Competition · AMC preparation · step 4 of 4

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 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

Set up cube coordinates

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

Measure the four sides

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

Spot the rhombus

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

Area(EJCI) = 1/2 · EC · JI

4STEP 4

Find the two diagonals

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)

Each diagonal length is one more 3D distance — same Pythagoras-in-coordinates idea as the sides.

The rhombus's two diagonals are the cube's space diagonal EC = s√(3) and the midline segment JI = s√(2), which has the length of a face diagonal.

▸ Why?

EC joins the opposite corners C and E of the cube, so it is the space diagonal. It is the hypotenuse of the right triangle △ CAE whose two perpendicular legs are the bottom-face diagonal CA (length s√(2)) and the vertical edge AE (length s), so EC² = (s√(2))² + s² = 3s² and EC = s√(3).

▸ Why?

The face diagonal CA and the edge AE meet at a right angle at A, and EC is the hypotenuse opposite that right angle, so the two legs' squares add up to EC².

Pythagorean theoremIn a right triangle the two legs' squares add up to the hypotenuse's square (a²+b²=c²).

▸ Why?

The leg CA has length s√(2) because it is itself the hypotenuse of a right triangle whose two perpendicular legs are edges of the square bottom face, each of length s, so CA² = s² + s² = 2s².

Pythagorean theoremIn a right triangle the two legs' squares add up to the hypotenuse's square (a²+b²=c²).

▸ Why?

JI has the length of a face diagonal, s√(2), because J and I are the midpoints of the two short sides FB and HD of the rectangle FBDH, and that midline is parallel to and the same length as the rectangle's long sides FH and BD.

▸ Why?

JI and the face diagonal BD point the same way and match end to end, so JI slides straight onto BD without stretching.

▸ Why?

A slide lays JI exactly on top of BD, and sliding never changes a length, so the two are equal.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

BD is the diagonal of a square face, cutting across the two perpendicular edges BC and CD, so BD² = s² + s² = 2s² and BD = s√(2).

▸ Why?

BD is the hypotenuse of the right triangle △ BCD whose two perpendicular legs are the edges BC and CD, so the legs' squares add up to BD².

Pythagorean theoremIn a right triangle the two legs' squares add up to the hypotenuse's square (a²+b²=c²).

8.G.B.8Visualize Spatial Relationships

5STEP 5

Divide by the face area

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

Area(EJCI) = 1/2(2√(3))(2√(2)) = 2√(6). R = 2√(6)/4 = √(6)/2. R² = 6/4 = 3/2 → (C)

Answer

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{\sqrt{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)!

Set up cube coordinates
Measure the four sides
Spot the rhombus
Find the two diagonals
Divide by the face area

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Set up cube coordinates

Measure the four sides

Spot the rhombus

Find the two diagonals

Divide by the face area

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