, 2021

The distance between $J(2,7)$ and $K(5,3)$ is equal to $\sqrt{(2-5)^2+(7-3)^2} = \sqrt{3^2+4^2} = 5$.

Therefore, if we consider $\triangle JKL$ as having base $JK$ and height $h$, then we want $\dfrac{1}{2} \cdot JK \cdot h \leq 10$ which means that $h \leq 10 \cdot \dfrac{2}{5} = 4$.

In other words, $L(r,t)$ can be any point with $0 \leq r \leq 10$ and $0 \leq t \leq 10$ whose perpendicular distance to the line through $J$ and $K$ is at most 4.

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The slope of the line through $J(2,7)$ and $K(5,3)$ is equal to $\dfrac{7-3}{2-5} = -\dfrac{4}{3}$.

Therefore, this line has equation $y - 7 = -\dfrac{4}{3}(x-2)$.

Muliplying through by 3, we obtain $3y - 21 = -4x + 8$ or $4x + 3y = 29$.

We determine the equation of the line above this line that is parallel to it and a perpendicular distance of 4 from it.

The equation of this line will be of the form $4x + 3y = c$ for some real number $c$, since it is parallel to the line with equation $4x + 3y = 29$.

To determine the value of $c$, we determine the coordinates of one point on this line.

To determine such a point, we draw a perpendicular of length 4 from $K$ to a point $P$ above the line.

Since $JK$ has slope $-\dfrac{4}{3}$ and $KP$ is perpendicular to $JK$, then $KP$ has slope $\dfrac{3}{4}$.

Draw a vertical line from $P$ and a horizontal line from $K$, meeting at $Q$.

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Since $KP$ has slope $\dfrac{3}{4}$, then $PQ:QK = 3:4$, which means that $\triangle KQP$ is similar to a 3-4-5 triangle.

Since $KP = 4$, then $PQ = \dfrac{3}{5}KP = \dfrac{12}{5}$ and $QK = \dfrac{4}{3}KP = \dfrac{16}{5}$.

Thus, the coordinates of $P$ are $\left(5 + \dfrac{16}{5}, 3 + \dfrac{12}{5}\right)$ or $\left(\dfrac{41}{5}, \dfrac{27}{5}\right)$.

Since $P$ lies on the line with equation $4x + 3y = c$, then $$c = 4 \cdot \dfrac{41}{5} + 3 \cdot \dfrac{27}{5} = \dfrac{164}{5} + \dfrac{81}{5} = \dfrac{245}{5} = 49$$ and so the equation of the line parallel to $JK$ and 4 units above it is $4x + 3y = 49$.

In a similar way, we find that the line parallel to $JK$ and 4 units below it has equation $4x + 3y = 9$. (Note that $49 - 29 = 29 - 9$.)

This gives us the following diagram:

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The points $L$ that satisfy the given conditions are exactly the points within the square, below the line $4x+3y=49$ and above the line $4x+3y=9$. In other words, the region $\mathcal{R}$ is the region inside the square and between these lines.

To find the area of $\mathcal{R}$, we take the area of the square bounded by the lines $x=0$, $x=10$, $y=0$, and $y=10$ (this area equals $10 \cdot 10$ or 100) and subtract the area of the two triangles inside the square and not between the lines.

The line with equation $4x+3y=9$ intersects the $y$-axis at $(0,3)$ (we see this by setting $x=0$) and the $x$-axis at $\left(\dfrac{9}{4},0\right)$ (we see this by setting $y=0$).

The line with equation $4x+3y=49$ intersects the line $x=10$ at $(10,3)$ (we see this by setting $x=10$) and the line $y=10$ at $\left(\dfrac{19}{4},10\right)$ (we see this by setting $y=10$).

The bottom triangle that is inside the square and outside $\mathcal{R}$ has area $\dfrac{1}{2}\cdot 3 \cdot \dfrac{9}{4} = \dfrac{27}{8}$.

The top triangle that is inside the square and outside $\mathcal{R}$ has horizontal base of length $10 - \dfrac{19}{4}$ or $\dfrac{21}{4}$ and vertical height of length $10 - 3$ or $7$, and thus has area $\dfrac{1}{2} \cdot \dfrac{21}{4} \cdot 7 = \dfrac{147}{8}$.

Finally, this means that the area of $\mathcal{R}$ is $$100 - \dfrac{27}{8} - \dfrac{147}{8} = 100 - \dfrac{174}{8} = 100 - \dfrac{87}{4} = \dfrac{313}{4}$$ which is in lowest terms since the only divisors of the denominator that are larger than 1 are 2 and 4, while the numerator is odd.

When we write this area in the form $\dfrac{300+a}{40 - b}$ where $a$ and $b$ are positive integers, we obtain $a = 13$ and $b = 36$, giving $a+b = 49$.