CEMC Hypatia, 2024

To determine the coordinates of $F$, we find the point of intersection of
the line through $A$ and $C$ and the line through $B$ and $E$.

The line through $A(0,0)$ and $C(12,12)$ has slope $\frac{12-0}{12-0}=1$.

Since it passes through $(0,0)$, this
line has equation $y=x$.

The line through $B(12,0)$ and $E(0,6)$ has slope $\frac{6-0}{0-12}=-\frac12$.

Since it passes through $(0,6)$, this
line has equation $y=-\frac12x+6$.

To determine the $x$-coordinate of
the point of intersection, $F$, we
solve $x=-\frac12x+6$, which gives
$\frac32x=6$ or $3x=12$, and so $x=4$.

Since $F$ lies on the line with
equation $y=x$, then the coordinates
of $F$ are $(4,4)$.
Solution 1:

Consider $\triangle AEF$ as having
base $AE=6$.

Then $\triangle AEF$ has height equal
to the perpendicular distance from $F$ to $AE$, which is 4, the $x$-coordinate of $F$.

The area of $\triangle AEF$ is thus
$\frac12\cdot6\cdot4=12$.

Solution 2:

We can determine the area of $\triangle AEF$ by subtracting the area of $\triangle AFB$ from the area of $\triangle AEB$.

Consider $\triangle AFB$ as having
base $AB=12$.

Then $\triangle AFB$ has height equal
to the perpendicular distance from $F$ to $AB$, which is 4, the $y$-coordinate of $F$.

The area of $\triangle AFB$ is thus
$\frac12\cdot12\cdot4=24$.

The area of $\triangle AEB$ is $\frac12\cdot12\cdot6=36$, and so the area
of $\triangle AEF$ is $36-24=12$.
To determine the area of quadrilateral $GDEF$, our strategy will be to subtract
the area of $\triangle AEF$ and the
area of $\triangle CDG$ from the area
of $\triangle ACD$.

We need to find the area of $\triangle CDG$ still, which means finding the coordinates of $G$.

We can find the coordinates of $G$ by
determining the intersection of the line through $A$ and $C$ with the given circle.

Thus, we proceed by finding the equation of the circle.

Since the circle has diameter $EB$,
then its centre is the midpoint of $EB$, which is $\left(\frac{0+12}{2},\frac{6+0}{2}\right)$
or $(6,3)$.

The diameter has length $EB=\sqrt{(12-0)^2+(0-6)^2}$ or $EB=\sqrt{180}$, which simplifies to $EB=6\sqrt{5}$.

Thus the radius of the circle is $r=\frac12\cdot6\sqrt{5}=3\sqrt{5}$, and so
the circle has equation $(x-6)^2+(y-3)^2=(3\sqrt{5})^2$ or $(x-6)^2+(y-3)^2=45$.

Suppose the $x$-coordinate of $G$ is $g$.

Since $G$ lies on the line with
equation $y=x$, then the coordinates
of $G$ are $(g,g)$.

The point $G$ also lies on the
circle, and thus the coordinates of $G$ satisfy the equation of the
circle.

That is, $(g-6)^2+(g-3)^2=45$, and
solving for $g$, we get $$\begin{align*} g^2-12g+36+g^2-6g+9&=45 \\ 2g^2-18g+45&=45\\ 2g^2-18g&=0\\ 2g(g-9)&=0\end{align*}$$ and so $g=0$ or $g=9$.

Since $G$ is distinct from $A$, then $g=9$ and $G$ has coordinates $(9,9)$.

We may now determine the area of $\triangle CDG$.

Consider $\triangle CDG$ as having
base $CD=12$.

Then $\triangle CDG$ has height equal
to the perpendicular distance from $G$ to $CD$, which is $12-9=3$, since $CD$ lies along the line $y=12$ and the $y$-coordinate of $G$ is 9.

The area of $\triangle CDG$ is thus
$\frac12\cdot12\cdot3=18$.

The area of $\triangle ACD$ is half
the area of square $ABCD$ or $\frac12\cdot12^2=72$.

From part (b), the area of $\triangle AEF$ is $12$, and so the area
of $GDEF$ is $72-18-12=42$.