2023 AMC 10B Problem 24

P21-25~7 minPrint

Coordinate Geometry Quadrilaterals & Polygon Areas Transformations & Symmetry Test small/specific values or special cases to find or verify the answer Exploit symmetry to reduce work or pair up objects

What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ , $0\le v\le1,$ and $0\le w\le1$ ?

Show solution without answering

Key insight

For each fixed w the points form a 2 by 1 rectangle. As w goes from 0 to 1, the rectangle slides along (-3, 4) and sweeps out a hexagon.

Solution

Three letters can move at once, which is hard to picture. So hold one of them still first.

1. Split the point into pieces.
$(2u - 3w,\ v + 4w) = (2u,\ v) + w\,(-3,\ 4).$
2. Hold $w$ still. With $w = 0$ , the points are $(2u, v)$ with $u$ and $v$ from $0$ to $1$ . That is the rectangle with corners $(0,0)$ , $(2,0)$ , $(2,1)$ , $(0,1)$ : width $2$ , height $1$ .
3. Now let $w$ grow. For any fixed $w$ , we get the same rectangle moved by $w\,(-3, 4)$ . As $w$ goes from $0$ to $1$ , the rectangle slides in a straight line until it has moved $3$ left and $4$ up. At $w = 1$ its corners are $(-3,4)$ , $(-1,4)$ , $(-1,5)$ , $(-3,5)$ .
4. Outline the swept region. It is everything the rectangle passes over. Its boundary is:
- the bottom and right sides of the starting rectangle: $(0,0) \to (2,0) \to (2,1)$ ,
- a slanted edge traced by the corner $(2,1)$ as it slides to $(-1,5)$ ,
- the top and left sides of the final rectangle: $(-1,5) \to (-3,5) \to (-3,4)$ ,
- a slanted edge traced by the corner $(0,0)$ as it slides to $(-3,4)$ .

So the region is a hexagon with corners $(0,0)$ , $(2,0)$ , $(2,1)$ , $(-1,5)$ , $(-3,5)$ , $(-3,4)$ . The other two corners of the rectangles, $(0,1)$ and $(-1,4)$ , end up inside.
5. Side lengths.

side	from	to	length
1	$(0,0)$	$(2,0)$	$2$
2	$(2,0)$	$(2,1)$	$1$
3	$(2,1)$	$(-1,5)$	$\sqrt{3^2 + 4^2} = 5$
4	$(-1,5)$	$(-3,5)$	$2$
5	$(-3,5)$	$(-3,4)$	$1$
6	$(-3,4)$	$(0,0)$	$\sqrt{3^2 + 4^2} = 5$

Perimeter. $2 + 1 + 5 + 2 + 1 + 5 = 16$ .

The answer is $\boxed{\textbf{(E)}\ 16}$ .

Why this works

Each letter that runs from $0$ to $1$ adds one straight "piece" of motion: $u$ gives a step of length $2$ , $v$ a step of length $1$ , $w$ a step of length $5$ . Sliding a shape along a step adds that step twice to the outline, once on each side. So the perimeter is twice $2 + 1 + 5$ , which is $16$ . Freezing one variable and watching the others is a good way to picture any region described this way.

Alternative approach

Plug in every combination of $u, v, w$ equal to $0$ or $1$ . That gives eight corner points: $(0,0)$ , $(2,0)$ , $(0,1)$ , $(2,1)$ , $(-3,4)$ , $(-1,4)$ , $(-3,5)$ , $(-1,5)$ . The region is the shape you get by stretching a rubber band around all eight. Plotting them shows $(0,1)$ and $(-1,4)$ inside the band, and the other six are the hexagon above. Add its six sides with the distance formula to get $16$ .

Concise solution

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

Write the point as a sum of three independent pieces:
$(2u - 3w,\ v + 4w) = u\,(2, 0) + v\,(0, 1) + w\,(-3, 4), \qquad u, v, w \in [0, 1].$
So the region is obtained by starting with the segment from $(0,0)$ to $(2,0)$ , sweeping it along $(0,1)$ (giving a $2 \times 1$ rectangle), and then sweeping that rectangle along $(-3, 4)$ .

Sweeping a convex polygon along a vector produces a convex polygon whose boundary consists of the original edges (each appearing once on the "front" and once on the "back") together with two copies of the sweep vector. Starting from the segment $(2,0)$ and sweeping twice, every one of the three generating vectors therefore appears exactly twice on the boundary. The region is a centrally symmetric hexagon with vertices
$(0,0),\ (2,0),\ (2,1),\ (-1,5),\ (-3,5),\ (-3,4),$
and its side vectors in order are $(2,0), (0,1), (-3,4), (-2,0), (0,-1), (3,-4)$ , with lengths $2, 1, 5, 2, 1, 5$ .

Perimeter: $2(2 + 1 + 5) = 16$ .

The answer is $\boxed{\textbf{(E)}\ 16}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)