# Problem of the Week Problem E and Solution Pi Hexagons

## Problem

Pi Day is an annual celebration of the mathematical constant $$\pi$$. Pi Day is observed on March $$14$$, since $$3$$, $$1$$, and $$4$$ are the first three significant digits of $$\pi$$.

Archimedes determined lower bounds for $$\pi$$ by finding the perimeters of regular polygons inscribed in a circle with diameter of length $$1$$. (An inscribed polygon of a circle has all of its vertices on the circle.) He also determined upper bounds for $$\pi$$ by finding the perimeters of regular polygons circumscribed in a circle with diameter of length $$1$$. (A circumscribed polygon of a circle has all sides tangent to the circle. That is, each side of the polygon touches the circle in one spot.)

In this problem, we will determine a lower bound for $$\pi$$ and an upper bound for $$\pi$$ by considering an inscribed regular hexagon and a circumscribed regular hexagon in a circle of diameter $$1$$.

Consider a circle with centre $$C$$ and diameter $$1$$. Since the circle has diameter $$1$$, it has circumference equal to $$\pi$$. Now consider the inscribed regular hexagon $$DEBGFA$$ and the circumscribed regular hexagon $$HIJKLM$$.

The perimeter of hexagon $$DEBGFA$$ will be less than the circumference of the circle, $$\pi$$, and will thus give us a lower bound for the value of $$\pi$$. The perimeter of hexagon $$HIJKLM$$ will be greater than the circumference of the circle, $$\pi$$, and will thus give us an upper bound for the value of $$\pi$$.

Using these hexagons, determine a lower and an upper bound for $$\pi$$.

Note: For this problem, you may want to use the following known results:

1. A line drawn from the centre of a circle perpendicular to a tangent line meets the tangent line at the point of tangency.

2. For a circle with centre $$C$$, the centres of both the inscribed and circumscribed regular hexagons will be at $$C$$.

## Solution

For the inscribed hexagon, draw line segments $$AC$$ and $$DC$$, which are both radii of the circle.

Since the diameter of the circle is 1, $$AC=DC=\frac{1}{2}$$. Since the inscribed hexagon is a regular hexagon with centre $$C$$, we know that $$\triangle ACD$$ is equilateral (a justification of this is provided at the end of the solution). Thus, $$AD = AC = \tfrac{1}{2}$$, and the perimeter of the inscribed regular hexagon is $$6\times AD = 6 \left(\tfrac{1}{2} \right)=3$$. Since the perimeter of this hexagon is less than the circumference of the circle, this gives us a lower bound for $$\pi$$. That is, this tells us that $$\pi >3$$.

For the circumscribed hexagon, draw line segments $$LC$$ and $$KC$$. Since the circumscribed hexagon is a regular hexagon with centre $$C$$, we know that $$\triangle LCK$$ is equilateral (a justification of this is provided at the end of the solution). Thus, $$\angle LKC = 60\degree$$. Drop a perpendicular from $$C$$, meeting $$LK$$ at $$N$$. We know that $$N$$ must be the point of tangency. Thus, $$CN$$ is a radius and so $$CN=0.5$$. In $$\triangle CNK$$, $$\angle NKC = \angle LKC=60\degree$$.

Since $$\angle CNK = 90\degree$$, \begin{aligned} \sin(\angle NKC)&= \frac{CN}{KC} \\ \sin(60^{\circ}) &= \frac{0.5}{KC}\\ \frac{\sqrt{3}}{2}&=\frac{0.5}{KC}\\ \sqrt{3}KC&=1\\ KC&=\frac{1}{\sqrt{3}} \end{aligned} But $$\triangle LCK$$ is equilateral, so $$LK = KC = \frac{1}{\sqrt{3}}$$.

Thus, the perimeter of the circumscribed hexagon is $$6\times LK=6 \times \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} \approx 3.46$$.

Since the perimeter of this hexagon is greater than the circumference of the circle, this gives us an upper bound for $$\pi$$. That is, this tells us that $$\pi < \frac{6}{\sqrt{3}}$$.

Therefore, the value for $$\pi$$ is between $$3$$ and $$\frac{6}{\sqrt{3}}$$ . That is, $$3<\pi<\frac{6}{\sqrt{3}}$$.

Extension: Archimedes used regular $$12$$-gons, $$24$$-gons, $$48$$-gons and $$96$$-gons to get better approximations for the bounds on $$\pi$$. Can you?

Equilateral triangle justification:

In the solutions, we used the fact that both $$\triangle ACD$$ and $$\triangle LCK$$ are equilateral. In fact, a regular hexagon can be split into six equilateral triangles by drawing line segments from the centre of the hexagon to each vertex, which we will now justify.

Consider a regular hexagon with centre $$T$$. Draw line segments from $$T$$ to each vertex and label two adjacent vertices $$S$$ and $$U$$.

Since $$T$$ is the centre of the hexagon, $$T$$ is of equal distance to each vertex of the hexagon. Since the hexagon is a regular hexagon, each side of the hexagon has equal length. Thus, the six resultant triangles are congruent. Therefore, the six central angles are equal and each is equal to $$\frac{1}{6}(360\degree) = 60\degree$$.

Now consider $$\triangle STU$$. We know that $$\angle STU = 60\degree$$. Also, $$ST= UT$$, so $$\triangle STU$$ is isosceles and $$\angle TSU = \angle TUS = \frac{180\degree - 60\degree}{2} = 60\degree$$.

Therefore, all three angles in $$\triangle STU$$ are equal to $$60\degree$$ and so $$\triangle STU$$ is equilateral. Since the six triangles in the hexagon are congruent, this tells us that all six triangles are all equilateral.