AMC 10 · 2007 · #8

Grade 10 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
inscribed-anglearc-measurerotation-isometrygcd symmetry-argumentidentify-subproblems ↑ Prerequisites: inscribed-anglearc-measure
📏 Long solution 💡 3 insights
Problem
Chords join each of twelve equally spaced marks to the mark five steps clockwise. The chain closes into a twelve-pointed star. Find the angle at one of its points.

Pick an answer.

(A)
20
(B)
24
(C)
30
(D)
36
(E)
60
How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram): once the clock is redrawn as a bare circle with twelve equally spaced points, the tip angle is visibly an angle whose corner sits on the circle and whose two sides are chords — which is the one configuration that has a theorem attached to it. Tool #15 (Organize Information in More Ways): relabel the hours as 0 through 11 and read "five clockwise" as adding 5 modulo 12; the drawing rule becomes arithmetic, and the question of whether the path closes on all twelve marks becomes a divisibility question. Tool #7 (Identify Subproblems): the problem is really two separate jobs, and only one of them is a computation. Job one is to show all twelve tip angles are equal, so that "the angle at each vertex" is a single number at all; job two is to measure one of them. Skipping job one and measuring a single tip is the gap in the quick argument — it proves one angle is 30^°, not that every angle is. Tool #16 (Change Focus): to check the answer by a genuinely different mechanism, stop measuring angles and instead walk around the star watching how the direction of travel turns. That route uses only rotation and supplementary angles, so it does not lean on the theorem the main route leans on.

1STEP 1

Twelve equal marks means 30-degree hours

Twelve equal marks make one step 30 degrees.

360^°/12 = 30^° per hour, arc spanning m hours = 30m degrees
2STEP 2

The path closes only after all twelve

The chain closes only after visiting all twelve.

5t ≡ 0 (mod 12), gcd(5,12) = 1 ⟹ 12 ∣ t ⟹ t = 12
3STEP 3

One rotation makes all twelve angles equal

A rotation makes every tip angle equal.

ρ₃₀^° : k ↦ k+1 ⟹ {k, k+5} ↦ {k+1, k+6} ⟹ ∠_k = ∠_k+1 for all k
4STEP 4

The tip is an inscribed angle on a two-hour arc

The tip is an inscribed angle on a two-step arc, giving 30.

frown{10 11 12} = 2 × 30^° = 60^°, ∠ 12-5-10 = 1/2 × 60^° = 30^°
5STEP 5

Same answer without the inscribed angle theorem

A turning argument confirms 30 without the inscribed angle theorem, choice (C).

ρ₁₅₀^°(12 5) = 5 10 → turn = 150^°; tip = 180^° - 150^° = 30^°; 12 × 150^° = 1800^° = 5 × 360^° ⟹ (C)
Answer
30
Put the marks on a unit circle, mark k at (cos(90^°-30k), sin(90^°-30k)). Then 5 = (1/2, -√(3)/2), 12 = (0,1) and 10 = (-√(3)/2, 1/2). The ray from 5 to 12 points along (-1/2, 1+√(3)/2), at 105^°; the ray from 5 to 10 points along (-√(3)/2-1/2, 1/2+√(3)/2), at 135^°. The gap is exactly 30^°, and running the same computation at all twelve marks gives 30^° twelve times. Two structural checks agree. First, twelve tips of 30^° sum to 360^°, which is the standard total for the points of a star of this kind. Second, the general shape of the argument gives 180^° - (360^° k)/n for the tip of an n-mark star stepping by k; on the familiar five-pointed star (n=5, k=2) that returns the well-known 180^° - 144^° = 36^°, so the formula is calibrated against something already known, and here it gives 180^° - 150^° = 30^°. The size also passes an eyeball test: a five-hour chord spans a 150^° arc, nearly a diameter, so the two chords at a tip both run almost straight across the circle and must meet in a narrow spike — 30^° fits, 60^° would be far too blunt. The wrong choices are readable. (E) 60 is the arc itself, the classic slip of forgetting to halve. (D) 36 is the tip of the ordinary five-pointed star — right family, wrong star. (A) 20 and (B) 24 correspond to 18 and 15 equal parts of a circle, and no arc on a twelve-mark circle has either measure.
💡Key takeaway

A corner sitting on a circle sees the far arc at half size, so the star's point is half the two-hour gap it looks across — thirty degrees — and one thirty-degree turn of the clock proves every other point matches it.

  • Twelve equal marks means 30-degree hours
  • The path closes only after all twelve
  • One rotation makes all twelve angles equal
  • The tip is an inscribed angle on a two-hour arc
  • Same answer without the inscribed angle theorem

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