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Problem of the Week
Problem D and Solution
A New Pi Plate

Problem

A pi-plate is in the shape of a regular hexagon with side length \(8\), and has an interesting pattern on it. From each vertex in the hexagon, a circular arc with centre at the vertex and radius \(8\) is drawn. This results in a flower-like pattern inside the hexagon, which is then shaded. Determine the total area of the shaded region.

The flower-like pattern consists of 6 identical petals. Each petal extends from the centre of the hexagon to one of the 6 vertices. The 6 petals are shaded.

Solution

We label the centre of the hexagon \(O\) and consider two adjacent vertices, \(A\) and \(B\). We draw in \(OA\) and \(OB\).

Line segments OA and OB each divide a petal in the flower-like pattern in half.

We note that a regular hexagon is composed of six congruent equilateral triangles (the proof of this is left up to the student). Thus, \(\triangle AOB\) is an equilateral triangle with side length \(8\).

To determine the total shaded area, we can consider the area of one-half of each ’petal’ of the flower, and then multiply this area by \(12\), since there are \(12\) half-petals in the diagram. To determine the area of the region between the arc through \(B\) and \(O\) and the line segment \(OB\), we can calculate the area of the sector \(AOB\), with centre \(A\) and radius \(AB\), and subtract the area of \(\triangle AOB\).

We first determine the area of \(\triangle AOB\). Construct altitude \(OT\). Since \(\triangle AOB\) is equilateral, it follows that \(OT\) bisects \(AB\).

Since \(AB = 8\), it follows that \(AT = 4\). By the Pythagorean Theorem in \(\triangle ATO\), \(AO^2 = AT^2 + OT^2\). Therefore, \(8^2 = 4^2+OT^2\), and \(OT^2 = 64 - 16 = 48\) follows. Since \(OT>0\), we have \(OT=\sqrt{48}\).

Therefore, the area of \(\triangle AOB\) is \(\dfrac{(AB)\times (OT)}{ 2} = \dfrac{8\times \sqrt{48}}{2} = 4\sqrt{48} = 16\sqrt{3}.\)

We now determine the area of the sector \(AOB\). Since \(\angle OAB = 60\degree\), the area of sector \(AOB\) is \(\frac{1}{6}\) of the area of a circle with radius \(8\). Thus, the area of the sector is equal to \(\frac{1}{6} \pi (AB)^2 = \frac{1}{6} \pi (8)^2 = \frac{32}{3} \pi.\)

Therefore, the area of the shaded region between the arc through \(B\) and \(O\) and the line segment \(OB\) is \(\left(\frac{32}{3} \pi- 16\sqrt{3} \right) \text{ units}^2\).

The original pi-plate has \(12\) of these regions, so the entire shaded area is \(12 \times (\frac{32}{3} \pi- 16\sqrt{3}) = 128\pi - 192\sqrt{3} \approx 69.57 \text{ units}^2\).