AMC 10 · 2002 · #19

Grade 8 algebra
Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known
systems-of-equationssymmetric-polynomialssigned-square-root polynomial-substitutionidentify-subproblems ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 2 insights
Problem
Positive real numbers a, b, c satisfy a(b+c) = 152, b(c+a) = 162, and c(a+b) = 170. Find the value of the product abc.

Pick an answer.

(A)
672
(B)
688
(C)
704
(D)
720
(E)
750
How to solve
Strategy Organize Information in More Ways

As written, the system looks nonlinear and unfriendly. Tool #15 (Organize Information in More Ways) says to re-read it after expanding: a(b+c) = ab + ac, and likewise for the others, so every equation is a sum of two of the three quantities ab, bc, ca. Tool #4 (Introduce a Variable) makes that official — treat ab, bc, ca as three brand-new unknowns, and the system becomes an ordinary linear one where each equation omits exactly one unknown. That shape has a standard trick: add everything, then subtract each equation to isolate the missing piece. Tool #13 (Convert to Algebra) finishes the job, because the product of the three pairwise products is (abc)², so one square root delivers the answer. The point is that we never need a, b, c separately.

1STEP 1

Expand to see the real unknowns

Expanding shows only the pairwise products appear, two per equation.

ab + ca = 152, ab + bc = 162, bc + ca = 170
2STEP 2

Add all three equations

Adding all three counts each product twice, giving their total as 242.

2(x + y + z) = 484 ⟹ x + y + z = ab + bc + ca = 242
3STEP 3

Subtract to isolate each product

Subtracting each original equation from that total recovers 90, 80 and 72.

bc = 90, ca = 80, ab = 72
4STEP 4

Multiply the three products

Multiplying all three puts each letter in twice, so (abc)² = 518,400.

(abc)² = (ab)(bc)(ca) = 72 · 90 · 80 = 518,400
5STEP 5

Take the positive square root

Since all are positive, the positive root gives 720, choice (D).

abc = √(518,400) = √(720²) = 720 (D)
Answer
720
Everything above shows only that IF such numbers exist THEN abc = 720, so confirm they exist. Recovering the letters, a² = (ab)(ca)/bc = (72 · 80)/90 = 64, and positivity gives a = 8, then b = 72/8 = 9 and c = 80/8 = 10. Test all three original equations: 8(9+10) = 152, 9(10+8) = 162, 10(8+9) = 170. Every one holds, and 8 · 9 · 10 = 720. One warning about a tempting shortcut: since abc = 72c, it is tempting to say abc must be a multiple of 72 and pick the only such choice. That step secretly assumes c is a whole number, which the problem never states — a, b, c are only required to be positive reals, so such a divisibility argument is not supported by the givens even though the answer happens to come out that way.
💡Key takeaway

When the same combinations keep showing up, give them names — here the three pairwise products turn a scary nonlinear system into an easy linear one.

  • Expand to see the real unknowns
  • Add all three equations
  • Subtract to isolate each product
  • Multiply the three products
  • Take the positive square root

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