AMC 10 · 2003 · #9

Grade 8 geometry-2d
reflection-symmetrycoordinate-geometrysymmetry-argument systematic-enumerationextreme-principle ↑ Prerequisites: reflection-symmetry
📏 Long solution 💡 2 insights
Problem
A set S of points in the xy-plane survives four moves unchanged: reflection across the x-axis, reflection across the y-axis, reflection across the line y=x, and a half-turn about the origin. The point (2,3) belongs to S. Find the smallest number of points S can possibly contain.

Pick an answer.

(A)
1
(B)
2
(C)
4
(D)
8
(E)
16
How to solve
Strategy Draw a Diagram

The four symmetries are geometric moves, so tool #1 (Draw a Diagram) turns them into something you can see: plot (2,3), then plot every mirror image the rules demand. Tool #2 (Make a Systematic List) keeps that generation honest, because the question is a count and a missed point or a double-counted point ruins it. Tool #14 (Extreme Principle) names the real job: the word "smallest" means two separate claims have to be settled, a floor (every legal S must contain at least these points) and a ceiling (this particular set of points is itself legal). Getting the forced points is only half the problem; showing nothing more is forced is the other half.

1STEP 1

Write each symmetry as a coordinate rule

Each move is a coordinate rule; the half-turn is just the two axis flips combined.

(a,b)↦(a,-b), (a,b)↦(-a,b), (a,b)↦(b,a), (a,b)↦(-a,-b)
2STEP 2

Chase the point through the rules

Chasing the given point through the rules forces all of (±2,±3) and (±3,±2).

(2,3) → (±2,±3) and (±3,±2)
3STEP 3

Confirm the eight points are all different

Since 2 and 3 differ and neither is zero, those are eight distinct points.

2 ≠ 3, 2 ≠ 0, 3 ≠ 0 → |{(±2,±3),(±3,±2)}|=8
4STEP 4

Check the eight points close up

Testing every rule on that set produces nothing new, so it is genuinely closed.

T={(±2,±3),(±3,±2)}: (a,b)↦(-a,b), (a,-b), (b,a), (-a,-b) all stay inside T
5STEP 5

Count and conclude

Bound and construction meet, so the smallest possible size is 8, choice (D).

2 × 2 × 2 = 8 → (D)
Answer
8
The four sign patterns applied to (2,3) alone already force 4 points, and the swap across y=x doubles that to 8 because 2 ≠ 3 makes (3,2) a new point. So 1, 2, and 4 are all impossible, which kills (A), (B), and (C). At the other end, the eight-point set was checked rule by rule and produced nothing new, so no legal set is forced past 8 and (E) 16 is too large. Only (D) survives, and it survives for a reason rather than by elimination. A quick picture agrees: the eight points sit at the corners of two nested rectangles that together form a shape as symmetric as a square.
💡Key takeaway

Apply every symmetry rule to the point you were given until nothing new appears, then check that the list you ended with obeys all the rules itself, and that list is the smallest set possible.

  • Write each symmetry as a coordinate rule
  • Chase the point through the rules
  • Confirm the eight points are all different
  • Check the eight points close up
  • Count and conclude