AMC 10 · 2009 · #11

Grade 6 patterngeometry-2d
pattern-recognitionsequences-arithmeticsystematic-enumeration easier-related-problempattern-recognition ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Each figure wraps the previous one in a new square ring one unit larger on each side. Find how many pieces the twentieth figure holds.

Pick an answer.

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

The first few counts come straight off the figures.

F₁ = 1, F₂ = 5, F₃ = 13
2STEP 2

Size of each new ring

Each new ring's size grows by a fixed step.

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

Add up all the rings into a formula

Adding all the rings gives a closed formula.

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

Evaluate at n = 20

Evaluating it gives 761, choice (C).

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