AMC 10 · 2009 · #22

Grade 8 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Sum of Primitive Solids

area-trianglesvolume-rectangular-prismsurface-areacoordinate-geometry identify-subproblems ↑ Prerequisites: area-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A cube of cake with edge 2 inches is iced on its four sides and top, but not the bottom. Seen from straight above the cake is a 2-by-2 square, and M is the midpoint of the left edge. Two vertical cuts split the cake into three pieces: one runs from the top-right corner of the square to M, and the other starts at the bottom-right corner and meets the first cut at a right angle. Triangle B, the top of the piece that keeps the whole right edge, holds c cubic inches of cake and s square inches of icing. What is c + s?

Pick an answer.

(A)

$\frac{24}{5}$

(B)

$\frac{32}{5}$

(C)

$8+\sqrt5$

(D)

$5+\frac{16\sqrt5}{5}$

(E)

$10+5\sqrt5$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The single question c + s hides several smaller jobs: find the area of triangle B, turn that into the volume of a prism, then add up only the iced faces. Break it into those subproblems. The area of B needs the two legs of a right triangle, which a labeled diagram plus the Pythagorean theorem deliver. Visualizing the 3D piece tells us which faces are iced.

1STEP 1

Read the top view

Name the right corners Q1, Q2 and P where the cuts meet. Triangle B is Q1-Q2-P, right-angled at P, hypotenuse Q1-Q2 = 2.

△ B = △ Q₁ Q₂ P, ∠ P = 90°, Q₁Q₂ = 2

2STEP 2

Area of the helper triangle

First take triangle Q1-Q2-M: base Q1-Q2 = 2 with M sitting 2 to the left, so its area is 2.

[△ Q₁ Q₂ M] = 1/2 · 2 · 2 = 2

3STEP 3

Find the perpendicular leg

Q1-M = √5 by Pythagoras. Reading that same area 2 with Q1-M as base gives the perpendicular cut Q2P = 4√5/5.

Q₁M = √(2² + 1²) = √(5); 1/2 · √(5) · Q₂P = 2 → Q₂P = 4/√5 = 4√5/5

4STEP 4

Area of triangle B

Pythagoras on B gives the other leg Q1P = 2√5/5, so its area is 1/2 · 2√5/5 · 4√5/5 = 4/5.

Q₁P = √(2² - (4/√5)²) = √(4 - 16/5) = √(4/5) = 2/√5; [△ B] = 1/2 · 2/√5 · 4/√5 = 4/5

5STEP 5

Volume of the piece

The vertical cuts make the piece a prism: triangle B on top, height 2, so c = 4/5 × 2 = 8/5.

c = [△ B] · 2 = 4/5 · 2 = 8/5

6STEP 6

Add the icing and finish

Only the top (4/5) and the outer wall Q1-Q2 (2 × 2 = 4) are iced, so s = 24/5 and c + s = 32/5, choice (B).

s = 4/5 + (2 · 2) = 4/5 + 4 = 24/5; c + s = 8/5 + 24/5 = 32/5

Answer

32/5

The answer 32/5 = 6.4 is modest, which fits a small wedge of a cube whose total volume is only 8 and whose iced surface is 20. The volume 8/5 = 1.6 is well under the whole 8, and the icing 24/5 = 4.8 is a small slice of the 20 square inches of total icing. The messy square roots inside the calculation cancel, leaving a clean fraction, which matches an answer choice exactly.

💡Key takeaway

Slice the hard question into pieces: find the little triangle's area, stack it into a prism for volume, then add icing only on the crust you didn't cut through.

Read the top view
Area of the helper triangle
Find the perpendicular leg
Area of triangle B
Volume of the piece
Add the icing and finish