AMC 10 · 2018 · #4
Grade 8 geometry-2dPick an answer.
AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each face of a box is a rectangle whose area is the product of two of the three edge lengths. Tool #1 (Draw a Diagram) makes that concrete: sketch the box, label the edges a, b, c, and read off that the three faces meeting at a corner have areas ab, ac, bc. Tool #13 (Convert to Algebra) turns the three given areas into three equations ab = 24, ac = 48, bc = 72. Tool #7 (Identify Subproblems) gives the clean path through them: divide two equations to eliminate a variable, solve the resulting single equation, then back-substitute. No need to find a big product or guess — the equations unwind directly.
Name the edges a, b, c; each face is the product of two edges, so the three distinct areas give ab = 24, ac = 48, bc = 72.
A box face is just a rectangle, and its area is the product of the two edge lengths that form it.
6.G.A.4Convert To AlgebraDivide ac = 48 by ab = 24 to cancel the shared edge a: 48/24 = 2, so c/b = 2, giving c = 2b.
Dividing two face areas that share an edge cancels that edge and leaves the ratio of the other two.
6.RP.A.3Identify SubproblemsSubstitute c = 2b into bc = 72: 2b² = 72, so b² = 36, giving positive b = 6, c = 12.
Once everything is written through one edge, you get b² = 36, and taking the positive square root pins down b.
8.EE.A.2Convert To AlgebraFrom ab = 24 with b = 6, a = 4. The three edges 4, 6, 12 sum to 4 + 6 + 12 = 22, answer (B).
With one edge known, every other edge falls out by a single division, and the sum is just addition.
8.EE.C.8Convert To AlgebraEach box face is the product of two edges, so dividing two faces cancels their shared edge and the rest falls out.