Let $A, B$, and $C$ be points on a line with $B$ between $ A$ and $C,$ so that $BC <AB$. Build up the squares $ABDE$ and $BCFG $ on the same side of the line.
(a) Calculate the ratio $\frac {EF} {AG}.$
(b) Calculate the sum of $\angle BAG + \angle GFE.$
(c) Prove that the $AG, EF$ and $DC $ lines compete on a single point.
Fake solution: Because we're given no other information on the location of B, we can assume the values of (a) and (b) are invariant for point B. Let B be the point approaching the midpoint of AC. (a) EF is two sidelengths, AG is a diagonal, so $\frac{EF}{AG}=\frac{2}{\sqrt2}$
(b) $\angle BAG, \angle GFE$ are $45^o, 0$ respectively. So, $\angle BAG + \angle GFE=45^o$
(c) $AG, DC$ are diagonals which have a common endpoint, which is on $EF$.
But as it is an Olympics issue, there must be a way to solve
Nice approach: Homothety.
Let $ \rho$ be the transformation of a $45^\circ$ rotation about $B$ and scaling by $ \sqrt{2}$ about $B$.
Let $\phi$ be the transformation of a $90^\circ$ rotation about $B$
(a) $\rho(BAG) = BEF$.
Hence $\frac{ EF}{AG} = \sqrt{2}$.
(b) Since $\rho BAG = BEF$, hence $\angle BAG = \angle BEF$.
$ \angle BAG + \angle GFE = \angle BEF + \angle GFE = 360^\circ - \angle EBG - \angle BGF = 360^\circ - 45^\circ - 270^\circ = 45^ \circ $
(c) Let the foot of the perpendicular from $B$ to $AG$ be $I$.
Let the foot of the perpendicular from $B$ to $CD$ be $J$. Since$ \phi (BAG) = BCD$, $BI$ = $BJ$.
Also, $AG$ and $CD$ are perpendicular. Let them intersect at $H$
Hence $BIHJ$ is a square, and $\rho (I) = H$.
Since $I$ lies on $AG$, it follows that $\rho(I) = H$ lies on $\rho(AG) = EF$.
Thus, $EF$ passes through $H$, so these 3 lines are congruent.
Note: We can also conclude that
1) $BH \perp EF$, and
2) $H$ is the intersection of the circumcircles about both squares.
Brute force:
(a) Let $AB = a, BC = b$.
Then $AG^2 = AB^2 + BG^2 = a^2 + b^2$ and
$EF^2 = DG^2 + AC^2 = (a-b)^2 + (a+b)^2 = 2 (a^2 + b^2)$.
Hence $\frac{EF}{AG} = \sqrt{2}$.
(b) $ \tan \angle BAG = \frac{b}{a}$.
$\tan \angle GFE = \frac{ a-b}{ a+b}$.
$ \tan ( \angle BAG + \angle GFE ) = \frac{ \frac{b}{a} + \frac{a-b}{a+b} } { 1 - \frac{b}{a} \times \frac{a-b}{a+b}} = 1 $.
Show that these angles do not sum up to more than $90^\circ$ to conclude that they sum up to $45^\circ$.
(c) Let $B$ be the origin. We have
$A = (-a, 0)$, $G = (0,b)$,
$E = (-a,a), F = (b, b)$,
$C = (b,0)$, $D = (0,a)$,
Show that $ ( \frac{ab(a-b)}{a^2+b^2}, \frac{ab(a+b)}{a^2+b^2})$ lie on all 3 lines.