AMC 10 · 2017 · #22

Grade 8 geometry-2d

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

equilateral-triangletangent-circlescircular-sectorarea-triangles area-differencethirty-sixty-ninety-triangle ↑ Prerequisites: equilateral-triangle

📏 Long solution 💡 3 insights

Problem

An equilateral triangle ABC has its two sides AB and AC just touching a circle, meeting it at the corner points B and C. Find what fraction of the triangle's area sits outside the circle.

Pick an answer.

(A)

$\frac{4\sqrt{3}\pi}{27}-\frac{1}{3}$

(B)

$\frac{\sqrt{3}}{2}-\frac{\pi}{8}$

(C)

$\frac{1}{2}$

(D)

$\sqrt{3}-\frac{2\sqrt{3}\pi}{9}$

(E)

$\frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shape that lies inside the triangle but outside the circle is curved and awkward to measure directly, so Tool #7 (Identify Subproblems) splits the job into pieces we already know how to handle: the whole triangle, a circular sector, and a plain triangle inside that sector. Tool #1 (Draw a Diagram) and the tangent-equals-perpendicular fact pin down the circle's center and the key 120° central angle. Tool #4 (Introduce a Variable) fixes a convenient side length so every area becomes a number. Finally Tool #16 (Change Focus / Count the Complement) is the cleanest finish: instead of chasing the curved region, find the area inside both shapes, then subtract from the triangle.

1STEP 1

Set up the picture

Set side = 1; since AB, AC are tangent, OB ⊥ AB and OC ⊥ AC, and symmetry puts O on the bisector of angle A, so OB = OC = r.

AB = AC = BC = 1, OB ⊥ AB, OC ⊥ AC

2STEP 2

Find the radius

In right triangle ABO, ∠BAO = 30°, so the 30-60-90 ratio gives r = 1/√(3) and r² = 1/3.

∠ BAO = 30°, r = OB = AB/√(3) = 1/√(3), r² = 1/3

3STEP 3

Get the central angle BOC

In quadrilateral ABOC the angles are 90° + 90° + 60°, and since a quadrilateral totals 360°, ∠BOC = 120°.

∠ BOC = 360° - 90° - 90° - 60° = 120°

4STEP 4

Area shared by circle and triangle

The shared piece is the segment cut by chord BC: the sector (one-third of the circle) minus triangle BOC = π/9 - √3/12.

segment = π/9 - √(3)/12

5STEP 5

Subtract to get the outside area

Outside area = whole triangle √3/4 minus the shared segment = √3/3 - π/9.

√3/4 - (π/9 - √3/12) = √3/3 - π/9

6STEP 6

Form the fraction and simplify

Divide the outside area by √3/4: (√3/3 - π/9)/(√3/4) = 4/3 - 4√3 π/27, which is choice (E).

(√3/3 - π/9)/√3/4 = 4/3 - 4√(3) π/27 → (E)

Answer

4/3-4√(3)π/27

A numeric check: 4/3 ≈ 1.333 and 4√3 π/27 ≈ 4(1.732)(3.1416)/27 ≈ 0.806, so the fraction is about 0.527. That is just over half, which fits the picture: the circle bulges only a little way up from the base BC, leaving most of the tall triangle uncovered. A fraction between 0 and 1 that is a bit above 1/2 is exactly what we should expect.

💡Key takeaway

Split the curved region into a pie wedge and a triangle, find what the circle and triangle share, then subtract that from the whole triangle.

Set up the picture
Find the radius
Get the central angle BOC
Area shared by circle and triangle
Subtract to get the outside area
Form the fraction and simplify