AMC 10 · 2019 · #25

Grade 10 geometry-2d
cyclic-quadrilateralinscribed-angleangle-sum-trianglerecursive-sequenceexponents pattern-recognitioneasier-related-problemextreme-principle ↑ Prerequisites: cyclic-quadrilateralangle-sum-triangle
📏 Long solution 💡 4 insights
Problem
Start with a triangle whose angles are 59.999, 60, and 60.001 degrees. Drop all three altitudes and keep only their feet — those three feet form the next triangle. Repeat forever. Find the smallest step at which the triangle has an angle bigger than 90 degrees.

Pick an answer.

(A)
10
(B)
11
(C)
13
(D)
14
(E)
15
How to solve
Strategy Look for a Pattern

Tracking actual points for 15 rounds is hopeless, but the question only asks about angles. Tool #4 (Introduce a Variable): name the angles x_n, y_n, z_n so each round becomes a number update instead of a construction. Tool #1 (Draw a Diagram): the two right angles at an altitude foot put four points on one circle, which is what converts the picture into an angle equation. Tool #5 (Look for a Pattern): that equation, x_n = 180° - 2x_n-1, is the same every round. Tool #9 (Easier Related Problem): instead of the angle itself, follow only its gap from 60° — the gap obeys the much simpler rule "multiply by -2". Tool #14 (Extreme Principle): the triangle turns obtuse the first moment the largest gap crosses 30°, so the whole problem collapses to one inequality about powers of 2.

1STEP 1

See what one round builds

Look at the new triangle one round builds.

A₁A₀ ⊥ B₀C₀, B₁B₀ ⊥ A₀C₀, C₁C₀ ⊥ A₀B₀
2STEP 2

Name the angles, not the points

Name the angles, not the points.

x_n + y_n + z_n = 180°; △ A_nB_nC_n obtuse ⇔ max(x_n, y_n, z_n) > 90°
3STEP 3

Two right angles make a circle

Two right angles put four points on one circle.

∠ A₀A₁B₀ = ∠ A₀B₁B₀ = 90° → A₀, B₀, A₁, B₁ concyclic → ∠ A₀A₁B₁ = ∠ A₀B₀B₁ = 90° - x₀
4STEP 4

One rule, repeated every round

The same rule repeats every round.

x_n = 180° - 2x_n-1, y_n = 180° - 2y_n-1, z_n = 180° - 2z_n-1
5STEP 5

Follow the gap from 60 degrees

The gap from sixty doubles each round.

d_n = x_n - 60° = -2 d_n-1 ⟹ d_n = (-2)ⁿ d₀, so x_n = 60° + (-2)ⁿ (x₀ - 60°)
6STEP 6

Plug in the three starting gaps

Plug in the three starting gaps.

{ 60°, 60° - 0.001 · 2ⁿ, 60° + 0.001 · 2ⁿ }
7STEP 7

Confirm the rule stayed legal

Confirm the rule stayed legal.

n = 14: 0.001 · 2¹⁴ = 16.384 → (43.616°, 60°, 76.384°) — acute
8STEP 8

Find the first round past 90

The first round past ninety degrees is 15.

0.001 · 2ⁿ > 30 ⇔ 2ⁿ > 30000; 2¹⁴ = 16384, 2¹⁵ = 32768 → n = 15
Answer
15
Check the very first round directly: x₁ = 180 - 2(59.999) = 60.002, 180 - 2(60) = 60, 180 - 2(60.001) = 59.998. The gaps went from (-0.001, 0, +0.001) to (+0.002, 0, -0.002) — doubled and flipped sign, exactly as the formula d_n = (-2)ⁿ d₀ predicts, and they still add to 180°. Now sanity-check the answer against the choices. At n = 13 the gap is 0.001 × 8192 = 8.192, largest angle 68.192°; at n = 14 it is 16.384, largest angle 76.384°; at n = 15 it is 32.768, largest angle 92.768°. So choices (A) 10, (B) 11, (C) 13 and (D) 14 all still give acute triangles, and 15 is the first failure. Every angle stays positive throughout (27.232° at n = 15), so no triangle degenerates before then, and the sum is 27.232 + 60 + 92.768 = 180 ✓.
💡Key takeaway

Each round replaces every angle x by 180° - 2x, which leaves 60° alone and doubles the distance from 60° while flipping its sign — so a gap of 0.001° needs 0.001 × 2ⁿ > 30, and the first time that happens is n = 15.

  • See what one round builds
  • Name the angles, not the points
  • Two right angles make a circle
  • One rule, repeated every round
  • Follow the gap from 60 degrees
  • Plug in the three starting gaps
  • Confirm the rule stayed legal
  • Find the first round past 90