, 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 $(-3,49)$.