AMC 10 · 2015 · #25

Grade 7 geometry-3d
Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration
volume-rectangular-prismsurface-arealinear-diophantine bound-inequality-then-enumerate ↑ Prerequisites: surface-areavolume-rectangular-prism
📏 Long solution 💡 4 insights
Problem
A box has whole-number edges a ≤ b ≤ c. Its volume abc equals its surface area 2(ab+bc+ca) as a number. Count how many ordered triples (a,b,c) make this happen.

Pick an answer.

(A)
$\; 4$
(B)
$\; 10$
(C)
$\; 12$
(D)
$\; 21$
(E)
$\; 26$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question asks "how many", so the goal is a complete, no-overlap count (Tool #2, Make a Systematic List). To make the list finite, first translate volume and surface area into the equation abc=2(ab+bc+ca) and divide to get 1a\frac{1}{a}+1b\frac{1}{b}+1c\frac{1}{c}=12\frac{1}{2} (Tool #4, Introduce a Variable). The Extreme Principle (Tool #14) then traps the smallest edge a in a tiny range {3,4,5,6}, and each value of a becomes a separate subproblem (Tool #7) solved by listing factor pairs.

1STEP 1

Write the volume

A box with edges a, b, c has volume equal to the product abc.

V = abc
2STEP 2

Write the surface area and equate

Three pairs of identical faces give surface area 2(ab+bc+ca); set it equal to the volume abc.

abc = 2(ab+bc+ca)
3STEP 3

Divide by abc

Divide both sides by abc; each right-hand term drops one variable, leaving 1a\frac{1}{a}+1b\frac{1}{b}+1c\frac{1}{c}=12\frac{1}{2}.

1 = 2(1c\frac{1}{c}+1a\frac{1}{a}+1b\frac{1}{b}) → 1a\frac{1}{a}+1b\frac{1}{b}+1c\frac{1}{c}=12\frac{1}{2}
4STEP 4

Bound the smallest edge a

Since a carries the largest fraction, 12\frac{1}{2} ≤ 3a\frac{3}{a} and 1a\frac{1}{a} stays under 12\frac{1}{2}, trapping a in {3, 4, 5, 6}.

12\frac{1}{2} ≤ 3a\frac{3}{a} → a ≤ 6, 1a\frac{1}{a} ＜ 12\frac{1}{2} → a ≥ 3
5STEP 5

Case a = 3

With a=3, 1b\frac{1}{b}+1c\frac{1}{c}=16\frac{1}{6} factors to (b-6)(c-6)=36; its factor pairs give 5 triples.

1b\frac{1}{b}+1c\frac{1}{c}=16\frac{1}{6} → (b-6)(c-6)=36
6STEP 6

Case a = 4

With a=4, 1b\frac{1}{b}+1c\frac{1}{c}=14\frac{1}{4} factors to (b-4)(c-4)=16; its factor pairs give 3 triples.

1b\frac{1}{b}+1c\frac{1}{c}=14\frac{1}{4} → (b-4)(c-4)=16
7STEP 7

Case a = 5

With a=5, 1b\frac{1}{b}+1c\frac{1}{c}=310\frac{3}{10} forces b=5 and c=10, giving the single triple (5,5,10).

1b\frac{1}{b}+1c\frac{1}{c}=310\frac{3}{10} → (b,c)=(5,10)
8STEP 8

Case a = 6

With a=6, 1b\frac{1}{b}+1c\frac{1}{c}=13\frac{1}{3} forces b=6 and c=6, giving the single cube (6,6,6).

1b\frac{1}{b}+1c\frac{1}{c}=13\frac{1}{3} → (b,c)=(6,6)
9STEP 9

Add the cases

The four cases never overlap, so add: 5+3+1+1=10 ordered triples, giving (B).

5 + 3 + 1 + 1 = 10 → (B)
Answer
10
10 is one of the listed choices. The cube (6,6,6) checks out directly: its volume is 216 and its surface area is 6 · 6² = 216, equal as required. The smallest edge a stayed inside {3,4,5,6}, so no case was skipped, and the counts 5,3,1,1 all came from complete factor-pair lists. The total 10 matches choice (B).
💡Key takeaway

Turn "volume equals surface area" into 1a\frac{1}{a}+1b\frac{1}{b}+1c\frac{1}{c}=12\frac{1}{2}, squeeze the smallest edge into a few values, then count factor pairs in each case.

  • Write the volume
  • Write the surface area and equate
  • Divide by abc
  • Bound the smallest edge a
  • Case a = 3
  • Case a = 4
  • Case a = 5
  • Case a = 6
  • Add the cases

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