AMC 10 · 2025 · #12

Grade 8 geometry-2d

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

equilateral-trianglearea-circlessimilar-figuresarea-regular-hexagon identify-subproblems ↑ Prerequisites: area-triangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Three shapes each hold some equal-sized circles that touch each other snugly: an equilateral triangle holds 3 circles, a rhombus with a 60° angle holds 2 circles, and a regular hexagon holds 6 circles. For each shape, form the ratio (total area of its circles) ÷ (area of the shape); call these T, R, and H. Decide how T, R, and H compare in size.

Pick an answer.

(A)

T=H=R

(B)

H<R=T

(C)

H=R<T

(D)

H<R<T

(E)

H<T<R

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Comparing three different shapes at once is overwhelming, so Tool #7 (Identify Subproblems) cuts each shape into equilateral triangles that each hold exactly one circle. In the rhombus and hexagon that circle is the triangle's incircle, so both reduce to copies of a single base unit: one equilateral triangle with its inscribed circle. Tool #9 (Solve an Easier Related Problem) computes that one base ratio once and reuses it for both R and H. Tool #1 (Draw a Diagram) supplies the cutting lines and the corner geometry. Because the triangle's three disks pack differently (each hugs a corner, not the incircle), its ratio must be found separately, and Tool #3 (Eliminate Possibilities) confirms only one answer choice fits the comparison.

1STEP 1

Cut the rhombus and hexagon into triangles

The 60° rhombus splits into 2 equilateral triangles, the hexagon into 6 — and each disk is exactly its own triangle's incircle.

rhombus = 2 △, hexagon = 6 △, each disk = incircle of its △

2STEP 2

Same unit means R equals H

Both are copies of one tile — triangle plus incircle — and repeating a tile scales top and bottom alike, so H=R.

R=2 A_incircle/2 A_△=A_incircle/A_△, H=6 A_incircle/6 A_△=A_incircle/A_△ → H=R

3STEP 3

Measure the base unit's triangle and incircle

For side s the area is √3/4 s², and since the center is the centroid the incircle radius is a third of the height: r=s/2√3.

A_△=√3/4s², r=s/2√3

4STEP 4

The base ratio (this is H = R)

Divide the incircle area π s²/12 by √3/4 s²; the s² cancels and H=R is fixed at √3π/9≈0.605.

H=R=(π s²/12)/√3 s²/4=π/3√3=√3 π/9≈0.605

5STEP 5

Find the disk radius inside the triangle T

A corner disk of radius ρ sits 2ρ from its 60° vertex, and neighbors touching forces s=2ρ(1+√3), so ρ=s/(2(1+√3)).

s=2ρ (1+√3) → ρ=s/(2(1+√3))

6STEP 6

Compute the triangle's ratio T

Divide the total disk area 3πρ² by √3/4 s² and use (1+√3)²=4+2√3: T=3π/(6+4√3)≈0.729.

T=3πρ²/√3/4s²=3π/(√3 (1+√3)²)=3π/(6+4√3)≈0.729

7STEP 7

Compare and pick the answer

Compare: H=R≈0.605 against T≈0.729, so H=R ＜ T — the only choice saying that is (C).

H=R≈0.605 ＜ T≈0.729 → H=R ＜ T=(C)

Answer

H=R ＜ T

Look at the pictures: the triangle looks the fullest of circle, the hexagon and rhombus look equally full but a bit emptier — matching H=R ＜ T. The exact numbers confirm it: H=R=√3π/9≈0.605 and T=3π/(6+4√3)≈0.729, and 0.605 ＜ 0.729. Both stay safely below 1, as any area-of-circles-inside-a-shape ratio must. The equality H=R is exact (not a coincidence of rounding) because both shapes are literally tiled by the same triangle-plus-incircle unit.

💡Key takeaway

A 60° rhombus and a regular hexagon are both made of the same equilateral-triangle-with-one-circle tile, so their fill ratios match; the triangle packs three circles tighter, so it wins: H=R ＜ T.

Cut the rhombus and hexagon into triangles
Same unit means R equals H
Measure the base unit's triangle and incircle
The base ratio (this is H = R)
Find the disk radius inside the triangle T
Compute the triangle's ratio T
Compare and pick the answer