, 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$2688+336+896+112=4032$.