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 cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A cube of cake with edge 2 inches is iced on its four sides and top (not the bottom). Looking straight down, the top is a 2-by-2 square. Two vertical cuts split the cake into three pieces. One cut runs from a top corner to M, the midpoint of a far edge; the other runs from the neighboring corner and meets the first cut at a right angle. The piece whose top view is triangle B holds c cubic inches of cake and carries s square inches of icing. Find c + s.

Givens: The cake is a cube with edge length 2 inches.; Icing covers the four vertical sides and the top only.; Top view is a 2x2 square; one cut goes from a corner to M, the midpoint of an edge.; The second cut leaves the adjacent corner and meets the first cut at a right angle at point P.; Triangle B is the top of the target piece and has its right angle at P.

Unknowns: c = volume of the piece under triangle B (cubic inches); s = icing area on that piece (square inches); c + s

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #17 Visualize Spatial Relationships

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.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.1 Step 1

Read the top view

Name the two right-hand corners Q1 (top) and Q2 (bottom), and let M be the midpoint of the left edge.
The first cut is the segment Q1-M.
The second cut leaves Q2 and hits Q1-M perpendicularly at P.
Triangle B is Q1-Q2-P, and its right angle sits at P.
Its longest side is the whole right edge of the square, Q1-Q2, which has length 2.

$$\triangle B = \triangle Q_1 Q_2 P,\quad \angle P = 90^\circ,\quad Q_1Q_2 = 2$$

💡 Labeling the corners turns a picture into a triangle whose sides I can measure.

#7 Identify Subproblems 6.G.A.1 Step 2

Area of the helper triangle

To get triangle B's legs, first study triangle Q1-Q2-M.
Its base Q1-Q2 is the right edge, length 2, and M sits 2 units to the left of that edge, so the height is 2.
Its area is one half base times height.

$$[\triangle Q_1 Q_2 M] = \tfrac{1}{2}\cdot 2 \cdot 2 = 2$$

💡 A triangle's area is easiest when the base is a side you already know and the height is a clean horizontal distance.

#4 Introduce a Variable 8.G.B.7 Step 3

Find the perpendicular leg

The segment Q1-M is the distance from the corner to the midpoint: horizontal run 2, vertical rise 1, so by the Pythagorean theorem its length is the square root of 5.
Now read the same triangle Q1-Q2-M using Q1-M as the base.
The height to that base is exactly the perpendicular cut Q2-P.
Setting the two area formulas equal gives Q2-P.

$$Q_1M = \sqrt{2^2 + 1^2} = \sqrt{5};\quad \tfrac{1}{2}\cdot \sqrt{5}\cdot Q_2P = 2 \;\Rightarrow\; Q_2P = \frac{4}{\sqrt5} = \frac{4\sqrt5}{5}$$

💡 The same triangle has one area, so measuring it two ways pins down the unknown height.

#7 Identify Subproblems 8.G.B.7 Step 4

Area of triangle B

Triangle B is right-angled at P, with hypotenuse Q1-Q2 = 2 and one leg Q2-P known.
The other leg Q1-P comes from the Pythagorean theorem.
The area of a right triangle is half the product of its two legs.

$$Q_1P = \sqrt{2^2 - \left(\tfrac{4}{\sqrt5}\right)^2} = \sqrt{4 - \tfrac{16}{5}} = \sqrt{\tfrac{4}{5}} = \frac{2}{\sqrt5};\quad [\triangle B] = \tfrac{1}{2}\cdot \frac{2}{\sqrt5}\cdot \frac{4}{\sqrt5} = \frac{4}{5}$$

💡 Once both legs of a right triangle are known, its area is just half their product.

#17 Visualize Spatial Relationships 7.G.B.6 Step 5

Volume of the piece

Every cut is vertical, so the piece is a prism: triangle B on top, the same triangle on the bottom, height 2.
Volume is base area times height.

$$c = [\triangle B]\cdot 2 = \frac{4}{5}\cdot 2 = \frac{8}{5}$$

💡 A straight vertical cut just stacks the top shape all the way down, so volume is area times height.

#7 Identify Subproblems 7.G.B.6 Step 6

Add the icing and finish

Icing sits only on original outer surfaces.
The top face is triangle B, area 4/5.
Of the three vertical faces, only Q1-Q2 was an outside wall of the cube; it is 2 wide and 2 tall, so its icing is 4.
The two cut faces are fresh, so they have no icing.
Add the top and the side, then add c to get the answer.
So s = 4/5 + 4 = 24/5, and c + s = 8/5 + 24/5 = 32/5, which is choice (B).

$$s = \frac{4}{5} + (2\cdot 2) = \frac{4}{5} + 4 = \frac{24}{5};\quad c + s = \frac{8}{5} + \frac{24}{5} = \frac{32}{5}$$

💡 Only the crust you didn't slice through keeps its icing.

[1] #1 6.G.A.1 Name the two right-hand corners Q1 (top) and Q2 (bottom), and let M be the midpo

[2] #7 6.G.A.1 To get triangle B's legs, first study triangle Q1-Q2-M. Its base Q1-Q2 is the ri

[3] #4 8.G.B.7 The segment Q1-M is the distance from the corner to the midpoint: horizontal run

[4] #7 8.G.B.7 Triangle B is right-angled at P, with hypotenuse Q1-Q2 = 2 and one leg Q2-P know

[5] #17 7.G.B.6 Every cut is vertical, so the piece is a prism: triangle B on top, the same tria

[6] #7 7.G.B.6 Icing sits only on original outer surfaces. The top face is triangle B, area 4/5

Review

Reasonableness: 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.

Alternative: Place coordinates with the square corners at (\pm 1, \pm 1) and M at (-1,0). Drop the foot of the perpendicular from (1,-1) onto the line through (1,1) and (-1,0); it lands at P = (1/5, 3/5). The two legs measure 2/\sqrt5 and 4/\sqrt5, giving area 4/5 directly, then the same volume and icing tally reproduce 32/5.

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of triangle B and the helper triangle from base and height.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the diagonal cut length and both legs of right triangle B.)
7.G.B.6 Solve real-world problems involving area, surface area, and volume (Turning triangle B into a prism volume and summing the iced top and side areas.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

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

Review

CCSS standards used (min grade 8)

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