A circle with centre $O$ is inscribed in square $A B C D$ having side length 60 . Point $E$ is the midpoint of $A D$. Line segments $A C$ and $B E$ intersect the top half of the circle at points $F$ and $G$ respectively, and they intersect each other at point $H$. What is the total area of the shaded regions?

![](https://cdn.mathpix.com/cropped/2024_04_17_2dbee4c747fee2855e37g-3.jpg?height=466&width=461&top_left_y=1908&top_left_x=886)
We will suppose $O$ is at the origin, which gives rise to coordinates $A(-30,30), B(30,30)$, $C(30,-30)$, and $D(-30,-30)$. Additionally, we will have $E(-30,0)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-20.jpg?height=643&width=873&top_left_y=318&top_left_x=686)

The line passing through $A$ and $C$ also passes through the origin and has slope $m=\frac{-30-30}{30-(-30)}=\frac{-60}{60}=-1$, and hence has equation $y=-x$.

The line through $B$ and $E$ has slope $m=\frac{30-0}{30-(-30)}=\frac{30}{60}=\frac{1}{2}$ and passes through $(30,30)$, so its equation is $y=\frac{1}{2} x+b$ for some real number $b$.

By substituting $(x, y)=(30,30)$, we have $30=\frac{1}{2}(30)+b$ or $b=15$. Thus, the line passing through both $B$ and $E$ has equation $y=\frac{1}{2} x+15$.

We can now compute the coordinates of points $G$ and $H$.

Since $H$ the intersection of line segments $A C$ and $B E$, the $x$-coordinate of $H$ must satisfy $-x=\frac{1}{2} x+15$ or $-15=\frac{3}{2} x$, which can be solved to get $x=-10$.

Since $H$ lies on the line $y=-x$, this means $H$ has coordinates $(-10,10)$.

The circle has diameter equal to the side length of the square, which is given to be 60 . Therefore, its radius is $\frac{60}{2}=30$, which means the equation of the circle is $x^{2}+y^{2}=30^{2}$.

Points $G$ and $E$ are the points of intersection with the circle and the line $y=\frac{1}{2} x+15$, which means their $x$-coordinates can be found by substituting $\frac{1}{2} x+15$ in the equation of the circle for $y$ and solving for $x$.

That is, the $x$-coordinates of $E$ and $G$ are the solutions to the equation

$$
x^{2}+\left(\frac{1}{2} x+15\right)^{2}=900
$$

which can be expanded to get

$$
x^{2}+\frac{1}{4} x^{2}+15 x+225=900 \text {. }
$$

Multiplying through by 4 and rearranging, we get

$$
5 x^{2}+60 x-2700=0
$$

and after dividing through by 5 we have

$$
x^{2}+12 x-540=0 \text {. }
$$

We know that the $x$-coordinate of $E$ is a root, which means $x+30$ must be a factor of the polynomial on the right, which leads to

$$
(x+30)(x-18)=0 .
$$

Thus, the other solution is $x=18$, so the $x$-coordinate of $G$ must be 18 . Substituting into $y=\frac{1}{2} x+15$, we get that $y=\frac{18}{2}+15=24$, so $G$ has coordinates $(18,24)$.

We next observe that the shaded region bound by the circle and the line segments $A F$ and $A B$ is congruent to the region bound by the circle, $A F$, and $A E$.

Thus, the shaded region is equal to the sum of the areas of $\triangle H G O$ and $\triangle A H E$.

If we take $A E$ as the base of $\triangle A H E$, then its height will be the horizontal distance from $H$ to line segment $A E$. This is equal to the difference between the $x$ coordinates of $A$ and $H$, or $-10-(-30)=20$.

Thus, the area of $\triangle A H E$ is

$$
\frac{1}{2} \times A E \times 20=10 A E=10(30)=300
$$

To compute the area of $\triangle H G O$, we will subtract the area of $\triangle E H O$ from the area of $\triangle E G O$. Taking $E O$ as a common base, the heights of $\triangle E H O$ and $\triangle E G O$ are the $y$-coordinates of $H$ and $G$, respectively. These values are 10 and 24 .

The length of $E O$ is $0-(-30)=30$, so the area of $\triangle H G O$ is

$$
\begin{aligned}
\frac{1}{2} \times E O \times 24-\frac{1}{2} \times E O \times 10 & =\frac{1}{2} \times E O \times(24-10) \\
& =\frac{1}{2}(30)(14) \\
& =210 .
\end{aligned}
$$

Therefore, the total area of the shaded regions is $300+210=510$.

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)