I have been doing some ukmt problems for fun but this one has seemed to stumbled me (Question 8 from the 2015/2016 seniour maths challenge group round http://www.furthermaths.org.uk/docs/1Group%20for%20teams.pdf). I get that if I name the sides of each square $a,b,c,d,e$ starting from the biggest square clockwise I get that $a^2+b^2+c^2+d^2+e^2=2016$. We want to find the area of the pond, which is its side squared call it $\mu$. By Pythagorous $\mu^2=b^2+c^2$ and by similar triangles $\mu=d$. Then, $a^2+\mu^2+\mu^2+e^2=2016$ which implies that $a^2+2\mu^2+e^2=2016$. But from here I am stuck, I can't seem to get the sides of the two biggest squares in terms of $\mu$. I tried making triangles by considering the diagonals going through the two and second two smallest squares but with no successes. I checked the answer sheet and it said that the value of $\mu^2=288$ from that it follows that $a^2+e^2=5\mu$ but even knowing what I need to get I still can't do the problem. Any help would be greatly appriciated
Interesting Geometry Problem UKMT 2015/2016
contest-mathgeometry
Related Solutions
To make explicit Peter Taylor's lower bound: the corners of the square are $90^\circ$. If you don't split one of them, you have a $90^\circ$ angle. If you do, the largest small angle you can get is $45^\circ$. Then if the other two angles of that triangle are equal, they are $67.5^\circ$ This does not show that there is such a construction. I have found several constructions that result in a smaller square.
You'll want something like this:
i.e. where the side of the inscribed square goes through the corner points of the little cut out squares.
Draw some helpful little lines:
And you find that the area is the middle $6\times 6$ square plus exactly half of the area of all the black and blue triangles (since for every blue triangle there is a corresponding black triangle). So, the area is:
$$36 + \frac{4 \cdot 2 \cdot 6}{2}=36+24=60$$
Here's a graphical assist from @Blue:
Best Answer
This actually reminded me of a video by Mathologer - the image of the ponds and lawn in particular. The video in question is here, and the similarities become very apparent roughly sixteen minutes in.
Sadly, I've never tried to prove these facts myself, and they're offered without proof in the video, so this answer is incomplete in that respect. Sorry.
Consider a triangle of side lengths $A,B,C$ (need not be a right triangle). Attach a square to each of the triangle's sides. Then connect the top-right vertex of one square to the top-left of the next going clockwise. This creates side lengths of $X,Y,Z$, as below, and
$$X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2)$$
Let our $A,B,C$ triangle be a right triangle now with hypotenuse $C$. Repeat the setup above. We get the below:
It turns out that the area of the yellow square is equal to that we added to $C$ (i.e. $Z^2 = C^2$). Moreover, the yellow square has one fifth the area of the two orange squares (i.e. $5Z^2 = X^2+Y^2$).
How shall we apply these to your problem? Label the edges in the same manner as I have above (hypotenuse of the smallest right triangle is $C$, the $A$ and $C$ squares eventually form edge $Y$, etc.):
We seek $Z^2$, and know $A^2+B^2+C^2+X^2+Y^2=2016$.
However, we know $X^2+Y^2 = 5Z^2$, and $Z^2=C^2$ and thus
$$A^2+B^2+6Z^2=2016 \tag 1$$
We return to our first relation for arbitrary center triangles:
$$X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) = 3A^2 + 3B^2 + 3C^2$$
We apply the $Z^2 = C^2$ and $X^2+Y^2=5Z^2$ relations and simplify, solving for $A^2+B^2$:
$$Z^2 = A^2 +B^2 \tag 2$$
Combining $(1),(2)$ is trivial and gives us
$$7Z^2 = 2016 \implies Z^2 = \frac{2016} 7 = 288$$