2000 AMC 10 Problem 5

P1-10~2 minPrint

Triangles: Area & Pythagorean Quadrilaterals & Polygon Areas Similar & Congruent Triangles Use an invariant, parity, or coloring argument Test small/specific values or special cases to find or verify the answer

Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?

(a) the length of the segment $MN$

(b) the perimeter of $\triangle PAB$

(d) the area of trapezoid $ABNM$

Show solution without answering

Key insight

P moves along a line parallel to AB, so its height above AB never changes. Since MN is always half of AB, only the perimeter can change.

Solution

Check the four quantities one at a time.

(a) Length of $MN$ . $M$ is halfway along $PA$ and $N$ is halfway along $PB$ . So triangle $PMN$ is a copy of triangle $PAB$ shrunk to half size, and $MN$ is half of $AB$ . Since $A$ and $B$ do not move, $MN$ does not change.
(b) Perimeter of triangle $PAB$ . Picture sliding $P$ very far to the right. The sides $PA$ and $PB$ become very long, while $AB$ stays the same. So the perimeter changes.
(c) Area of triangle $PAB$ . Area $= \tfrac12 \cdot \text{base} \cdot \text{height}$ . Use $AB$ as the base; it does not change. The height is the distance from $P$ down to line $AB$ . $P$ is on a line parallel to $AB$ , and two parallel lines stay the same distance apart everywhere, so the height does not change. The area does not change.
(d) Area of trapezoid $ABNM$ . The trapezoid is the big triangle $PAB$ with the small triangle $PMN$ cut off the top. The small triangle is half as wide and half as tall, so its area is $\tfrac14$ of the big one. The trapezoid is always $\tfrac34$ of the big triangle's area, which does not change. So the trapezoid's area does not change.

Only one quantity, the perimeter, changes.

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

Why this works

Sliding the top corner of a triangle along a line parallel to its base keeps both the base and the height the same, so the area stays fixed. Anything built from the midpoints is a fixed fraction of the big triangle, so it stays fixed too. Only the slanted sides are free to change, and moving $P$ to an extreme position makes that easy to see.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(B)}\ 1}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)