AMC 10 · 2011 · #22

Grade 8 geometry-2dpattern
recursive-sequencesequences-geometricpolygon-inequality pattern-recognition ↑ Prerequisites: sequences-geometric
📏 Medium solution 💡 3 insights
Problem
Each triangle's incircle touch points become the next triangle's sides, until none exists. Find the last perimeter.

Pick an answer.

(A)
$\frac{1509}{8}$
(B)
$\frac{1509}{32}$
(C)
$\frac{1509}{64}$
(D)
$\frac{1509}{128}$
(E)
$\frac{1509}{256}$
How to solve
Strategy Look for a Pattern

Listing all the triangles by hand would be slow, so Tool #5 (Look for a Pattern) is the anchor: work out how one triangle turns into the next and find the rule that repeats. Tool #4 (Introduce a Variable) makes that rule clean by naming the middle side length, which exposes that the perimeter simply halves each step and the sides keep the same shape. Then the only question is when the shrinking sides get too lopsided to form a triangle, and Tool #14 (Extreme Principle) pins down that last surviving step by pushing the triangle inequality to its breaking point.

1STEP 1

Turn tangency points into new side lengths

The touch points give the next sides directly.

AD = s-a, BE = s-b, CF = s-c
2STEP 2

See the shape stays the same and halves

The three sides stay consecutive and simply halve.

m-1, m, m+1 ⟶ m/2-1, m/2, m/2+1
3STEP 3

Write the middle side after n steps

So the perimeter has a closed formula.

m_n = 2012/(2^ n-1), P_n = 6036/(2^ n-1)
4STEP 4

Find when the triangle stops existing

It stops when the middle side drops past 2.

(m-1)+m > m+1 ⟹ m > 2
5STEP 5

Solve for the last surviving triangle

The last one has perimeter 1509/128, choice (D).

2^ n-1 < 1006 → n = 10; P₁₀ = 6036/512 = 1509/128
Answer
1509/128
The perimeters halve cleanly: 6036, 3018, 1509, … down to P₁₀ = 1509/128 ≈ 11.79. The very next value P₁₁ ≈ 5.89 would give sides near 0.96, 1.96, 2.96, where 0.96 + 1.96 = 2.92 < 2.96 — the two short sides can't reach across, so no triangle, confirming the stop at T₁₀. Every answer choice has numerator 1509 = 3 × 503 and differs only by the power of two, so the whole problem reduces to finding the right exponent; 128 = 2⁷ matches (D).
💡Key takeaway

Each new triangle's perimeter is exactly half the last one's, so keep halving until the three sides get too lopsided to close up — the last one that still closes is the answer.

  • Turn tangency points into new side lengths
  • See the shape stays the same and halves
  • Write the middle side after n steps
  • Find when the triangle stops existing
  • Solve for the last surviving triangle