, 2025

In rectangle $ABCD$, $AD$ is parallel to the $y$-axis and soopposite side $BC$ is also parallel to the $y$-axis.

This means that sides $AD$ and $BC$ are vertical, so $A$ and $D$ each have the same $x$-coordinate, $10$, and $B$ and $C$ also have the same $x$-coordinate, $20$.

Similarly, $AB$ is parallel to the
$x$-axis and so $CD$ is also parallel to the $x$-axis.

Thus, sides $AB$ and $CD$ are horizontal, so $A$ and $B$ have the same $y$-coordinate, $15$, and $C$ and $D$ also have the same $y$-coordinate, $27$.

Therefore, the coordinates of $B$ are
$(20,15)$, and the coordinates of
$D$ are $(10,27)$, as shown.

[[IMAGE0]]

Since $AB=20-10=10$ and $BC=27-15=12$, the area of rectangle $ABCD$ is $10\times12=120$.
Point $E$ lies on $AD$ as shown and thus has $x$-coordinate $10$.

Point $E$ lies on the line with
equation $y=-\frac32x+39$, and so
when $x=10$, the $y$-coordinate is $y=-\frac32(10)+39=24$.

Point $F$ lies on $AB$ and thus has $y$-coordinate $15$.

Point $F$ lies on the line with
equation $y=-\frac32x+39$, and so
when $y=15$, we get $15=-\frac32x+39$ or $-24=-\frac32x$, and so $x=24\times\frac23=16$. Point $E$ has coordinates $(10,24)$, and so $EA=24-15=9$.

Point $F$ has coordinates $(16,15)$, and so $AF=16-10=6$.

[[IMAGE1]]

The area of $\triangle EAF$ is
$\frac12(EA)(AF)=\frac12(9)(6)=27$.

The area of pentagon $BCDEF$ is the
area of $ABCD$ minus the area of
$\triangle EAF$, which is $120-27=93$.
Solution 1:

Point $G$ lies on $AD$ as shown and thus has $x$-coordinate $10$.

Point $G$ lies on the line with
equation $y=mx+b$, and so when $x=10$, the $y$-coordinate is $y=10m+b$.

Point $H$ lies on $AB$ as shown and thus has $y$-coordinate $15$.

Point $H$ lies on the line with
equation $y=mx+b$, and so when $y=15$, we get $15=mx+b$ or $15-b=mx$.

Therefore, $x=\dfrac{15-b}{m}$.

Point $G$ has coordinates $(10,10m+b)$, and so $GA=10m+b-15$

Point $H$ has coordinates $\left(\dfrac{15-b}{m},15\right)$, and so
$AH=\dfrac{15-b}{m}-10$.

[[IMAGE2]]

The area of $\triangle GAH$ is
$\dfrac12(GA)(AH)=\dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right)$.

Setting the area of $\triangle GAH$
equal to $-\dfrac{8}{m}$ and
simplifying, we get $$\begin{align*} \dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right) &= -\dfrac{8}{m}\\ 2m\times\dfrac12(10m+b-15)\left(\dfrac{15-b}{m}-10\right) &= -\dfrac{8}{m} \times2m \ \ \ \ \text{ (since } m\neq0)\\ (10m+b-15)\left(m\times\left(\dfrac{15-b}{m}-10\right)\right) &= -16\\ (10m+b-15)(15-b-10m) &= -16\\ (10m+b-15)^2 &= 16\\ 10m+b-15 &= \pm4\end{align*}$$ and so $10m+b=11$ or $10m+b=19$.

Since $10m+b$ is the $y$-coordinate of $G$, and $G$ lies between $A(10,15)$ and $D(10,27)$, then $15<10m+b<27$ and so $10m+b\neq11$.

Both $m$ and $b$ are integers with $m<0$ and $b<50$.

Suppose that $m=-1$, the largest
possible value of $m$.

In this case, $10(-1)+b=19$ and so
$b=29$. When $(m,b)=(-1,29)$, we confirm that the $x$-coordinate of $H$ is $x=\dfrac{15-b}{m}=\dfrac{15-29}{-1}=14$,
and so $H$ lies between $A$ and $B$.

If $m=-2$, then $10(-2)+b=19$ and so $b=39$. When $(m,b)=(-2,39)$, the $x$-coordinate of $H$ is $x=\dfrac{15-39}{-2}=12$, and so $H$ lies between $A$ and $B$.

If $m=-3$, then $10(-3)+b=19$ and so $b=49$. When $(m,b)=(-3,49)$, the $x$-coordinate of $H$ is $x=\dfrac{15-49}{-3}=\dfrac{34}{3}$, and so
$H$ lies between $A$ and $B$.

If $m\leq-4$, then $b=19-10m\geq19-10(-4)=59$ which is not
possible since $b<50$.

The ordered pairs of integers $(m,b)$
with $b<50$, and for which the
area of $\triangle GAH$ is equal to
$-\dfrac{8}{m}$, are $(-1,29), (-2,39)$ and $(-3,49)$.

Solution 2:

Point $G$ lies on $AD$ as shown and thus has $x$-coordinate $10$.

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

Since $G$ lies between $A$ and $D$, then $15<g<27$ and $GA=g-15$.

Point $H$ lies on $AB$ as shown and thus has $y$-coordinate $15$.

[[IMAGE3]]

Suppose the $x$-coordinate of
$H$ is $h$.

Since $H$ lies between $A$ and $B$, then $10<h<20$ and $AH=h-10$.

The area of $\triangle GAH$ is $\dfrac12(GA)(AH)=\dfrac12(g-15)(h-10)$.

The area of $\triangle GAH$ is also
equal to $-\dfrac{8}{m}$, and so
$\dfrac12(g-15)(h-10)=-\dfrac{8}{m}$
or $(g-15)(h-10)=-\dfrac{16}{m}$.

The line through $G(10,g)$ and $H(h,15)$ has slope $\dfrac{g-15}{10-h}$.

The line through $G$ and $H$ has equation $y=mx+b$ and thus slope $m$.

Equating slopes, we get $\dfrac{g-15}{10-h}=m$.

Using the equations $\dfrac{g-15}{10-h}=m$ and $(g-15)(h-10)=-\dfrac{16}{m}$, along with
the property that if $p=q$ and $r=s$, then $p\times r=q\times s$, we get the
following: $$\begin{align*} \dfrac{g-15}{10-h}\times(g-15)(h-10)&=m\times -\dfrac{16}{m}\\ -\dfrac{g-15}{h-10}\times(g-15)(h-10)&=m\times -\dfrac{16}{m}\\ (g-15)^2&=16 \ \ \ (\text{since both } m \text{ and } h-10 \text{ are not 0})\\ g-15&=\pm4\end{align*}$$ and so $g=15+4=19$ or $g=15-4=11$.

Since $15<g<27$, then $g=19$.

The line with equation $y=mx+b$
passes through $G(10,19)$ and so
$19=10m+b$ or $b=19-10m$.

Both $m$ and $b$ are integers with $m<0$ and $b<50$.

Suppose that $m=-1$, the largest
possible value of $m$.

In this case, $b=19-10(-1)=29$. When
$(m,b)=(-1,29)$, we use the equation
of the line $y=mx+b$ to confirm that
the $x$-coordinate of $H(h,15)$ is $h=\dfrac{15-b}{m}=\dfrac{15-29}{-1}=14$,
and so $H(14,15)$ lies between $A(10,15)$ and $B(20,15)$, as required.

If $m=-2$, then $b=19-10(-2)=39$. When $(m,b)=(-2,39)$, we can similarly show that
$h=12$ and so $H$ lies between $A$ and $B$.

If $m=-3$, then $b=19-10(-3)=49$. When $(m,b)=(-3,49)$, we get $h=\dfrac{34}{3}$ and so $H$ lies between $A$ and $B$.

If $m\leq-4$, then $b=19-10m\geq19-10(-4)=59$ which is not
possible since $b<50$.

The ordered pairs of integers $(m,b)$
with $b<50$, and for which the
area of $\triangle GAH$ is equal to
$-\dfrac{8}{m}$, are $(-1,29), (-2,39)$ and $(-3,49)$.