Picking Coordinates

We have $\overrightarrow{PQ} \; = \; \left(\begin{array}{c}39 \\ -52 \end{array}\right) \hspace{2cm} \overrightarrow{QR} \; =\; \left(\begin{array}{c} 85 \\ 204 \end{array}\right) \hspace{2cm} \overrightarrow{PR} \; = \; \left(\begin{array}{c} 124 \\ 152 \end{array} \right)$ Since the highest common factor of $39$ and $-52$ is $13$ , there are $14$ points with integer coordinates on the line segment $PQ$ , namely the points $(-24+3k,17-4k)$ for $k = 0,1,2,\ldots,13$ . Similarly, the highest common factor of $85$ and $204$ is $17$ , so there are $18$ points with integer coordinates on the line segment $QR$ , and since the highest common factor of $124$ and $152$ is $4$ , there are $5$ points with integer coordinates on the line segment $PR$ . Thus there are (we have counted the triangle vertices twice) $E \; = \; 14 + 18 + 5 - 3 \; = \; 34$ points with integer coordinates that lie on the perimeter of $PQR$ . Since $\left(\begin{array}{c} 39 \\ -52 \\ 0 \end{array}\right) \; \wedge \; \left(\begin{array}{c} 124 \\ 152 \\ 0 \end{array}\right) \; = \; \left(\begin{array}{c} 0 \\ 0 \\ 12376 \end{array} \right)$ the triangle $PQR$ has area $\tfrac12 \times 12376 = 6188$ square units.

Using Pick's Theorem , there are $I$ points with integer coefficients lying inside the triangle $PQR$ , where $6188 \; = \; I + \tfrac12E - 1$ so that $I = 6172$ . Thus there are a total of $I+E = \boxed{6206}$ points with integer coefficients lying either inside the triangle $PQR$ or on the perimeter of $PQR$ .

If you construct a rectangle $ARCD$ containing $PQR$ with width $124$ units and height $204$ units, then this rectangle has $205\times 125$ lattices points on or inside it.

Firstly, since $\textrm{gcd}(85,204)=17$ , there are $18$ lattice points on $QR$ . Then the number of lattice points on or inside right-triangle $RCQ$ excluding those on $QR$ is $\dfrac{205\times 86-18}{2}$ (this is obtained by reflecting $C$ over $QR$ , thus forming a rectangle).

Secondly, since $\textrm{gcd}(39,52)=13$ , there are $14$ lattice points on $PQ$ . Then the number of lattice points on or inside right-triangle $PDQ$ excluding those on $PQ$ is $\dfrac{40\times 53-14}{2}$ (obtained by reflecting $D$ over $PQ$ , thus forming a rectangle).

Lastly, since $\textrm{gcd}(152,124)=4$ , there are $5$ lattice points on $PR$ . Then the number of lattice points on or inside right-triangle $PAR$ excluding those on $PR$ is $\dfrac{153\times 125-5}{2}$ (obtained in the aforementioned manner).

So the number of lattice points on or inside triangle $PQR$ is $205\times 125-\dfrac{205\times 86-18}{2}-\dfrac{40\times 53-14}{2}-\dfrac{153\times 125-5}{2}=\boxed{6206}.$

Note that this solution is more inefficient, but does not use Pick's Theorem.

Let's draw the lines passing through the vertices of the rectangle, namely

$\displaystyle y_{PR}=\frac{38x}{31}+\frac{1439}{31} \\ \displaystyle y_{PQ}=-\frac{4x}{3}-15 \\ \displaystyle y_{QR}=\frac{12x}{5}-71$

Now, let be $x_0 \in [-24,15] \subset \mathbb{Z}$ . The point $(x_0, \lfloor y_{PR}(x_0)\rfloor )$ is the point, right under $(x_0, y_{PR}(x_0) )$ , which coordinates are integers. Similarly, the point $(x_0, \lceil y_{PR}(x_0)\rceil )$ is the point, right above $(x_0, y_{PR}(x_0) )$ , which coordinates are, again, integers. Hence, the set $A=\{\textbf{a}=(x,y) | x=x_0, \hspace{3pt} y_{PR}(x_0) \leq y \leq y_{PQ}(x_0) \} \subset \mathbb{Z}^2$ has cardinality

$\displaystyle |A|=\sum_{x_0=-24}^{15} (\lfloor y_{PR}(x_0)\rfloor-\lceil y_{PQ}(x_0)\rceil+1)=\sum_{x_0=-24}^{15} \bigg(\Bigl\lfloor \frac{38x_0}{31}+\frac{1439}{31} \Bigl\rfloor-\Bigl\lceil -\frac{4x_0}{3}-15 \Bigr\rceil+1\bigg)=2004$

We do the same for $x_1 \in [16,100]$ , getting $B=\{\textbf{b}=(x,y) | x=x_0, \hspace{3pt} y_{PR}(x_0) \leq y \leq y_{RQ}(x_0) \} \subset \mathbb{Z}^2$ . So,

$\displaystyle |B|=\sum_{x_1=16}^{100} (\lfloor y_{PR}(x_1)\rfloor-\lceil y_{RQ}(x_1)\rceil+1)=\sum_{x_1=16}^{100} \bigg(\Bigl\lfloor \frac{38x_1}{31}+\frac{1439}{31}\Bigl\rfloor-\Bigl\lceil \frac{12x_1}{5}-71\Bigr\rceil+1\bigg)=4202$

Eventually,

$\displaystyle |A|+|B|=2004+4202=\boxed{6206}$