# Solution to the [[Shaded Spikes in a Hexagon]] Puzzle +-- {.image} [[ShadedSpikesinaHexagon.png:pic]] > What fraction is shaded? The hexagon is regular, with equally spaced dots around its perimeter. =-- ## Solution by Properties of a [[Regular Hexagon]] and the [[Area of a Triangle]] Consider the diagram labelled as below. +-- {.image} [[ShadedSpikesinaHexagonLabelled.png:pic]] =-- From the properties of a [[regular hexagon]], the triangles with apex at $O$ and side one of those of the hexagon are all [[equilateral triangles|equilateral]]. Each has area $\frac{1}{6}$th of that of the hexagon. Let $h$ be the height of one equilateral triangle, so the length of $O B$, and let $b$ be the base, so the length of $A C$. Then the area of the hexagon is $6 \times \frac{1}{2} b h = 3 b h$. As the points $L$ and $D$ are the [[midpoints]] of their sides, the length of $L D$ is half way between that of $A C$, namely $b$, and of $K E$, namely $2 b$. So $L D$ has length $\frac{3}{2} b$. Similarly, the height of $A$ above the line $L D$ is $\frac{1}{2} h$, so the area of triangle $A D L$ is $\frac{1}{2} \times \frac{3}{2} b \times \frac{1}{2} h = \frac{3}{8} b h$. The length of $L F$ is $2 h$ and of $K L$ is $\frac{1}{2} b$, so the area of triangle $L F K$ is $\frac{1}{2} \times \frac{1}{2} b \times 2 h = \frac{1}{2} b h$. The height of $H$ above $K J$ is half of the height of $G$ (and of $O$) so is $\frac{1}{2} h$. Since $K J$ has length $\frac{1}{2} b$ this means that triangle $K H J$ has area $\frac{1}{2} \times \frac{1}{2} b \times \frac{1}{2} h = \frac{1}{8} b h$. The total shaded area is therefore: $$ \frac{3}{8} b h + \frac{1}{2} b h + \frac{1}{8} b h = b h $$ Since the area of the hexagon is $3 b h$ this means that one third of the hexagon is shaded. ## Solution by Properties of a [[Regular Hexagon]] and [[Dissection]] Consider the diagram labelled as below. +-- {.image} [[ShadedSpikesinaHexagonLabelledv2.png:pic]] =-- The (orange) shaded areas are cut and reassembled as follows. 1. Triangle $K H J$ is flipped and translated to $A B D$, where it fits exactly. 2. Triangle $M D N$ is rotated round to fit in $B C M$. 3. Triangle $O F P$ is rotated round to fit in $Q P K$. 4. Triangle $O Q K$ is translated to fit in $N O L$. This leaves the shaded area being [[parallelogram]] $K A C O$ which consists of two of the six [[equilateral triangles]] that make up the [[hexagon]]. The shaded area is therefore one third of the total.