subdivided hexagon iii solution in Notes

# Solution to the [[Subdivided Hexagon III]] Puzzle

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[[SubdividedHexagonIII.jpeg:pic]]

> What’s the total area of this regular hexagon?
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## Solution by Properties of a [[Regular Hexagon]], [[Similar Triangles]], and Solutions of [[Quadratic Equations]]

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[[SubdividedHexagonIIILabelled.jpeg:pic]]
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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:

\[
\label{first}
\frac{h^2}{a^2} \left( 6 + \sqrt{3} a^2 \right) = 2
\]

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

$$
\frac{6}{a} + h \left(\frac{6}{a^2} + \sqrt{3} \right) = \sqrt{3} a
$$

This rearranges to:

\[
\label{eqh}
h = \frac{ \sqrt{3} a - \frac{6}{a} }{\frac{6}{a^2} + \sqrt{3}} = \frac{ a(\sqrt{3} a^2 - 6)}{6 + \sqrt{3} a^2}
\]

Putting this into Equation \eqref{first} gives:

$$
\left(\frac{\sqrt{3} a^2 - 6}{6 + \sqrt{3} a^2}\right)^2
 \left( 6 + \sqrt{3} a^2 \right) = 2
$$

which rearranges as follows:

$$
\begin{aligned}
(\sqrt{3} a^2 - 6)^2 &= 2(6 + \sqrt{3} a^2 ) \\
3 a^4 - 12 \sqrt{3} a^2 + 36 &= 12 + 2 \sqrt{3} a^2 \\
3 a^4 - 14 \sqrt{3} a^2 + 24 &= 0 \\
\end{aligned}
$$

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 \eqref{eqh}, 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]]


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[[SubdividedHexagonIIIRotated.jpeg:pic]]
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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:

$$
\frac{y + 1}{x + \sqrt{3}} = \frac{1 + 1}{p + \sqrt{3}} = \frac{2}{p + \sqrt{3}}
$$

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

$$
\frac{y - 2}{x} = \frac{1 - 2}{\sqrt{3}} = - \frac{1}{\sqrt{3}}
$$

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:

$$
\frac{q + 1}{ \sqrt{3}(2 - q) + \sqrt{3}} = \frac{2}{p + \sqrt{3}}
$$

which rearranges to:

$$
q = \frac{5 \sqrt{3} - p}{3\sqrt{3} + p}
$$

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:

$$
p + \sqrt{3} = \frac{3}{2} (q- 1)(\sqrt{3} - p) = \frac{3( \sqrt{3} - p)(\sqrt{3} - p)}{3 \sqrt{3} + p}
$$

This rearranges as follows:

$$
\begin{aligned}
(p + \sqrt{3})(3 \sqrt{3} + p) &= 3(\sqrt{3} - p)(\sqrt{3} - p) \\
p^2 + 4 \sqrt{3} p + 9 &= 9 - 6 \sqrt{3} p + 3 p^2 \\
0 &= 2p^2 - 10 \sqrt{3} p \\
&= 2 p (p - 5 \sqrt{3})
\end{aligned}
$$

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 $6 \times 3 = 18$.