AMC 10 · 2009 · #15

Grade 6 patterngeometry-2d
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence
pattern-recognitionsequences-arithmeticsystematic-enumeration easier-related-problempattern-recognition ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A sequence of figures is built from diamonds. F₁ is a single diamond, F₂ has 5, and F₃ has 13. For every n at least 3, the figure Fₙ is made from the previous figure Fₙ₋₁ by surrounding it with a new square frame that carries one more diamond on each side than the outer square of Fₙ₋₁ had. How many diamonds are in F₂₀?

Pick an answer.

(A)
401
(B)
485
(C)
585
(D)
626
(E)
761

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Reaching F₂₀ by drawing all twenty figures is hopeless, so the move is to see how the count climbs from one figure to the next. The figure and the building rule tell you exactly how big each new ring is. Once that ring size follows a clean pattern, add the rings up into a single formula and evaluate it at n = 20 instead of drawing anything.

1STEP 1

Read the counts from the figure

Each figure is the one before it with a fresh square ring wrapped outside, so the counts read F₁ = 1, F₂ = 5, F₃ = 13.

F₁ = 1, F₂ = 5, F₃ = 13
2STEP 2

Size of each new ring

Fₙ's new frame has n diamonds per side, so it adds 4n - 4 — 4n spots less the 4 shared corners; check F₃: 5 + 8 = 13.

ring(n) = 4n - 4; F₃ = 5 + (4 · 3 - 4) = 5 + 8 = 13
3STEP 3

Add up all the rings into a formula

Stacking every ring onto F₁ = 1 gives 1 + 4(1 + 2 + … + (n-1)), and that triangular sum folds up into Fₙ = 2n(n-1) + 1.

F_n = 1 + 4(1 + 2 + … + (n-1)) = 1 + 4·((n-1)n)/2 = 2n(n-1) + 1
4STEP 4

Evaluate at n = 20

Substituting n = 20 gives 2 · 20 · 19 + 1 = 760 + 1 = 761 diamonds, choice (E).

F₂₀ = 2 · 20 · 19 + 1 = 760 + 1 = 761
Answer
761
The count grows like 2n squared, so near n = 20 it should be a bit under 2 times 400 = 800; 761 sits right there, while 401 and 485 are far too small for a figure this deep and 626 does not match the 2n(n-1)+1 pattern that the earlier figures obey. The formula reproduces every known value: n = 1 gives 1, n = 2 gives 5, n = 3 gives 13, so trusting it at n = 20 giving 761 is sound.
💡Key takeaway

When a shape grows by a bigger border each step, add the borders up into one formula instead of drawing every stage.

  • Read the counts from the figure
  • Size of each new ring
  • Add up all the rings into a formula
  • Evaluate at n = 20

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