AMC 10 · 2011 · #25

Grade 8 geometry-2dpattern
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
recursive-sequencesequences-geometricpolygon-inequality pattern-recognition ↑ Prerequisites: sequences-geometric
📏 Medium solution 💡 3 insights
Problem
Start with triangle T₁ whose side lengths are 2011, 2012, 2013. For any triangle Tₙ = △ABC, let its incircle touch sides AB, BC, AC at D, E, F; the next triangle Tₙ₊₁ has side lengths AD, BE, CF, but only when those three lengths actually form a triangle. The sequence stops at the last triangle that exists. Find that last triangle's perimeter.

Pick an answer.

(A)
$\frac{1509}{8}$
(B)
$\frac{1509}{32}$
(C)
$\frac{1509}{64}$
(D)
$\frac{1509}{128}$
(E)
$\frac{1509}{256}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Tangents from a vertex to the incircle are equal, so with semiperimeter ss the next triangle's sides are s−as-a, s−bs-b, s−cs-c.

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

See the shape stays the same and halves

Sides m−1m-1, mm, m+1m+1 become m2−1\frac{m}{2}-1, m2\frac{m}{2}, m2+1\frac{m}{2}+1 — still 1 apart, with the middle and the perimeter both exactly halved.

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

Write the middle side after n steps

Halving from m1=2012m_1=2012 gives mn=20122n−1m_n=\frac{2012}{2^{n-1}}, so every triangle's perimeter is Pn=60362n−1P_n=\frac{6036}{2^{n-1}}.

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

Find when the triangle stops existing

The two short sides must beat the longest: (m−1)+m>m+1(m-1)+m \gt m+1, so the triangle survives exactly while m>2m \gt 2.

(m-1)+m ＞ m+1 ⟹ m ＞ 2
5STEP 5

Solve for the last surviving triangle

mn>2m_n \gt 2 means 2n−1<10062^{n-1} \lt 1006, so n=10n=10 is the last triangle and P10=1509128P_{10}=\frac{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

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