AMC 10 · 2008 · #25

Grade 11 algebra
complex-numbersrecursive-sequencerotation-isometry work-backwardspattern-recognition ↑ Prerequisites: complex-numbers
📏 Long solution 💡 2 insights
Problem
A list of points is grown by one fixed rule mixing the two coordinates. The hundredth point is given. Find the sum of the first point's coordinates.

Pick an answer.

(A)
$-\frac{1}{2^{97}}$
(B)
$-\frac{1}{2^{99}}$
(C)
0
(D)
$\frac{1}{2^{98}}$
(E)
$\frac{1}{2^{96}}$
How to solve
Strategy Work Backwards

The finish is given and the start is wanted, so tool #11 (Work Backwards) sets the direction. Stepping back 99 times by hand is hopeless, so tool #15 (Organize Information in More Ways) does the real work first: stop carrying a_n and b_n as two separate numbers and pack the pair into the single complex number a_n+b_ni. The two-part rule then collapses into one multiplication, and 99 steps become one power. Tool #9 (Solve an Easier Related Problem) shrinks that power: instead of attacking (√(3)+i)⁹⁹, cube the multiplier and watch it turn into 8i. Tool #5 (Look for a Pattern) names the three-step cycle that results, but the cycle here is proved by that one cubing rather than read off a list of sample terms - which matters, because a pattern spotted in the first few terms carries no promise about the hundredth.

1STEP 1

Pack the pair into one number

Packing the pair turns the rule into a multiplication.

z_n = a_n + b_ni ⟹ z_n+1 = (√(3)+i) z_n
2STEP 2

One fixed multiplier, one formula

A fixed multiplier gives a closed form with one power.

z_n = (√(3)+i)ⁿ⁻¹z₁ ⟹ z₁₀₀ = (√(3)+i)⁹⁹z₁
3STEP 3

Cube the multiplier

Cubing the multiplier gives something very simple.

(√(3)+i)² = 2+2√(3)i, (√(3)+i)³ = 8i
4STEP 4

Turn 99 into 33 cubes

The large power then collapses into a single term.

(√(3)+i)⁹⁹ = (8i)³³ = 2⁹⁹i ⟹ z₁₀₀ = 2⁹⁹i z₁
5STEP 5

Check the start is forced

The multiplier is nonzero, so the start is forced.

2⁹⁹i ≠ 0 ⟹ z₁ = z₁₀₀/2⁹⁹i exists and is the only solution
6STEP 6

Divide and add the coordinates

Dividing and adding gives 1/2⁹⁸, choice (D).

z₁ = (4-2i)/2⁹⁹ → a₁ = 1/2⁹⁷, b₁ = -1/2⁹⁸ → a₁+b₁ = 1/2⁹⁸ → (D)
Answer
1/2⁹⁸
Run the sequence forward from the answer instead of backward. With z₁ = (4-2i)/2⁹⁹, the formula gives z₁₀₀ = 2⁹⁹i · (4-2i)/2⁹⁹ = i(4-2i) = 2+4i, which is the point (2,4) the problem states. A size check agrees on its own: |√(3)+i| = √(3+1) = 2, so every step doubles the distance from the origin and 99 steps multiply it by 2⁹⁹. Since |z₁₀₀| = √(2²+4²) = 2√(5), the start must sit at distance 2√(5)/2⁹⁹ = √(5)/2⁹⁸ from the origin, and the found point does, since (1/2⁹⁷)² + (1/2⁹⁸)² = (4+1)/2¹⁹⁶ = 5/2¹⁹⁶. That same scale explains the choice list: anything of size 1/2⁹⁶ is four times too big, ruling out (E), and (C) would need a₁ = -b₁, which forces z₁ = a₁(1-i) and hence z₁₀₀ = 2⁹⁹a₁(1+i), a point whose two coordinates are equal - but (2,4) has unequal coordinates. So 1/2⁹⁸, choice (D), survives every check.
💡Key takeaway

When one rule keeps stirring two numbers together, pack the pair into a single number - then a hundred steps become one multiplication you can undo.

  • Pack the pair into one number
  • One fixed multiplier, one formula
  • Cube the multiplier
  • Turn 99 into 33 cubes
  • Check the start is forced
  • Divide and add the coordinates