AMC 10 · 2007 · #23
Grade 9 number-theorygeometry-2dPick an answer.
Tool #13 (Convert to Algebra) does the main work: the sentence 'area equals 3 times perimeter' becomes 1/2ab = 3(a+b+√(a²+b²)), and everything after that is bookkeeping on one equation. Tool #4 (Introduce a Variable) makes the squaring short by naming the sum S=a+b and the product P=ab. Tool #15 (Organize Information in More Ways) rewrites the result as a single product (a-12)(b-12)=72, which converts an equation into a factoring question. Tool #3 (Eliminate Possibilities) is the step that actually decides the count: the square root has to be removed by squaring, squaring cannot distinguish c from -c, and one factorization of 72 survives the algebra while describing a triangle with a negative hypotenuse. Tool #11 (Work Backwards) then rebuilds a genuine triangle from each surviving factorization, so the answer rests on sufficiency and not only on necessity. Tool #2 (Make a Systematic List) finishes by pairing off the divisors of 72.
Write the condition as one equation
The condition becomes one equation with a root in it.
The whole sentence collapses into one equation once the hypotenuse is written as √(a²+b²).
9.A-CED.A.2Convert To AlgebraIsolate the root and record its price
Isolating the root records a positivity gate to spend later.
Squaring cannot tell c from -c, so the sign must be written down before it is thrown away.
9.A-CED.A.3Organize Information In More WaysSquare, using the Pythagorean theorem
Squaring with the Pythagorean theorem gives a linear relation.
Naming the sum and the product turns a messy squaring into one short line.
8.G.B.7Introduce A VariableTurn the equation into a product
Rearranging turns it into a product of two brackets.
Adding the one right constant collapses a scattered equation into a single product.
9.A-SSE.A.2Organize Information In More WaysSpend the gate on the negative branch
The gate kills the whole negative branch.
The one small candidate the algebra offers is the triangle whose hypotenuse came out negative.
7.EE.B.4Eliminate PossibilitiesCheck the surviving pairs are real triangles
The survivors really are right triangles meeting the condition.
Running the algebra backwards turns 'it must look like this' into 'this really is a triangle'.
6.G.A.1Work BackwardsPair off the divisors of 72
Pairing the divisors gives six pairs.
Divisors of 72 come in partner pairs, and since 72 is not a square nobody is their own partner.
The divisors of the number pair off, and since it is not a perfect square nobody is their own partner.
▸ Why?
Each divisor multiplies with exactly one partner to give the number back.
▸ Why?
A number is its own partner only when it is a perfect square, which this prime recipe rules out.
Count triangles, not ordered pairs
Counting triangles rather than ordered pairs gives 6, choice (A).
Swapping the two legs is a flip of the same triangle, not a new one.
8.G.A.2Eliminate PossibilitiesArea equal to 3 times perimeter is the same as saying the inscribed circle has radius 6, and that turns the whole question into counting the factor pairs of 72 — after throwing out the one pair whose hypotenuse comes out negative.
- Write the condition as one equation
- Isolate the root and record its price
- Square, using the Pythagorean theorem
- Turn the equation into a product
- Spend the gate on the negative branch
- Check the surviving pairs are real triangles
- Pair off the divisors of 72
- Count triangles, not ordered pairs