Notes
subdivided hexagon iii solution

Solution to the Subdivided Hexagon III Puzzle

Solution by Properties of a Regular Hexagon, Similar Triangles, and Solutions of Quadratic Equations

Label the points as in the diagram above, where $I$ is such that $G I$ is perpendicular to $H F$.

Let the side length of the hexagon be $a$, then from the area of a regular hexagon, the total area of the hexagon is $\frac{3 \sqrt{3}}{2} a^2$.

Let line segment $B H$ have length $b$. Then from the area of the purple triangle, $a b = 6$.

Let line segment $G I$ have length $h$. Using lengths in an equilateral triangle, $I F$ has length $\sqrt{3} h$. Triangles $H B C$ and $H I G$ are similar, so line segment $H I$ has length $\frac{h}{a} \times b = \frac{6 h}{a^2}$.

The area of triangle $H F G$ is then $\frac{1}{2} h \left( \frac{6 h}{a^2} + \sqrt{3} h\right)$, so:

Then, from the lengths in a regular hexagon, line segment $B F$ has length $\sqrt{3} a$, so:

This rearranges to:

Putting this into Equation (1) gives:

which rearranges as follows:

Putting $c = \sqrt{3} a^2$ turns this into $c^2 - 14 c + 24 = 0$ which has solutions $c = 12$ and $c = 2$.

From Equation (2), if $c = 2$ then $h = -\frac{1}{2} h$, which is negative, so this isn’t a solution to the original problem. Hence $c = 12$ and so the area of the hexagon is $\frac{3 \sqrt{3} a^2}{2} = \frac{3 \times 12}{2} = 18$.

Solution by Equations of Straight Lines, Area of a Triangle, and Properties of a Regular Hexagon

The version of the diagram above is rotated so that $A$ is vertically above the centre, $O$. Overlay a coordinate system so that $O$ is the origin, $O A$ is along the vertical axis, and $A$ is at the point $(0,2)$ (note that this is likely to be a different scale than the one that gives the areas).

The point $H$ appears to also lie on the vertical axis. This is what will be demonstrated.

In this coordinate system, the line $B F$ has equation $y = 1$, so $H$ has $y$-coordinate $1$. Let its $x$-coordinate be $p$.

Point $C$ has coordinates $(-\sqrt{3},-1)$, so the line through $C$ and $H$ has equation:

Point $A$ has coordinates $(0,2)$ and $F$ has coordinates $(\sqrt{3},1)$, so the line through $A$ and $F$ has equation:

This rearranges to $x = \sqrt{3}(2 - y)$.

Point $G$ is on the intersection of these lines. Let $q$ be its $y$-coordinate, then this satisfies:

which rearranges to:

The areas of the purple and orange triangles are, in this coordinate system, given by $\frac{1}{2} \times 2 \times (p + \sqrt{3})$ and $\frac{1}{2} \times (q - 1) \times (\sqrt{3} - p)$. Since the purple triangle is three times the area of the orange then:

This rearranges as follows:

So either $p = 0$ or $p = 5 \sqrt{3}$. The second of these is outside the hexagon, so we must have $p = 0$.

This means that $H$ lies on the line $O A$, so then the purple triangle is one sixth of the area of the hexagon, meaning that the hexagon has area $6 \times 3 = 18$.