Notes
five regular polygons solution

Solution to the Five Regular Polygons Puzzle

Solution by Congruent Triangles and Lengths and Angles in an Equilateral Triangle

In the labelled diagram above, $A G$ is the continuation of $A H$ so that $A G C$ is an equilateral triangle and $E$ is a point on the side of the blue square so that angle $E \hat{C} G$ is $60^\circ$.

Since $A H$ and $A B$ have the same length, as do $A G$ and $A C$, it must be that $H G$ and $B C$ have the same length. Then angle $C \hat{B} H$ is the exterior angle of an equilateral triangle so is $120^\circ$. Therefore, angles $H \hat{C} B$ and $B \hat{H} C$ add up to $180^\circ - 120^\circ = 60^\circ$ as the angles in a triangle add up to $180^\circ$. So also angles $B \hat{H} C$ and $G \hat{H} F$ add up to $180^\circ - 60^\circ - 60^\circ$ as angles at a point on a straight line also add up to $180^\circ$. Hence angles $H \hat{C} B$ and $F \hat{H} F$ are equal. Since $B C$ and $G H$ have the same length, and so do $H C$ and $F H$, it is therefore the case that triangles $F G H$ and $H B C$ are congruent.

This means that angle $H \hat{G} F$ is the same as angle $C \hat{B} H$ which is $120^\circ$, so angles $H \hat{G} F$ and $C \hat{A} G$ add up to $180^\circ$ which means that $G F$ is parallel to $A C$. This means that $G F E$ is a straight line. Since angle $A \hat{G} C$ is $120^\circ - 60^\circ = 60^\circ$, and angle $E \hat{C} G$ is $60^\circ$ by definition of $E$, triangle $G C E$ is equilateral and congruent to triangle $A C G$.

The length of $F D$ is the height of $C$ above $G E$, which is $\frac{\sqrt{3}}{2}$ times the length of $G C$, and this is the length of $A C$. So the length scale factor from the red square to the blue is $\frac{\sqrt{3}}{2}$ and therefore the area scale factor is $\frac{3}{4}$, making the area of the blue square $9$.

Solution by Invariance Principle

The green triangle can vary in size, giving two special configurations.

In this configuration, the purple triangle is perpendicular to the green, meaning that the green triangle with the white space makes half an equilateral triangle. The side length of the purple triangle is therefore $\frac{\sqrt{3}}{2}$ times the side length of the red square, and so the blue square has area $\frac{3}{4}$ that of the red square.

In this configuration, the green triangle has the same side length as the red square, as does the purple triangle. The blue square then has side length the height of the purple triangle, being $\frac{\sqrt{3}}{2}$ times that of the red square. This leads to an area of $9$ as before.