AMC 10 · 2003 · #17

Grade 8 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid
volume-spherevolume-conepercentage identify-subproblems ↑ Prerequisites: volume-spherevolume-cone
📏 Medium solution 💡 2 insights
Problem
A ball-shaped scoop of ice cream sits on top of a cone. The cone opening has the same diameter as the ball, so they share the same radius r. When the ball melts it shrinks to 75% of its frozen volume, and that melted liquid fills the cone exactly to the brim. Find how the cone's height h compares to its radius r, written as a ratio h:r.

Pick an answer.

(A)
2:1
(B)
3:1
(C)
4:1
(D)
16:3
(E)
6:1

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The problem is a chain of volumes with one shared measurement. Tool #4 (Introduce a Variable) is the anchor: name the shared radius r and the cone height h, and every quantity becomes an expression in r and h. Then Tool #7 (Identify Subproblems) breaks the work into three clean pieces — frozen sphere volume, then the 75% melted amount, then the cone volume — each a single plug-in of a given formula. Setting the melted amount equal to the cone volume gives one equation, and because both sides carry a factor of π r², that factor cancels and h falls out in terms of r. Tool #1 (Draw a Diagram) keeps the picture honest: it shows why the cone and sphere must share the same radius.

1STEP 1

Name the shared radius and height

Same diameter means one shared radius r; call the height h. Then the sphere is 4/3π r³ and the cone is 1/3π r² h.

V_sphere=4/3π r³, V_cone=1/3π r² h
2STEP 2

Melt the scoop down to 75%

Melting keeps 75%=3/4 of the frozen volume, and 3/4 cancels the 4/3 exactly, so the liquid measures π r³.

V_melted=3/4·4/3π r³=π r³
3STEP 3

Set melted volume equal to the cone

"Exactly fills" sets 1/3π r² h = π r³. Both sides carry π r², so divide it out and only h/3 = r is left.

1/3π r² h=π r³ ⟹ 1/3h=r
4STEP 4

Solve for h and read the ratio

Multiply h/3 = r by 3: h = 3r, so the cone stands three radii tall and h:r = 3:1, choice (B).

h=3r ⟹ h:r=3:1 → (B)
Answer
3:1
A tall, skinny cone makes sense here: a round scoop is a fat shape, and a cone only holds 1/3 of the box (cylinder) around it, so it needs extra height to match a ball's volume. Check the numbers directly with h=3r: cone volume =1/3π r²(3r)=π r³, which is exactly the melted volume π r³. The two match, so h:r=3:1 is right. A quick sanity bound: if h were only r (ratio 1:1) the cone would hold just 1/3π r³, far too little — so the ratio must be well above 1, and 3:1 fits while a giant 6:1 would overflow.
💡Key takeaway

When one melted volume has to exactly fill another shape, set the two volumes equal, cancel the parts they share, and the leftover tells you the missing measurement.

  • Name the shared radius and height
  • Melt the scoop down to 75%
  • Set melted volume equal to the cone
  • Solve for h and read the ratio

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