AMC 10 · 2024 · #14

Grade 8 geometry-2d

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

area-trianglesarea-circlesangle-sum-triangleisosceles-triangle area-differenceidentify-subproblems ↑ Prerequisites: area-trianglesarea-circlespythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

An equilateral triangle of height 24 has one side on line ℓ. A circle of radius 12 sits on the same side of ℓ as the triangle, tangent to ℓ and externally tangent to one slanted side of the triangle. The region squeezed between the triangle, the circle, and the line near the vertex where they meet has area a√(b) - cπ with b squarefree. Find a + b + c.

Pick an answer.

(A)

~72

(B)

~73

(C)

~74

(D)

~75

(E)

~76

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem describes the picture in words, so the first move is Tool #1 (Draw a Diagram): place line ℓ horizontally, drop the triangle on top, and put the circle in the wedge outside the triangle at the vertex C where one slanted side meets ℓ. The picture immediately shows that the target region is the wedge between two tangent segments and an arc. That target naturally splits into two clean subproblems (Tool #7): the kite-shaped quadrilateral formed by the vertex, the two tangent points, and the circle's center, minus the circular sector cut out of that quadrilateral. Each subproblem is a one-formula calculation, and subtracting gives the requested a√(b) - cπ.

1STEP 1

Sketch it: with tangent points E on ℓ, D on side CA, and center O, the target pocket at vertex C is quadrilateral ODCE minus sector ODE.

Target area = Area(ODCE) - Area(sector ODE)

2STEP 2

The interior angle at C is 60°, so the exterior wedge holding the circle is 180° - 60° = 120°; O lies on its bisector, giving ∠OCE = 60°.

∠ DCE = 180° - 60° = 120°, ∠ OCE =

\frac{1}{2}

(120°) = 60°

3STEP 3

The tangent radii give right angles, so △OEC is a 30-60-90 with OE = 12; then CE = 4√(3) and the kite ODCE has area 48√(3).

CE = 12/√(3) = 4√(3), Area(ODCE) = 2 ·

\frac{1}{2}

· CE · OE = 2 ·

\frac{1}{2}

· 4√(3) · 12 = 48√(3)

4STEP 4

The four angles of ODCE sum to 360°: two 90° right angles plus the 120° wedge leave central angle ∠DOE = 60°.

∠ DOE = 360° - 90° - 90° - 120° = 60°

5STEP 5

A sector of central angle θ° and radius r has area θ/360·π r²; with θ = 60 and r = 12 that is one sixth of the disk = 24π.

Area(sector ODE) =

\frac{60}{360}

π (12)² =

\frac{1}{6}

· 144π = 24π

6STEP 6

Subtract: target = 48√(3) - 24π, so a = 48, b = 3, c = 24 (b squarefree), and a + b + c = 75 → (D).

Target = 48√(3) - 24π → a + b + c = 48 + 3 + 24 = 75 → (D)

Answer

~75

Sanity check the size of the wedge: 48√(3) ≈ 48(1.732) ≈ 83.1, and 24π ≈ 75.4, so the target region has area roughly 7.7. That is a small sliver — exactly what you would expect for the thin pocket between a circle of radius 12 and a 60° vertex of a triangle, so the order of magnitude is right. Each tangent length CE = CD = 4√(3) ≈ 6.93, comfortably less than the radius 12, which matches the picture where the tangent points sit closer to C than O does. The triangle's height 24 never entered the calculation, consistent with the local nature of the region.

💡Key takeaway

Draw the picture, spot the kite plus pie-slice at the vertex, and the AMC 10 problem reduces to one 30-60-90 triangle and one $\frac{1}{6}$ -of-a-circle sector — Grade 7-8 geometry the whole way.

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