AMC 10 · 2018 · #4

Grade 8 geometry-2d
area-rectanglessystems-of-equationssurface-area convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectanglessystems-of-equations
📏 Medium solution 💡 2 insights
Problem
A rectangular box has edge lengths X, Y, and Z. Its six faces come in three matching pairs, with areas 24, 24, 48, 48, 72, 72 square units. We must find the sum X + Y + Z.

Pick an answer.

(A)
18
(B)
22
(C)
24
(D)
30
(E)
36

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

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.

1STEP 1

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.

ab = 24, ac = 48, bc = 72
2STEP 2

Divide ac = 48 by ab = 24 to cancel the shared edge a: 48/24 = 2, so c/b = 2, giving c = 2b.

ac/ab = 48/24 = 2 → c/b = 2 → c = 2b
3STEP 3

Substitute c = 2b into bc = 72: 2b² = 72, so b² = 36, giving positive b = 6, c = 12.

b · 2b = 72 → 2b² = 72 → b² = 36 → b = 6, c = 12
4STEP 4

From ab = 24 with b = 6, a = 4. The three edges 4, 6, 12 sum to 4 + 6 + 12 = 22, answer (B).

a = 24/6 = 4 → X + Y + Z = 4 + 6 + 12 = 22 → (B)
Answer
22
Plug the edges back into all three face areas: ab = 4 · 6 = 24, ac = 4 · 12 = 48, bc = 6 · 12 = 72 — all six faces match the list 24, 24, 48, 48, 72, 72. The sum 4 + 6 + 12 = 22 is answer (B). It sits sensibly among the choices: too-small (A) 18 would need shorter edges that can't reach area 72, and the larger choices would overshoot the 24 face.
💡Key takeaway

Each box face is the product of two edges, so dividing two faces cancels their shared edge and the rest falls out.