AMC 10 · 2024 · #19

Grade 11 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

rotation-isometrysine-area-formulaequilateral-triangletrigonometric-ratios identify-subproblemsconvert-to-algebrasymmetry-argument ↑ Prerequisites: equilateral-trianglearea-trianglestrigonometric-ratios

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An equilateral triangle ABC with side length 14 is spun about its own center by an angle θ, where 0 ＜ θ ＜ 60^°, landing on a second equilateral triangle DEF. Taking the six vertices in the order A, D, B, E, C, F gives a hexagon, and that hexagon has area 91√(3). Find tanθ.

Pick an answer.

(A)

$\frac{3}{4}$

(B)

$\frac{5\sqrt{3}}{11}$

(C)

$\frac{4}{5}$

(D)

$\frac{11}{13}$

(E)

$\frac{7\sqrt{3}}{13}$

How to solve

Strategy Identify Subproblems

The hexagon has no formula of its own, so Tool #7 (Identify Subproblems) does the real work: join the center O to all six vertices and the hexagon falls apart into six triangles that each have two sides of the same known length. Tool #17 (Visualize Spatial Relationships) is what makes that split legal — seeing that a rotation about the center slides all six vertices along one circle, so every one of those six sides is the circumradius R. Tool #4 (Introduce a Variable) names the six central angles in terms of θ, and Tool #13 (Convert to Algebra) turns each wedge into 1/2R²sin(angle) so the given area becomes one equation. Tool #15 (Organize Information in More Ways) finishes it: rewriting sinθ+sin(120^°-θ) as a single sine of a shifted angle collapses two unknown quantities into one.

1STEP 1

Put all six vertices on one circle

Both triangles share one circumcircle.

R=14/√(3)=14√(3)/3, R²=196/3

2STEP 2

Fan the hexagon out from the center

The hexagon splits into six wedges.

[ADBECF]=[OAD]+[ODB]+[OBE]+[OEC]+[OCF]+[OFA]

3STEP 3

Measure the six central angles

They alternate theta and 120 degrees minus theta.

∠ AOD=∠ BOE=∠ COF=θ, ∠ DOB=∠ EOC=∠ FOA=120^°-θ

4STEP 4

Turn each wedge into a sine

The area rides on a sum of two sines.

[ADBECF]=3/2R²(sinθ+sin(120^°-θ))=98(sinθ+sin(120^°-θ))

5STEP 5

Use the given area

The given area pins the sum of sines.

sinθ+sin(120^°-θ)=91√(3)/98=13√(3)/14

6STEP 6

Fold two sines into one

They fold into sine of theta plus thirty equal to thirteen fourteenths.

√(3)sin(θ+30^°)=13√(3)/14 ⟹ sin(θ+30^°)=13/14

7STEP 7

Recover the cosine, then the tangent

Pythagoras recovers the cosine.

cos(θ+30^°)=3√(3)/14, tan(θ+30^°)=13/3√(3)=13√(3)/9

8STEP 8

Undo the 30 degree shift

Undoing the shift gives five root three over eleven.

tanθ=(t-tan 30^°)/(1+ttan 30^°)=10√(3)/9/22/9=10√(3)/22=5√(3)/11 → (B)

Answer

5√(3)/11

Three independent checks all agree. First, size: as θ→ 0^° the two triangles merge and the hexagon shrinks to one triangle of area √(3)/4 · 14²=49√(3), while at θ=60^° the six points are evenly spaced and the hexagon is regular with area 6·1/2R²sin 60^°=98√(3). The given 91√(3) sits between them, and since √(3)sin(θ+30^°) is strictly increasing on 0 ＜ θ ＜ 60^° there is exactly one θ that works. Second, the implied sine and cosine of θ itself are sinθ=5√(3)/14 and cosθ=11/14, and these pass the Pythagorean test: 75/196+121/196=196/196=1. Their ratio is 5√(3)/11, matching (B). Third, numerically θ=arcsin13/14-30^°≈ 38.21^°, so 98(sin 38.21^°+sin 81.79^°)≈ 98(0.6186+0.9897)≈ 157.62, and 91√(3)≈ 157.62. Substituting the other four choices for tanθ produces hexagon areas that miss 91√(3) by about 1% to 3%, so (B) is the only fit.

💡Key takeaway

Spinning the triangle about its center keeps all six corners on one circle, so slice the hexagon from the center into six wedges — each wedge is 1/2R²sin(its angle), and the whole problem shrinks to one sine equation.

Put all six vertices on one circle
Fan the hexagon out from the center
Measure the six central angles
Turn each wedge into a sine
Use the given area
Fold two sines into one
Recover the cosine, then the tangent
Undo the 30^° shift