In the figure below, ABCDEF is a regular hexagon and \(\angle AOF = 90^{\circ}\). FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?
Show Hint
In a regular hexagon the long diagonal through F is parallel to ED. Work out the two legs of the right triangle AOF in terms of the side.
Step 1: Name the side and place the hexagon on coordinates:
Let each side of the regular hexagon be s. Put the centre of the hexagon at the origin with the long diagonal FC along the x-axis.
Then F = \((-s, 0)\), and A, which is the next vertex on the top side, is at \((-\tfrac{s}{2}, \tfrac{\sqrt{3}}{2}s)\).
Step 2: Locate O:
FO is parallel to ED, and in a regular hexagon the diagonal FC is parallel to ED, so O lies on the line FC. Also \(\angle AOF = 90^{\circ}\), so AO is perpendicular to FC.
Hence O is the foot of the perpendicular from A to the x-axis, which gives \(O = (-\tfrac{s}{2}, 0)\).
Step 3: Find the legs of the triangle:
\[ FO = -\tfrac{s}{2} - (-s) = \tfrac{s}{2}, \qquad AO = \tfrac{\sqrt{3}}{2}s \]
Check: \(FO^2 + AO^2 = \tfrac{s^2}{4} + \tfrac{3s^2}{4} = s^2 = FA^2\), as it should be.
Step 4: Area of triangle AOF:
\[ \text{Area}(AOF) = \tfrac{1}{2} \times \tfrac{s}{2} \times \tfrac{\sqrt{3}}{2}s = \tfrac{\sqrt{3}}{8}s^2 \]
Step 5: Area of the hexagon:
A regular hexagon of side s has area \(\tfrac{3\sqrt{3}}{2}s^2\).
Step 6: Take the ratio:
\[ \frac{\sqrt{3}s^2/8}{3\sqrt{3}s^2/2} = \frac{1}{8} \times \frac{2}{3} = \frac{1}{12} \]
So option (1) is correct. Option (2), 1/6, is twice this value. Option (3), 1/24, is half of it. Option (4), 1/18, is two thirds of it. None of them matches the area we found.
Final Answer:
The triangle AOF is one twelfth of the hexagon.
\[ \boxed{\text{Option (1): } \tfrac{1}{12}} \]
