Problem D and Solution

Pi Squares

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:

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

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

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

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.