What is the value of the ratio $\frac{\text{Area of the hexagon}}{\text{Area of the grey triangle}}$?
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For geometry problems involving regular polygons, decomposing the shape into smaller, simpler shapes (like equilateral triangles in a hexagon) is a very effective strategy. Also, using a coordinate system can simplify area and length calculations.
Step 1: Split the hexagon from its centre.
Join the centre \(O\) to every vertex. This splits the regular hexagon into 6 identical equilateral triangles, each of area \(t\), so the hexagon's area is \(6t\).
Step 2: Locate triangle ABD using these 6 pieces.
Since \(A\) and \(D\) are opposite vertices, \(O\) lies exactly on segment \(AD\), so triangle \(ABD\) splits into \(\triangle ABO\) and \(\triangle OBD\). \(\triangle ABO\) is exactly one of the 6 equal triangles, so its area is \(t\).
Step 3: Work out the area of \(\triangle OBD\).
Quadrilateral \(OBCD\) can be split either by diagonal \(OC\) into \(\triangle OBC + \triangle OCD = t + t = 2t\), or by diagonal \(BD\) into \(\triangle OBD + \triangle BCD\). Since \(\triangle BCD\) also works out to area \(t\) (base \(BC\), same height as one hexagon triangle), \(\triangle OBD = 2t - t = t\).
Step 4: Add the pieces and take the ratio.
\[
\text{Area of } \triangle ABD = t + t = 2t
\]
\[
\frac{\text{Area of hexagon}}{\text{Area of } \triangle ABD} = \frac{6t}{2t} = \boxed{3}
\]
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Approach Solution -2
A direct algebraic way to get the same triangle's area is to place the hexagon on coordinates and plug the three vertices straight into the standard area formula for a triangle, without needing any trapezoid or decomposition argument.
Put the hexagon's centre at the origin, with side length \(s\), so going around we get \(A(s, 0)\), \(B\left(\frac{s}{2}, \frac{s\sqrt{3}}{2}\right)\), \(C\left(-\frac{s}{2}, \frac{s\sqrt{3}}{2}\right)\), \(D(-s, 0)\), \(E\left(-\frac{s}{2}, -\frac{s\sqrt{3}}{2}\right)\), \(F\left(\frac{s}{2}, -\frac{s\sqrt{3}}{2}\right)\).
For any triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), the area is given by the shoelace formula:
\[
\text{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Plugging in \(A(s, 0)\), \(B\left(\frac{s}{2}, \frac{s\sqrt{3}}{2}\right)\), \(D(-s, 0)\):
\[
\text{Area}_{ABD} = \frac{1}{2}\left| s\left(\frac{s\sqrt{3}}{2} - 0\right) + \frac{s}{2}(0 - 0) + (-s)\left(0 - \frac{s\sqrt{3}}{2}\right) \right|
\]
\[
= \frac{1}{2}\left| \frac{s^2\sqrt{3}}{2} + 0 + \frac{s^2\sqrt{3}}{2} \right| = \frac{1}{2}\left(s^2\sqrt{3}\right) = \frac{s^2\sqrt{3}}{2}
\]
The hexagon's area (six equilateral triangles of side \(s\)) is \(\frac{3\sqrt{3}}{2}s^2\), so the ratio is
\[
\frac{\frac{3\sqrt{3}}{2}s^2}{\frac{\sqrt{3}}{2}s^2} = 3
\]
