Problem 764

Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2.

Since equilateral $\triangle PQR$ has perimeter 12, then its side length is 4.

Consider equilateral triangles with side length $2$.

Six of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4.

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Hide/Reveal Description of Triangles

Draw an equilateral triangle with the base at the bottom. This is an "up" triangle.
Flip this triangle upside down such that the vertex is at the bottom. This is a "down" triangle.
Adjoin three triangles side-by-side in the order up,down, up. This forms the top half of the hexagon.
Flip these three triangles upside down to form the bottom half of the hexagon.
In a second image, start again with the line of up, down, up triangles. Place a fourth triangle so that it shares the top edge of the down triangle. These four triangles form a new, larger, equilateral triangle.

Note that the six equilateral triangles around the centre of the hexagon give a total central angle of $6 \cdot 60^\circ = 360^\circ$ (a complete circle) and the three equilateral triangles along each side of the large equilateral triangle make a straight angle of $180^\circ$ (since $3 \cdot 60^\circ = 180^\circ)$.

Also, the length of each side of the hexagon is $2$ and the measure of each internal angle is $120^\circ$, which means that the hexagon is regular. Similarly, the triangle is equilateral.

Since the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is $4:6$ which is equivalent to $2:3$.
Since sector $AOB$ is $\frac{1}{6}$ of a circle with radius $18$, its area is $\frac{1}{6}(\pi\cdot 18^2)$ or $54\pi$.

For the line $AP$ to divide this sector into two pieces of equal area, each piece has area $\frac{1}{2}(54\pi)$ or $27\pi$.

We determine the length of $OP$ so that the area of $\triangle POA$ is $27\pi$.

Since sector $AOB$ is $\frac{1}{6}$ of a circle, then $\angle AOB = \frac{1}{6}(360^\circ) = 60^\circ$.

Drop a perpendicular from $A$ to $T$ on $OB$.

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The area of $\triangle POA$ is $\frac{1}{2}(OP)(AT)$.

$\triangle AOT$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle.

Since $AO = 18$, then $AT = \frac{\sqrt{3}}{2}(AO) = 9\sqrt{3}$.

For the area of $\triangle POA$ to equal $27\pi$, we have $\frac{1}{2}(OP)(9\sqrt{3}) = 27\pi$ which gives $OP = \dfrac{54\pi}{9\sqrt{3}} = \dfrac{6\pi}{\sqrt{3}} = 2\sqrt{3}\pi$.

(Alternatively, we could have used the fact that the area of $\triangle POA$ is $\frac{1}{2}(OA)(OP)\sin(\angle POA)$.)