AMC 10 · 2014 · #12

Grade 11 geometry-2d
arc-measurechord-perpendicular-from-centerlaw-of-cosinesarea-circles convert-to-algebraidentify-subproblems ↑ Prerequisites: arc-measurearea-circles
📏 Long solution 💡 3 insights
Problem
Two circles cross at the same two points, seeing the shared chord at different angles. Compare their areas.

Pick an answer.

(A)
2
(B)
$1+\sqrt3$
(C)
3
(D)
$2+\sqrt3$
(E)
4
How to solve
Strategy Introduce a Variable

Nothing in the problem has a size, so Tool #4 (Introduce a Variable) supplies the one length both circles must agree on: the common chord c=AB. Every other quantity gets written in terms of c, and c cancels at the end. Tool #1 (Draw a Diagram) makes the two centers, the two radii and the chord visible, which is what reveals that both centers sit on the perpendicular bisector of AB. Tool #7 (Identify Subproblems) splits the job into one identical subproblem per circle: given a chord and its central angle, find the radius. Tool #3 (Eliminate Possibilities) settles the part the question actually hangs on — the problem asks for larger over smaller, and there are two ways to hand out the arcs, so one of them has to be ruled out by proof rather than by guess.

1STEP 1

Find the one shared length

The chord is the one shared length.

AB=c, ∠ AO₁B=30°, ∠ AO₂B=60°
2STEP 2

Fold each triangle in half

Folding each triangle halves its angle.

AM=c/2, ∠ AO₁M=30°/2=15°, ∠ AO₂M=60°/2=30°
3STEP 3

Write each radius from the chord

Each radius follows from the chord and a sine.

ρ₁=c/(2sin 15°), ρ₂=c/(2sin 30°)
4STEP 4

Rule out the wrong assignment

The smaller angle belongs to the bigger circle.

sin 15° < sin 30° ⟹ ρ₁=c/(2sin 15°) > c/(2sin 30°)=ρ₂
5STEP 5

Turn areas into a radius ratio

The area ratio is the radius ratio squared.

πρ₁²/πρ₂²=(ρ₁/ρ₂)²=(( c/(2sin 15°) )/(c/(2sin 30°)))²=((sin 30°)/(sin 15°))²
6STEP 6

Get the exact value of sin 15

The angle addition formula gives an exact value.

sin 15°=sin 45°cos 30°-cos 45°sin 30°=√2/2·√3/2-√2/2·1/2=(√6-√2)/4
7STEP 7

Simplify to a clean number

Simplifying gives 2+√3, choice (D).

ρ₁/ρ₂=1/2/((√6-√2)/4)=2/(√6-√2)=(2(√6+√2))/4=(√6+√2)/2, ((√6+√2)/2)²=(8+4√3)/4=2+√3 → (D)
Answer
2+√3
Put numbers on the picture and check that it can actually be built. Take the chord to be AB=2, so the half-chord is 1. Then the smaller radius is ρ₂=1/(sin 30°)=2 and the larger is ρ₁=1/(sin 15°)≈ 3.8637, giving an area ratio of about (1.9319)²≈ 3.732, and 2+√3≈ 3.732 matches. The configuration is real, not just algebra: both centers lie on the perpendicular bisector of AB, at distances 1/(tan 15°)≈ 3.732 and 1/(tan 30°)≈ 1.732 from the midpoint, so the centers are ≈ 5.464 apart if they lie on opposite sides of AB and ≈ 2.000 apart if on the same side. Both distances lie strictly between ρ₁-ρ₂≈ 1.864 and ρ₁+ρ₂≈ 5.864, so in either arrangement the two circles genuinely cross at two points, and both central angles are below 180°, so 30° and 60° really are the minor arcs. The answer is the same in both arrangements, since it depends only on the two angles. Finally 2+√3≈ 3.73 sits between choices 3 and 4 and is bigger than 2, which fits a picture in which one center is more than twice as far from the chord as the other.
💡Key takeaway

The two circles share one chord, so each radius is fixed by the angle its center views that chord at; the wider 60° view means the smaller circle, and squaring the two views gives 2+√3.

  • Find the one shared length
  • Fold each triangle in half
  • Write each radius from the chord
  • Rule out the wrong assignment
  • Turn areas into a radius ratio
  • Get the exact value of sin 15
  • Simplify to a clean number