, 2025

We begin by placing $E(4,4)$
on $CD$, as shown.

[[IMAGE0]]

To determine the area of each of the three triangles, consider the
base of each to be $AB$.

Points $C$, $D$ and $E$ each have the same $y$-coordinate, $4$, and so each lies on the horizontal
line $y=4$.

The height of each of the three triangles is the vertical distance
between the line $y=4$ and the $x$-axis (the line through $A$ and $B$), which is $4$.

Since $AB=7$, then the area of
each of the three triangles is $\$\frac12\times 7\times4=14\$.$

The sum of the areas of $\$\triangle$
ABC$,$$\triangle$ ABD$, and
$\triangle ABE$ is $3\times14=42$.
$\triangle CDG$ has non-zero
area, so $G$ cannot lie on $CD$.

On $AB$, there are $8$ possible
locations for $G$.

These are the points $(k,0)$ for the
integers $0\leq k\leq 7$.

On $AD$, $(1,2)$ is the only additional point for
which the coordinates are both integers. Similarly, $(8,2)$ is the only additional possibility
for $G$ on $BC$. The points $(1,2)$ and $(8,2)$ are labelled $M$ and $N$ respectively, as shown.

[[IMAGE1]]

In total, there are $10$
possibilities for the point $G$.
Each such triangle has three vertices chosen from the $18$ points with integer coordinates on the
perimeter of $ABCD$, provided that
the three vertices are not all on the same line.

These $18$ points with integer
coordinates are the $8$ points on
$AB$ (including $A$ and $B$), the $8$ points on $CD$ (including $C$ and $D$), and the $2$ points $M(1,2)$ and $N(8,2)$.

Each such triangle is described by exactly one of the following
cases:

the number of triangle vertices at $M$ and $N$ is $0$
the number of triangle vertices at $M$ and $N$ is $1$
the number of triangle vertices at $M$ and $N$ is $2$

Case 1: the number of triangle vertices at $M$ and $N$ is $0$.

In this case, $2$ vertices are on
$AB$ and $1$ vertex is on $CD$, or vice versa.

Consider the triangles with $2$
vertices on $AB$ and 1 vertex on
$CD$.

Consider the base of each such triangle to lie along $AB$.

There is $1$ such triangle with
vertices at $A(0,0)$ and $B(7,0)$, and thus has base length 7.

There are $2$ such triangles with
base length $6$. One of these
triangles has vertices at $A(0,0)$
and $(6,0)$, and the other has
vertices at $(1,0)$ and $B(7,0)$.

Continuing in this way, there are $3$
triangles with base length $5$, $4$ triangles with base length $4$, $5$
triangles with base length $3$, $6$ triangles with base length $2$, and $7$ triangles with base length $1$.

Each of these triangles has height $4$ since all points on $CD$ are a vertical distance of $4$ from any base that lies along $AB$ (as in part (a)).

Suppose $Q$ is one such point on
$CD$ having integer
coordinates.

The sum of the areas of all triangles having $2$ vertices on $AB$ and $1$ vertex at $Q$ is $$\frac12\times4\times(1(7)+2(6)+3(5)+4(4)+5(3)+6(2)+7(1))=\frac12\times4\times84=168$$
Also, for each of these bases, there are $8$ possibilities for the third vertex that
lies on $CD$. These are the points
$(k,4)$ for integers $2\leq k\leq 9$.

Thus, the sum of the areas of all triangles having $2$ vertices on $AB$ and $1$ vertex on $CD$ is $168\times8=1344$.

In a similar way, the sum of the areas of all triangles having $2$ vertices on $CD$ and $1$ vertex on $AB$ is also $1344$, and so the sum of the areas of all
triangles in Case 1 is $1344\times2=2688$.

Case 2: the number of triangle vertices at $M$ and $N$ is $1$.

This case can be divided into the following two subcases:

$2$ vertices are on $AB$ (or $2$ vertices are on $CD$), and $1$ vertex is $M$ or $N$
$1$ vertex is on $AB$, $1$ vertex is on $CD$, and $1$ vertex is $M$ or $N$

Subcase 2(i): $2$
vertices on $AB$ (or $2$ vertices on $CD$), and $1$ vertex is $M$ or $N$.

Consider the triangles with $2$
vertices on $AB$ and $1$ vertex at either $M$ or $N$.

Consider the base of each such triangle to lie along $AB$.

The numbers and lengths of these bases are the same as in Case 1.

Each of these triangles has height $2$ since $M$ and $N$ are each a vertical distance of $2$ from any base that lies along $AB$.

The sum of the areas of all triangles having $2$ vertices on $AB$ and $1$ vertex at $M$ is $$\frac12\times2\times(1(7)+2(6)+3(5)+4(4)+5(3)+6(2)+7(1))=\frac12\times2\times84=84$$
Also, for each of these bases, the third vertex could also be $N$.

Thus, the sum of the areas of all triangles having $2$ vertices on $AB$ and $1$ vertex at either $M$ or $N$ is $84\times2=168$.

In a similar way, the sum of the areas of all triangles having $2$ vertices on $CD$ and $1$ vertex at either $M$ or $N$ is $168$, and so the sum of the areas of all
triangles in Subcase 2(i) is $168\times2=336$.

Subcase 2(ii): $1$ vertex is on $AB$, $1$ vertex is on $CD$, and $1$ vertex is $M$ or $N$.

Consider two fixed points with integer coordinates, point $P$ on $AB$, and point $Q$ on $CD$.

In the diagram, $\triangle PMQ$ and
$\triangle PNQ$ are two such
triangles described by Subcase 2(ii).

[[IMAGE2]]

The sum of the areas of $\$\triangle$
PMQ$and$$\triangle$ PNQ$ is
equal to the sum of the areas of $\$\triangle$
PMN$and$$\triangle$ QMN$.

Consider the base of both $\$\triangle$
PMN$and$$\triangle$ QMN$ to
be $MN$, which has length $7$. Then the height of each triangle is
$2$.

Thus, the sum of the areas of $\$\triangle$
PMN$and$$\triangle$ QMN$ is
$2\times\frac12\times2\times7=14$.

There are $8$ possible locations for
$P$ and $8$ possible locations for $Q$, and thus $8\times8=64$ different pairs of triangles
$PMN$ and $QMN$ whose areas have a sum of $14$. (You should confirm for yourself that
this is true even when $P$ is $(0,0)$ or $(7,0)$ and/or when $Q$ is $(2,4)$ or $(9,4)$.) The sum of the areas of all
triangles in Subcase 2(ii) is $14\times64=896$.

Case 3: the number of triangle vertices at $M$ and $N$ is $2$.

In this case, $1$ vertex is on
$AB$ (or on $CD$), $1$ vertex is $M$, and $1$ vertex is $N$.

Consider the triangles with vertices $M$, $N$, and $1$ vertex on $AB$.

Consider the base of each such triangle to be $MN=7$. Then each triangle has height $2$.

Since there are $8$ possible vertices
on $AB$, then the sum of all such
triangles is $\frac12\times2\times7\times8=56$. In a
similar way, the sum of the areas of all triangles having vertices $M$, $N$, and $1$ vertex on $CD$ is also $56$, and so the sum of the areas of all
triangles in Case 3 is $56\times2=112$.

The sum of the areas of all triangles whose vertices have integer
coordinates and lie on the perimeter of $ABCD$ is $2688+336+896+112=4032$.