Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #12

Grade 8 geometry-3d

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

similar-trianglesratio-proportionvolume-cylinder identify-subproblemseasier-related-problem ↑ Prerequisites: similar-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two upside-down (apex-down) right cones hold the SAME volume of liquid. The liquid surface in the narrow cone is a circle of radius 3 cm; in the wide cone it is a circle of radius 6 cm. Drop one identical marble (radius 1 cm) into each cone — both fully sink. The liquid level rises by some amount Δ_n in the narrow cone and Δ_w in the wide cone. Find the ratio Δ_n : Δ_w.

Pick an answer.

(A)

1:1

(B)

47:43

(C)

2:1

(D)

40:13

(E)

4:1

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Easier Related Problem

Tool #9 (Easier Problem) does the heavy lifting: imagine the cones are SO TALL that near the liquid surface they look like cylinders of radius 3 and 6. Then equal added volume V_m raises the narrow cylinder by V_m / (π · 3²) and the wide cylinder by V_m / (π · 6²) — ratio 36/9 = 4. Tool #1 (Draw) makes the cone-vs-cylinder picture explicit. Tool #7 (Subproblems) cleans up the rigorous version: same initial volume implies the narrow cone's height is 4 × the wide cone's height, and that same factor of 4 then reappears in the rise ratio. Tool #3 (Eliminate) confirms (E) against the five choices.

1STEP 1

Set up the two cones

Similar triangles: each cone's surface radius stays proportional to height, so r = $\frac{3}{h_n}$ · h (narrow) and r = $\frac{6}{h_w}$ · h (wide).

Narrow: r = 3/h_n · h. Wide: r = 6/h_w · h.

2STEP 2

Equate the two volumes

Equal initial volumes: $\frac{1}{3}$ π(3)² h_n = $\frac{1}{3}$ π(6)² h_w gives 9 h_n = 36 h_w, so h_n = 4 h_w — the narrow column is four times taller.

1/3π (3)² h_n = 1/3π (6)² h_w → h_n = 4 h_w

3STEP 3

Try the cylinder version first

Easier problem: treat each cone near its surface as a cylinder of radius 3 and 6; equal marble volume raises them in ratio $\frac{36}{9}$ = 4.

Δ_n^cyl/Δ_w^cyl = (V_m / (π · 9))/(V_m / (π · 36)) = 36/9 = 4

4STEP 4

Make the cone argument exact

Make it exact: each cone's volume in one height variable gives H_n³ - h_n³ = $\frac{V_m h_n²}{3π}$ , and likewise H_w³ - h_w³ = $\frac{V_m h_w²}{12π}$ .

H_n³ - h_n³ = (V_m h_n²)/3π, H_w³ - h_w³ = (V_m h_w²)/12π

5STEP 5

Read off the rise ratio

With h_n = 4 h_w both identities share one scale factor, so $\frac{H}{h}$ matches for both cones and $\frac{Δ_n}{Δ_w}$ = $\frac{h_n}{h_w}$ = 4.

Δ_n/Δ_w = (H_n - h_n)/(H_w - h_w) = h_n/h_w = 4

6STEP 6

Match against the choices

Match the ratio to the choices: 4 : 1 is (E); the others (1:1, 47:43, 2:1, 40:13) fail the equal-volume condition.

Δ_n : Δ_w = 4 : 1 → (E)

Answer

4:1

Two ways to feel that 4:1 is right. (a) The cylinder approximation (Tool #9) already gave exactly 4:1 — and the marble adds the same volume to each cone, so wider rim = smaller rise. The wide rim is twice the narrow rim, and area scales as radius squared (

\frac{6²}{3²}

= 4), so the wide cone needs 4 × less rise. (b) The narrow cone is four times TALLER for the same volume, and the rise scales proportionally with the original height — so 4:1 again. Both arguments converge, and the answer does NOT depend on the marble's actual volume — it could be any small object and the ratio stays 4:1.

💡Key takeaway

This AMC 10 problem only needs Grade 8 volume formulas and proportional reasoning you already know! Same liquid volume + a wider top means the narrow cone is 4 × taller. Drop the SAME marble into both — the wider rim spreads the rise out over 4 × the cross-section area, so the narrow cone's level jumps up 4 × as much. Ratio 4:1, answer (E).

Set up the two cones
Equate the two volumes
Try the cylinder version first
Make the cone argument exact
Read off the rise ratio
Match against the choices

A parent dashboard for the family lives at sensimlab.com.