Inscribing a hexagon in a circle

Task

Let $C$ be a circle with center $O$ and a diameter meeting $C$ in points $P$ and $S$ as shown below:

1. With straightedge and compass, show how to find a point $Q$ on $C$ so that triangle $OPQ$ is equilateral.
2. Repeating part (a) show how to find points $R, T, U$ on $C$ so that $PQRSTU$ is a regular hexagon.
3. Find the area of $PQRSTU$. How does it compare to the area of $C$?

IM Commentary

This task is primarily for instructive purposes but can be used for assessment as well. Parts (a) and (b) are good applications of geometric constructions using a compass and could be used for assessment purposes but the process is a bit long since there are six triangles which need to be constructed.

Solution

1. Since a regular hexagon divides the circle into six equal parts and there are $360$ degrees in the circle, each side of the regular hexagon should span a chord of $60$ degrees on the circle. A triangle with vertex $O$ and its other two vertices on the circle $C$ is an isosceles triangle since all radii of $C$ have the same length. So if the angle at $O$ measures $60$ degrees, the two base angles must also measure $60$ degrees making it an equilateral triangle. In order to construct our hexagon, we can start with segment $OP$ and successively build equilateral triangles from here.

   If we draw a circle of radius $|OP|$ centered at $P$ it will meet the circle $C$ in two points, labelled $Q$ and $U$ in the picture below:

   

   We have $|PO| = |PQ|$ since both are radii of the circle with center $P$ and radius $|OP|$. We also know that $|PO| = |QO|$ because both are radii of the circle with center $O$ and radius $|OP|$. Hence we have $$ |OP| = |PO| = |PQ| $$ and triangle $OPQ$ is equilateral. The same reasoning applies to triangle $OPU$.

2. If we construct a circle with center $S$ and radius $|OS|$, as in part (a), we find the following picture:

   

   The reasoning of part (a) applies to show that the two triangles $ROS$ and $SOT$ are both equilateral. Now angles $POQ$, $QOR$, and $ROS$ together form a line and so add up to $180$ degrees. Both $POQ$ and $ROS$ have been shown to be $60$ degree angles and so $QOR$ is also a $60$ degree angle. As in part (a), it follows that triangle $QOR$ is equilateral. The same reasoning applies to show that triangle $TOU$ is equilateral. We can now conclude that hexagon $PQRSTU$ is a regular hexagon as each of its six sides is congruent to the radius of circle $C$.

3. Hexagon $PQRSTU$ is made up of six congruent equilateral triangles so we need to find the area of one of these triangles. We will focus on triangle $OPQ$. Let $K$ be the midpoint of segment $OP$:

   

   Triangle $QKO$ is congruent to triangle $QKP$ by SSS: $OQ$ and $OP$ are radii of congruent circles, $|QK| = |QK|$ and $|OK| = |PK|$ since $K$ is the midpoint of segment $OP$. Since angles $OKQ$ and $PKQ$ are congruent and add up to $180$ degrees they are right angles. Thus lines $QK$ and $OP$ are perpendicular. We have $|OQ| = r$ since it is a radius of the circle with center $O$ and radius $r$. We also have $$ \frac{|OK|}{|OQ|} = \cos{KOQ} $$ and similarly $$ \frac{|QK|}{|OQ|} = \sin{KOQ}. $$ Angle $KOQ$ measures $60$ degrees since it is an angle in an equilateral triangle. So we have $\cos{KOQ} = \frac{1}{2}$ and $\sin{KOQ} = \frac{\sqrt{3}}{2}$. So we have $|OK| = \frac{r}{2}$ and $|QK| = \frac{\sqrt{3}r}{2}$. The area triangle $QOP$ is the same as the area of a rectangle of width $|OK|$ and height $|QK|$ which we have just seen is $$ \frac{\sqrt{3}r^2}{4}. $$ There are six of these equilateral triangles in the hexagon $PQRSTU$ so the area of this hexagon is $$ \frac{3\sqrt{3}r^2}{2}. $$ This is an area of about $2.60r^2,$ just under $83$ percent of the area of the circle.