Circles and Geometric Shapes | EOT

Question 21

Let $ABCD$ be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle \text{CDE} = \angle \text{ABC}$ , where $E$ is a point on the extension of side $CD$ ).

Solution

Understand the Question

Let the side length of the square be $s$ .
The quarter circle has radius $s$ , while the two semicircles constructed on adjacent sides have diameter $s$ and radius $r = \dfrac{s}{2}$ .
Region $A$ is the region of overlap between the two semicircles.
Region $B$ lies inside the quarter circle and outside both semicircles.
By calculating the area of each region using geometry and the principle of inclusion-exclusion, we show that $\text{Area}(A) = \text{Area}(B) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$ .

Step 1 · Define Dimensions and Radii

Let the side length of the square be $s$ .

Radius of the quarter circle: $R = s$
Diameter of each semicircle: $s$
Radius of each semicircle: $r = \dfrac{s}{2}$

Step 2 · Calculate Area of Region A

Diagram 1

Place the bottom-left corner of the square at $(0,0)$ .

The centers of the two semicircles are $\left(\dfrac{s}{2}, 0\right)$ and $\left(0, \dfrac{s}{2}\right)$ , each with radius $r = \dfrac{s}{2}$ .

The distance $d$ between their centers is:

\begin{aligned} d &= \sqrt{\left(\dfrac{s}{2} - 0\right)^2 + \left(0 - \dfrac{s}{2}\right)^2} \\[0.6em] &= \sqrt{\left(\dfrac{s}{2}\right)^2 + \left(-\dfrac{s}{2}\right)^2} \\[0.6em] &= \sqrt{\dfrac{s^2}{4} + \dfrac{s^2}{4}} \\[0.6em] &= \sqrt{\dfrac{2s^2}{4}} \\[0.6em] &= \dfrac{s}{\sqrt{2}} \end{aligned}

Using the formula for the overlap area of two identical circular disks of radius $r$ :

$\text{Area} = 2r^2 \cos^{-1}\left(\dfrac{d}{2r}\right) - \left(\dfrac{d}{2}\right)\sqrt{4r^2 - d^2}$

Evaluate $\dfrac{d}{2r}$ :

\begin{aligned} \dfrac{d}{2r} &= \dfrac{s/\sqrt{2}}{2(s/2)} \\[0.6em] &= \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} \end{aligned}

Since $\cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ , we have $\cos^{-1}\left(\dfrac{d}{2r}\right) = \dfrac{\pi}{4}$ .

Evaluate $\sqrt{4r^2 - d^2}$ :

\begin{aligned} \sqrt{4r^2 - d^2} &= \sqrt{4\left(\dfrac{s}{2}\right)^2 - \left(\dfrac{s}{\sqrt{2}}\right)^2} \\[0.6em] &= \sqrt{s^2 - \dfrac{s^2}{2}} \\[0.6em] &= \dfrac{s}{\sqrt{2}} \end{aligned}

Now compute $\text{Area}(A)$ :

\begin{aligned} \text{Area}(A) &= 2\left(\dfrac{s}{2}\right)^2 \left(\dfrac{\pi}{4}\right) - \left(\dfrac{s/\sqrt{2}}{2}\right)\left(\dfrac{s}{\sqrt{2}}\right) \\[0.6em] &= 2\left(\dfrac{s^2}{4}\right)\left(\dfrac{\pi}{4}\right) - \left(\dfrac{s}{2\sqrt{2}}\right)\left(\dfrac{s}{\sqrt{2}}\right) \\[0.6em] &= \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \end{aligned}

Step 3 · Calculate Area of Region B

Diagram 2

Area of the quarter circle:

$\text{Area}(\text{Quarter Circle}) = \dfrac{1}{4}\pi s^2$

Area of one semicircle of radius $\dfrac{s}{2}$ :

\begin{aligned} \text{Area}(\text{Semicircle}) &= \dfrac{1}{2}\pi \left(\dfrac{s}{2}\right)^2 \\[0.6em] &= \dfrac{\pi s^2}{8} \end{aligned}

Using the principle of inclusion-exclusion, the area inside the quarter circle and outside both semicircles is:

\begin{aligned} \text{Area}(B) &= \text{Area}(\text{Quarter Circle}) - \text{Area}(\text{Semicircle}_1) - \text{Area}(\text{Semicircle}_2) + \text{Area}(A) \\[0.6em] &= \dfrac{1}{4}\pi s^2 - \dfrac{\pi s^2}{8} - \dfrac{\pi s^2}{8} + \left(\dfrac{\pi s^2}{8} - \dfrac{s^2}{4}\right) \\[0.6em] &= \dfrac{\pi s^2}{4} - \dfrac{\pi s^2}{4} + \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \\[0.6em] &= \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \end{aligned}

Step 4 · Compare the Two Areas

From Step 2 and Step 3:

$\text{Area}(A) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

$\text{Area}(B) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

Therefore,

$\text{Area}(A) = \text{Area}(B)$

Answer

Both regions have equal area: $\text{Area}(A) = \text{Area}(B) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

Common Mistakes

Forgetting the Overlap (Inclusion-Exclusion): When subtracting both semicircles from the quarter circle to find Region $B$ , the overlapping Region $A$ is subtracted twice and must be added back once.
Incorrect Semicircle Radius: Using side length $s$ as the radius instead of $r = \dfrac{s}{2}$ , leading to an incorrect area of $\dfrac{\pi s^2}{2}$ for a semicircle.

Question 21

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