# Problem of the Week Problem D and Solution Pi Squares

## 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 square and a circumscribed square 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 square $$ABDE$$ and the circumscribed square $$FGHJ$$.

The perimeter of square $$ABDE$$ 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 square $$FGHJ$$ 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 squares, determine a lower bound and an upper bound for $$\pi$$.

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

1. For a circle with centre $$C$$, the diagonals of an inscribed square meet at $$90\degree$$ at $$C$$.

2. For a circle with centre $$C$$, the diagonals of a circumscribed square meet at $$90\degree$$ at $$C$$.

3. If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency.

## Solution

For the inscribed square $$ABDE$$, draw line segments $$AC$$ and $$BC$$. Both $$AC$$ and $$BC$$ are radii of the circle with diameter $$1$$, so $$AC=BC=0.5$$.

Since the diagonals of square $$ABDE$$ meet at $$90\degree$$ at $$C$$, it follows that $$\triangle ACB$$ is a right-angled triangle with $$\angle ACB=90\degree$$. We can use the Pythagorean Theorem to find the length of $$AB$$. \begin{aligned} AB^2&=AC^2+BC^2\\ &=(0.5)^2 + (0.5)^2\\ &=0.25 + 0.25\\ &=0.5 \end{aligned} Therefore, $$AB=\sqrt{0.5}$$, since $$AB > 0$$.

Since $$AB$$ is one of the sides of the inscribed square, the perimeter of square $$ABDE$$ is equal to $$4\times AB =4\sqrt{0.5}$$. This gives us a lower bound for $$\pi$$. That is, we know $$\pi >4\sqrt{0.5}\approx 2.828$$.

For the circumscribed square, let $$M$$ be the point of tangency on side $$FJ$$ and let $$N$$ be the point of tangency on $$GH$$. Draw radii $$CM$$ and $$CN$$. Since $$M$$ is a point of tangency, we know that $$\angle FMC = 90\degree$$, and thus $$CM$$ is parallel to $$FG$$. Similarly, $$CN$$ is parallel to $$FG$$.

Thus, $$MN$$ is a straight line segment, and since it passes through $$C$$, the centre of the circle, $$MN$$ must also be a diameter of the circle. Thus, $$MN=1$$. Also, $$FMNG$$ is a rectangle, so $$FG = MN=1$$ and the perimeter of square $$FGHJ$$ is equal to $$4\times FG = 4(1)=4$$. This gives us an upper bound for $$\pi$$. That is, we know $$\pi <4$$.

Therefore, a lower bound for $$\pi$$ is $$4\sqrt{0.5}\approx 2.828$$ and an upper bound for $$\pi$$ is 4. That is, $$4\sqrt{0.5} < \pi < 4$$.

Note: Since we know that $$\pi \approx 3.14$$, these are not the best bounds for $$\pi$$. Archimedes used regular polygons with more sides to get better bounds. In the Problem of the Week E problem, we investigate using regular hexagons to get better bounds.