AMC 10 · 2009 · #17

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta
geometric-series-infinitequadratic-equationssystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: geometric-series-infinite
📏 Medium solution 💡 3 insights
Problem
Two different geometric series share a first term and each sums to its own ratio. Find the sum of the two ratios.

Pick an answer.

(A)
0
(B)
$\frac {1}{2}$
(C)
1
(D)
$\frac {1 + \sqrt {5}}{2}$
(E)
2
How to solve
Strategy Convert to Algebra

The sentence 'the sum of the series equals its own ratio' is a verbal condition, so the first move is to turn it into an equation with the infinite geometric sum formula. That produces two equations that both start with the same unknown a. Trying to solve for a and each ratio separately is a trap, because a is never given. The productive move is to compare the two equations instead of solving either one: the shared a cancels, and the difference that is left factors as a difference of squares. Finally a concrete first term is tested, because a rule about a pair of series is only meaningful if such a pair can actually exist.

1STEP 1

Pin down the range of each ratio

Having a sum restricts both ratios.

a ＞ 0, 0 ＜ r₁ ＜ 1, 0 ＜ r₂ ＜ 1
2STEP 2

Turn each series into an equation

Each series becomes one equation.

a/(1-r₁) = r₁ ⟹ a = r₁(1-r₁) = r₁ - r₁², a/(1-r₂) = r₂ ⟹ a = r₂ - r₂²
3STEP 3

See that both ratios feed one expression

Both feed the same expression in their own ratio.

r₁ - r₁² = a = r₂ - r₂², r₁ ≠ r₂
4STEP 4

Subtract, then factor the difference

Subtracting and cancelling gives the sum as 1.

r₁ - r₂ = r₁² - r₂² = (r₁ - r₂)(r₁ + r₂) ⟹ 1 = r₁ + r₂
5STEP 5

Confirm such a pair can exist

An explicit pair confirms 1, choice (C).

a = 3/16: r² - r + 3/16 = (r - 1/4)(r - 3/4) = 0, a/(1 - 1/4) = 1/4, a/(1 - 3/4) = 3/4, 1/4 + 3/4 = 1
Answer
1
Two independent checks agree. First, the bound from step 1 puts each ratio strictly between 0 and 1, so their sum lies strictly between 0 and 2; that alone eliminates the choices 0 and 2 without any algebra. Second, the derivation used no numeric value of a, so the sum must be the same for every allowed first term, and testing two of them confirms it: a equal to three sixteenths gives ratios one fourth and three fourths, and a equal to one fifth gives ratios near 0.2764 and 0.7236; both pairs add to 1. The relation also predicts that the two ratios are mirror images about one half, since r₂ = 1 - r₁, and both test pairs match that. Everything is consistent with (C).
💡Key takeaway

When two different numbers make the same expression give the same value, subtract the two equations: the shared unknown vanishes and what is left hands you their sum.

  • Pin down the range of each ratio
  • Turn each series into an equation
  • See that both ratios feed one expression
  • Subtract, then factor the difference
  • Confirm such a pair can exist

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