Problem 462

Solution 1:

The opposite sides of $ABCD$ are
parallel, and so $ABCD$ is a
parallelogram. The opposite sides of a parallelogram are equal in
length, and so $AB=CD$ and $AD=BC$.

Since the value of $r+s+t$ is equal
to a constant, we may choose any location for $B(0,r)$ provided that it satisfies the
given condition $r<0$. We choose
$r=-2$, so that the $y$-coordinate of $B(0,r)$ is equal to the $y$-coordinate of $A(-3,-2)$, and so $AB$ is a horizontal line segment as
shown.

[[IMAGE0]]

In this case, the length of $AB$
is equal to the positive difference between the $x$-coordinates of $A$ and $B$, which is $0-(-3)=3$.

$CD$ is parallel to $AB$, and so $CD$ must also be a horizontal line
segment. Thus, points $C(6,10)$ and
$D(s,t)$ have equal $y$-coordinates, and so $t=10$. Further, $CD$ has the same length as $AB$, and so $6-s=3$ or $s=3$.

Therefore, the value of $r+s+t=-2+3+10=11$.

Solution 2:

The opposite sides of $ABCD$ are
parallel, and so $ABCD$ is a
parallelogram. The opposite sides of a parallelogram are equal in
length, and so $AB=CD$ and $AD=BC$.

Since $AB$ and $CD$ are parallel and equal in length, then
the vertical distance between $A$ and
$B$ must equal the vertical distance
between $C$ and $D$, and the horizontal distance between
$A$ and $B$ must equal the horizontal distance
between $C$ and $D$.

[[IMAGE1]]

The vertical distance between two points is equal to the non-negative
difference between their $y$-coordinates, and so $-2-r=t-10$ (assuming $r\leq-2$ and $t\geq10$ as in the diagram).

Simplifying, we get $-2+10=t+r$ and
so $t+r=8$.

The horizontal distance between two points is equal to the non-negative
difference between their $x$-coordinates, and so $0-(-3)=6-s$ (assuming $s<6$ as in the diagram).

Simplifying, we get $0+3=6-s$ or
$s=6-3$ and so $s=3$.

Therefore, the value of $r+s+t=(r+t)+s=8+3=11$. From Solution 1, we
note that $r=-2$, $s=3$, $t=10$ are values satisfying the given
conditions and for which $r+s+t=11$.