Notes
four squares xi solution

Solution to the Four Squares XI Puzzle

Solution by Symmetry and Lengths in Squares

Consider the diagram as labelled above, in which the line segment $S F$ is perpendicular to $A B$, and $S G$ is perpendicular to $B C$.

Rotating the upper blue square anti-clockwise $90^\circ$ about $S$ brings it to lie on top of the lower blue square, with $A$ brought to $Q$. Therefore, applying the same rotation to the line segment $A F$ means that it starts at $Q$ and lies along $Q C$, similarly rotating $S F$ starts at $S$ and lies along $S G$. Putting these together means that $F$ rotates to $G$ and so triangle $S F A$ rotates to triangle $S G Q$.

Hence line segments $S F$ and $S G$ are the same length which means that $S$ lies on the diagonal $B D$ of the black square. This in turn means that line segments $S A$ and $S C$ have the same length, so the side length of the green square is the length of the diagonal of the blue.

Therefore, by properties of squares, the green square has area twice that of the blue, so has area $24$.

Solution by Similar Triangles and Lengths in Squares

With the diagram labelled and annotated as above, consider triangle $A P B$. This is isosceles as $P$ is the midpoint of the line segment $A Q$.

Now consider triangle $A S C$. This is obtained from triangle $A P B$ by a rotation $45^\circ$ anti-clockwise about $A$ together with a scaling by $\sqrt{2}$. To see this, note that this combined transformation is what takes an adjacent side of a square to an opposite one, and therefore takes $P$ to $S$ and $B$ to $C$ while leaving $A$ where it is.

Therefore, triangle $A S C$ is isosceles and so $S C$ has the same length as $S A$. From this, the area of the green triangle is seen to be $24$.

Solution by Pythagoras' Theorem

With the points labelled as above, line segments $P I$ and $P J$ are perpendicular to $A B$ and $B C$ respectively. As before, $S G$ is perpendicular to $B C$, and $R K$ and $R H$ are perpendicular to $S G$ and $B C$ respectively.

Let $A P$ have length $a$, $A I$ have length $b$, and $P I$ have length $c$. Then from Pythagoras' theorem:

Triangles $A P I$ and $A Q B$ are similar as both are right-angled and share an angle at $A$. Since $A Q$ has twice the length of $A P$, the scale factor is $2$ and so $A B$ has length $2 b$ and $B Q$ has length $2 c$.

Triangle $Q R H$ is a rotation of triangle $A P I$, so line segment $Q H$ has length $b$. Triangle $S R K$ is a translation of triangle $A P I$, so line segment $K R$ has length $c$, meaning that line segment $G H$ also has length $c$.

To find the length of line segment $G C$, consider it as the length of $B C$ with $B Q$ and $Q H$ removed, and then $G H$ put back in. This is $2 b - 2 c - b + c = b - c$.

Now consider triangle $S G C$. Let $d$ be the length of $S C$, then $G C$ has length $b - c$ from above, and $S G$ has length $b + c$, so using Pythagoras’ theorem:

Then $a^2 = 12$, since that is the area of a blue square, so $d^2 = 24$ and the area of the green square is $24$.

Solution by Angle at the Centre is Twice the Angle at the Circumference

In the above diagram, the circle has centre $S$ and passes through point $A$. Since the line segment $S Q$ has the same length as $S A$, this circle also passes through $Q$.

Angle $A \hat{S} Q$ is $90^\circ$, as it is the angle between the two diagonals of the squares. Angle $A \hat{C} Q$ is $45^\circ$ as it is the angle between the diagonal and a side of the black square. Therefore, by the converse to the angle at the centre is twice the angle at the circumference, point $C$ also lies on the circle. Therefore line segment $S C$ has the same length as $S A$.

So the side length of the green square is the same as the length of the diagonal of the blue square. Hence the area of the green square is twice that of the blue, so is $24$.

Solution by Invariance Principle

The point where the blue squares meet the bottom edge of the black square can slide along the line through $B$ and $C$ (including outside the square) and there are multiple configurations that yield straightforward solutions.

In this configuration, the point $Q$ is at $B$, so the blue squares are vertical and the green diagonal.

In this configuration, the point $Q$ is at $C$, so the blue squares are diagonal and the green is vertical.