2000 AMC 10 Problem 5

Go through the four quantities one at a time.

(a) $M$ and $N$ are midpoints, so $\triangle PMN$ is a half-scale copy of $\triangle PAB$ and $MN = \tfrac12 AB$. Since $A$ and $B$ are fixed, $MN$ never changes.

(b) Imagine dragging $P$ far to the right along its line. Both $PA$ and $PB$ grow without bound while $AB$ stays put, so the perimeter changes.

(c) The area is $\tfrac12 \cdot AB \cdot h$, where $h$ is the distance from $P$ to line $AB$. That distance is the gap between two parallel lines, which is constant, so the area does not change.

(d) Trapezoid $ABNM$ is $\triangle PAB$ with $\triangle PMN$ removed, and the small triangle has $\tfrac14$ of the big one's area (linear ratio $\tfrac12$). So the trapezoid is always $\tfrac34$ of a constant area, which is constant.

Exactly one quantity, the perimeter, changes.

The answer is $\boxed{\textbf{(B)}\ 1}$.

Moving a vertex parallel to the opposite side is the classic area-preserving motion: the base and height are both untouched. Midpoints tie $MN$ and the small triangle to the big one by a fixed ratio, so anything determined by $AB$ and the height is frozen. Lengths of the slanted sides are the only free quantities, and pushing $P$ to an extreme position makes that visible.

• Believing the area of triangle PAB changes because the triangle looks different as P moves, forgetting that base AB and its height are both fixed.
• Thinking $MN$ changes because $M$ and $N$ move; the points move, but their separation stays $\tfrac12 AB$.
• Counting the trapezoid as changing since its slanted sides $AM$ and $BN$ change, even though its area does not.