with
many lovely images. We are sorry that we couldn't include all of
the solutions in full! The answers were:
The area of the shaded area is
$$
\frac{\pi}{3}+1-\sqrt{3}\approx 0.31517
$$
The area of the largest square which can be inscribed is
$$
2-\sqrt{3}\approx 0.267949
$$
The error is about $15\%$.
InNyeong from Nanjing International School
placed the shape on a coordinate grid to express all of the lines
in algebraic form and solved for the resulting intersections
numerically, with a beautifully written up solution:
Ammar Ali followed a similar approach to
InNyeon. although his solution was entirely algebraic. He found the
equations of the curves and then solved algebraically, finding the
coordinates for the points of intersection to be
$$
\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\quad \left(\frac{1}{2},
\frac{\sqrt{3}}{2}\right)\quad \left(1-\frac{\sqrt{3}}{2},
\frac{1}{2}\right)\quad \left(\frac{1}{2},
1-\frac{\sqrt{3}}{2}\right)
$$
The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is
given by
$$
d=\sqrt{\left(x_1-x_2)^2+(y_1-y_2)^2\right)}
$$
By using this formula we can find the squared distance $L^2$
between neighbouring points:
$$
L^2 = 2-\sqrt{3}
$$
This is the area of the largest possible square that fits in the
centre.
Ammar then proceeded with calculating the
integrals of the curves exactly, finding the correct expression for
the areas. We liked this method a lot because it could be
generalisable to curves other than circles (such as ellipses). To
perform the calculation, Amman used the knowledge of the points of
intersection, integrals and the additive properties of
areas
Yike provided an interesting solution based
upon the following modified image:
Trigonometry shows that $\lambda = 60^\circ$ and $\phi = 30^\circ$.
The cosine rule on triangle $OP_1P_2$ gives
$Q=(P_1P_2)^2=2-\sqrt{3}$, which is also the area of the largest
inscribed square (since it is the side-length squared!)
To find the exact area $C$, note that the area of triangle $P_1P_2B
= 0.5 r^2\sin(\lambda -\phi)=0.5 \sin(30^\circ)=0.25$.
So,
$$
C= Q+4\left(\frac{(\lambda-\phi)\pi r^2}{360^\circ}-0.25\right) =
2-\sqrt{3}+4\left(\frac{\pi}{12}-0.25\right)
=1-\sqrt{3}+\frac{\pi}{3}
$$
Bugboy sent a classic geometric proof based
on areas of sectors of circles, a very similar approach to that
used by Roy from Nanjing International School:
The largest square has vertices at the 4 points where two arcs meet
within the square. Because the arcs are 1 unit, the same as the
side length of the square, the point where arcs meet and the
opposite side of the square form an equilateral triangle. Therefore
the distance between the point and the far edge via the
perpendicular bisector (which also goes through the opposite corner
of the inside square) is$\frac{\sqrt{3}}{2}$. The distance between
a corner and the opposite side, plus the distance between the
opposite corner and the correspondingly opposite side is
$2\frac{\sqrt{3}}{2}=\sqrt{3}$. Each $\frac{\sqrt{3}}{2}$ line is
made of two parts: the distance between the two corners of the
inside square, and then the distance to the side of the outer
square. We have two of each in $\sqrt{3}$. The entire distance from
one side to another of the outer square is $1$, and this consists
of 2 end sections and 1 section between the two corners. Therefore
$\sqrt{3} -1$ is just the one diagonal length of the inner square.
The diagonal of any square is the side length multiplied by
$\sqrt{2}$, so in this case the side is $\frac{\sqrt{3}
-1}{\sqrt{2}}$. Square that for the area of the inner square:
$\frac{4-2\sqrt{3}}{2} =2-\sqrt{3}$. In the actual shape, we missed
out the segments on the edges of the square. The segment they are
part of has an angle of $30$ degrees (proof: the angle from the two
sides that meet at the corner where the sector is coming from to
the points at the end of the segment are $30$ degrees, as the other
side of the $90$ degree angle is part of an equilateral, as shown
earlier. Hence it's $60$ degrees, and the angles on either side of
the sector angle are $30$ degrees, leaving $90-30-30=30$ degrees
for the sector angle) so the area of the sector is $\pi\times
1^2\times 30/ 360=\frac{\pi}{12}$. The triangle within the sector
that complements the segment to make the sector has two sides
length 1 with included angle of $30$. Hence the area is
$\frac{1}{2}\times 1\times 1\times \sin30=\frac{1}{4}$. Therefore
each segment is $\frac{\pi}{12}-\frac{1}{4}$. There are 4 segments,
so the total area is $\frac{\pi}{3}-1$. Therefore the total area of
the shape is
$\frac{\pi}{3}-1+2-\sqrt{3}=1+\frac{\pi}{3}-\sqrt{3}$.
Finally, Scott from Pencoed Comprehensive
extracted the key geometrical information very clearly in a
good-sized and clear diagram (reduced in size here), so essential
to any work in geometry:
