AMC 8 · 2018 · #24

Grade 8 geometry-3d

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

pythagorean-theoremspatial-visualizationarea-trianglesformula-substitution identify-subproblemsarea-difference ↑ Prerequisites: pythagorean-theoremspatial-visualization

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A cube ABCDEFGH has C and E as opposite vertices (a space-diagonal pair). J is the midpoint of edge FB and I is the midpoint of edge HD. Let R be the ratio of the area of the quadrilateral cross-section EJCI to the area of one face of the cube. Find R².

Pick an answer.

(A)

$\frac{5}{4}$

(B)

$\frac{4}{3}$

(C)

$\frac{3}{2}$

(D)

$\frac{25}{16}$

(E)

$\frac{9}{4}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Create a Physical Representation

Tool #10 (Physical Representation): a 3D cube cross-section is much easier to grasp by holding a cube (a tissue box works) and tracing E → J → C → I → E with a finger. Doing this reveals two key facts before any calculation: (a) all four sides of EJCI look equal (it is a rhombus), and (b) the two diagonals are EC (the cube's space diagonal) and JI (a face-diagonal-length segment through the middle). Tool #7 (Identify Subproblems) then turns the area question into three small, separate pieces: (1) show EJCI is a rhombus, (2) find the two diagonals' lengths, (3) plug into the rhombus area formula $\frac{1}{2}$ d₁ d₂ and divide by the face area. Tool #17 (Visualize Spatially) supports recognizing that J and I sit on opposite vertical edges (one on the FB pillar, one on the HD pillar), so JI is parallel to and equal in length to the face diagonal BD of the bottom face.

1STEP 1

Fix edge length s = 2 and put C at the origin, so E = (2, 2, 2), J = (0, 2, 1), I = (2, 0, 1) land on whole-number coordinates.

C=(0,0,0), E=(2,2,2), J=(0,2,1), I=(2,0,1)

2STEP 2

The 3D distance formula gives all four sides EJ = JC = CI = IE = √(5), so EJCI is equilateral.

EJ = √(2²+0²+1²) = √(5), JC = √(0²+2²+1²) = √(5), CI = √(2²+0²+1²) = √(5), IE = √(0²+2²+1²) = √(5)

3STEP 3

Equal sides make EJCI a rhombus, so its area is half the product of the diagonals EC and JI.

Area(EJCI) =

\frac{1}{2}

· EC · JI

4STEP 4

The space diagonal EC = 2√(3), and JI = 2√(2) since JI lies flat and equals the bottom face's diagonal.

EC = 2√(3), JI = 2√(2)

5STEP 5

Rhombus area = 2√(6); dividing by the face area 4 gives R = $\frac{√(6)}{2}$ , and squaring collapses the radical to R² = $\frac{3}{2}$ .

Area(EJCI) =

\frac{1}{2}

(2√(3))(2√(2)) = 2√(6). R =

\frac{2√(6)}{4}

\frac{√(6)}{2}

. R² =

\frac{6}{4}

\frac{3}{2}

→ (C)

Answer

\frac{3}{2}

R² =

\frac{3}{2}

means the cross-section has area

\frac{3}{2}

times a face — about 1.22 times larger linearly (R =

\frac{√(6)}{2}

≈ 1.225). That feels right: the cross-section is a rhombus that slices diagonally through the cube from corner C up to corner E, so it should be visibly larger than a single face but not enormously so. Choice (E)

\frac{9}{4}

would mean the cross-section is 2.25× a face, which is too big for a region bounded by edge-midpoints; choice (A)

\frac{5}{4}

would mean barely larger than a face, which under-counts the stretch from the space-diagonal direction.

\frac{3}{2}

sits in the right window.

💡Key takeaway

This AMC 8 problem only needs Grade 8 distance-by-Pythagoras-in-coordinates that you already know — even in 3D, distance is just √((Δ x)² + (Δ y)² + (Δ z)²)!

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