Notes
two squares overlapping a square ii solution

Solution to the Two Squares Overlapping a Square Puzzle

Solution by Similar Triangles

The starting point of this strategy is to consider the rectangle $A J L M$. The triangles $C H J$ and $C I F$ are congruent, so the lengths of $J H$ and $I F$ are the same. The side of the longer square is therefore the same as the side of the shorter square plus the length of $J H$. As $E J$ is the side of the shorter square without $J H$, the segment $A J$ has length twice the side of the shorter square. The area of $A J L M$ is therefore double that of the smaller square, hence $24$.

Comparing this with the larger square, we have rectangle $E J L I$ too much and $M I F G$ too little. The areas of these rectangles are given by $F I \times G F$ and $E J \times I E$. We can replace $I E$ by $A M$ and $G F$ by $A G$.

Triangles $C I F$ and $A M N$ are congruent, so the lengths of $F I$ and $M N$ are the same. The lengths of $F P$ and $A J$ are the same, so $E P$ and $E J$ have the same length, then as triangles $A E P$ and $A G K$ are congruent, so $E J$ and $G K$ have the same length. So we can replace $F I$ by $M N$ and $E J$ by $G K$. The areas of the two rectangles are therefore equal to $A M \times G K$ and $A G \times M N$.

The triangles $A M N$ and $A G K$ are similar, meaning that $A M : M N = A G : G K$. Rearranging this shows that $A M \times G K = A G \times M N$ and so that the two rectangles $M I F G$ and $E J L I$ have the same area. This means that the green square has the same area as the rectangle $A J L M$, which is $24$.

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

Consider the circle centred at the point $E$ that passes through $A$ and $F$ (as $A E F G$ is a square, the lengths of $E A$ and $E F$ are equal so the circle does pass through both).

The angle $F \hat{E} A$ is a right-angle, as it is the corner of a square. The angle $F \hat{C} A$ is $45^\circ$, as it is between a diagonal and side of a square. Therefore, by the result that the angle at the centre is twice the angle at the circumference, $C$ lies on the circle centred at $E$. The segment $E C$ is therefore a radius of this circle and so the length of $E C$ is the same as that of $F E$. As $E C$ is a diagonal of the smaller square, the square with side $E C$ has area twice that of the smaller square and so the area of the green square is $2 \times 12 = 24$.