AMC 10 · 2018 · #10

Grade 8 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid
volume-rectangular-prismcoordinate-geometryspatial-visualization identify-subproblemsarea-difference ↑ Prerequisites: volume-rectangular-prism
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A box (rectangular parallelepiped) has bottom face ABCD and top face EFGH, with E,F,G,H directly above A,B,C,D. The edges are AB=3, BC=1, and the vertical edge CG=2. Point M is the midpoint of the top edge FG. Find the volume of the pyramid whose base is the slanted rectangle BCHE and whose apex is M.

Pick an answer.

(A)
1
(B)
$frac{4}{3}$
(C)
$frac{3}{2}$
(D)
$frac{5}{3}$
(E)
2

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships): the base BCHE is a slanted plane, so its area and the apex height are awkward to read directly. Placing the box on xyz-coordinates turns every edge into a clean number and lets us see how the tilted base sits inside the box. Tool #1 (Draw a Diagram): assigning coordinates to all eight corners makes M and the base concrete. Tool #7 (Identify Subproblems): instead of fighting the slanted pyramid head-on, we notice the base BCHE slices the box exactly in half, giving a triangular prism, and the pyramid is that prism minus two small corner tetrahedra — each piece is easy.

1STEP 1

Put the box on coordinates

Put A at the origin; the edge lengths turn every corner into clean coordinates, and the whole box has volume 6.

V_box = 3 · 1 · 2 = 6
2STEP 2

The base slices the box in half

Through the box's center, the base splits it into two equal halves; M's half is a triangular prism of volume 3 (ends BEF, CHG).

V_prism = 1/2 V_box = 3
3STEP 3

Area of a triangular end

The triangular end BEF is a right triangle with legs 3 (across) and 2 (down), so its area is 3.

[BEF] = 1/2· 3 · 2 = 3
4STEP 4

Cut off two corner tetrahedra

The prism equals the pyramid plus two corner tetrahedra; each M-BEF has base 3 and height 1/2, so its volume is 1/2.

V_M-BEF = 1/3· 3 · 1/2 = 1/2
5STEP 5

Subtract to get the pyramid

The wanted pyramid is the prism with both corner tetrahedra removed: 3 minus 1 leaves volume 2.

V_M-BCHE = 3 - 2·1/2 = 3 - 1 = 2 → (E)
Answer
2
The pyramid fills 2 of the prism's 3 units, i.e. two-thirds of the half-box, which is reasonable for a pyramid that spans the full slanted face and reaches the opposite edge. A direct check confirms it: the base BCHE has sides 1 (edge BC) and √(13) (edge BE, since √(3²+2²)=√(13)), so its area is √(13); the perpendicular distance from M to that plane works out to 6/√(13), giving 1/3·√(13)·6/√(13)=2 — the √(13) cancels and the answer is exactly 2=(E). The smaller choices (A) 1 through (D) 5/3 would each need the pyramid to fill less than two-thirds of the half-box, which the decomposition rules out.
💡Key takeaway

The slanted base cuts the box exactly in half into a prism of volume 3; shave off the two corner tetrahedra (1/2 each) and the pyramid that's left has volume 2.

  • Put the box on coordinates
  • The base slices the box in half
  • Area of a triangular end
  • Cut off two corner tetrahedra
  • Subtract to get the pyramid

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