AMC 10 · 2015 · #20
Grade 6 geometry-2dPick an answer.
AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #4 (Introduce a Variable) names the two sides a and b so that A+P becomes a single expression ab+2a+2b. The key move is a factoring trick: adding 4 turns that messy sum into the clean product (a+2)(b+2), so the whole problem reduces to a question about factor pairs. Tool #3 (Eliminate Possibilities) then fits perfectly because there are only five answer choices — a finite set — and we can test each one to see whether it survives. Tool #13 (Convert to Algebra) is the bridge that rewrites the geometry words 'area' and 'perimeter' into symbols we can manipulate.
Name the sides, write A+P
Naming the sides a and b turns area and perimeter into one algebraic expression for A+P.
Letting the sides be a and b turns the rectangle into one expression you can reshape.
6.EE.A.2Use Matrix LogicAdd 4 and factor
Add 4 to complete the sum into the product (a+2)(b+2), so A+P equals that product minus 4.
Borrowing a +4 snaps a stubborn sum into a tidy product you can factor.
6.EE.A.3Use Matrix LogicTurn it into a factor-pair test
Testing each choice shows 100, 104, 106, and 108 all split into two factors of N+4 that are at least 3.
Both factors must be ≥ 3 because the smallest side is 1, so 1+2=3.
4.OA.B.4Eliminate PossibilitiesThe one that fails
106=2×53 leaves no factor pair both at least 3, so no rectangle gives A+P=102.
When N+4 is twice a prime, its only factors are too small to be real sides.
4.OA.B.4Eliminate PossibilitiesArea plus perimeter is (a+2)(b+2)-4, so a number is reachable only if adding 4 gives two factors both at least 3.
- Name the sides, write A+P
- Add 4 and factor
- Turn it into a factor-pair test
- The one that fails