AMC 8 · 2018 · #24

Grade 8 geometry-3d

Archetype Compound Area by Decomposition / Difference

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of segments $\overline{FB}$ and $\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$

$\textbf{(A) } \frac{5}{4} \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{25}{16} \qquad \textbf{(E) } \frac{9}{4}$

Pick an answer.

(A)

$\frac{5}{4}$

(B)

$\frac{4}{3}$

(C)

$\frac{3}{2}$

(D)

$\frac{25}{16}$

(E)

$\frac{9}{4}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A cube $ABCDEFGH$ has $C$ and $E$ as opposite vertices (a space-diagonal pair). $J$ is the midpoint of edge $\overline{FB}$ and $I$ is the midpoint of edge $\overline{HD}$. Let $R$ be the ratio of the area of the quadrilateral cross-section $EJCI$ to the area of one face of the cube. Find $R^2$.

Givens: $ABCDEFGH$ is a cube, with $C$ and $E$ at opposite corners; $J$ is the midpoint of edge $\overline{FB}$; $I$ is the midpoint of edge $\overline{HD}$; $R = \dfrac{\text{Area}(EJCI)}{\text{area of one face of the cube}}$; Answer choices: (A) $\tfrac{5}{4}$, (B) $\tfrac{4}{3}$, (C) $\tfrac{3}{2}$, (D) $\tfrac{25}{16}$, (E) $\tfrac{9}{4}$

Unknowns: The value of $R^2$

Understand

Plan

Primary tool: #10 Create a Physical Representation

Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships

Tool #10 (Physical Representation): a 3D cube cross-section is much easier to grasp by holding a cube (a tissue box works) and tracing $E \to J \to C \to I \to E$ 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 $\overline{EC}$ (the cube's space diagonal) and $\overline{JI}$ (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 $\tfrac{1}{2} d_1 d_2$ 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 $\overline{JI}$ is parallel to and equal in length to the face diagonal $\overline{BD}$ of the bottom face.

Execute — Answer: C

#10 Create a Physical Representation 5.G.A.2 Step 1

Set up a convenient model.
Let the cube's edge length be $s = 2$ (so midpoints land on whole numbers).
Place $C$ at the origin and orient the cube with edges along the $x$, $y$, $z$ axes so $E$ — the vertex opposite $C$ — sits at $(2, 2, 2)$.
Then $B = (0, 2, 0)$, $D = (2, 0, 0)$, $F = (0, 2, 2)$, $H = (2, 0, 2)$, $J = $ midpoint of $\overline{FB} = (0, 2, 1)$, and $I = $ midpoint of $\overline{HD} = (2, 0, 1)$.

$$C=(0,0,0),\; E=(2,2,2),\; J=(0,2,1),\; I=(2,0,1)$$

💡 Putting the cube on coordinate axes is the Grade 5 "plot points to model a real-world figure" idea — it turns 3D vision into arithmetic.

#7 Identify Subproblems 8.G.B.8 Step 2

Find the four side lengths of $EJCI$ using the 3D distance formula (Pythagoras in space): $d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$.
Computing each side: $EJ$ from $(2,2,2)$ to $(0,2,1)$, $JC$ from $(0,2,1)$ to $(0,0,0)$, $CI$ from $(0,0,0)$ to $(2,0,1)$, $IE$ from $(2,0,1)$ to $(2,2,2)$.

$EJ = \sqrt{2^2+0^2+1^2} = \sqrt{5}$,$\;\;$ $JC = \sqrt{0^2+2^2+1^2} = \sqrt{5}$,$\;\;$ $CI = \sqrt{2^2+0^2+1^2} = \sqrt{5}$,$\;\;$ $IE = \sqrt{0^2+2^2+1^2} = \sqrt{5}$

💡 Distance between two points in coordinates is Grade 8's Pythagorean theorem, extended to 3D by squaring each axis difference.

#7 Identify Subproblems 6.G.A.1 Step 3

All four sides equal $\sqrt{5}$, so $EJCI$ is a rhombus.
The area of a rhombus is half the product of its diagonals: $\text{Area} = \tfrac{1}{2} d_1 d_2$.
The diagonals are $\overline{EC}$ (connecting the opposite vertices of the cube) and $\overline{JI}$ (connecting the two midpoints).

$$\text{Area}(EJCI) = \tfrac{1}{2} \cdot EC \cdot JI$$

💡 Knowing the rhombus area equals half the product of diagonals is a Grade 6 "area of a special quadrilateral" fact.

#17 Visualize Spatial Relationships 8.G.B.8 Step 4

Compute $EC$ — the cube's space diagonal — from $C=(0,0,0)$ to $E=(2,2,2)$: $EC = \sqrt{2^2+2^2+2^2} = \sqrt{12} = 2\sqrt{3}$.
Compute $JI$ from $J=(0,2,1)$ to $I=(2,0,1)$: $JI = \sqrt{2^2+(-2)^2+0^2} = \sqrt{8} = 2\sqrt{2}$.
(Note: the $z$-coordinates match, so $JI$ lives in a horizontal plane and equals the face diagonal of the bottom — a nice sanity check.)

$$EC = 2\sqrt{3}, \quad JI = 2\sqrt{2}$$

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

#7 Identify Subproblems 8.EE.A.2 Step 5

Plug the two diagonals into the rhombus area formula, then divide by the face area $s^2 = 2^2 = 4$ to get $R$.
Finally, square it.

$\text{Area}(EJCI) = \tfrac{1}{2}(2\sqrt{3})(2\sqrt{2}) = 2\sqrt{6}$. $\;\;R = \dfrac{2\sqrt{6}}{4} = \dfrac{\sqrt{6}}{2}$. $\;\;R^2 = \dfrac{6}{4} = \dfrac{3}{2} \;\Rightarrow\; \textbf{(C)}$

💡 Squaring $\tfrac{\sqrt{6}}{2}$ collapses the radical: $(\sqrt{6})^2 = 6$ — Grade 8 square-root manipulation.

[1] #10 5.G.A.2 Set up a convenient model. Let the cube's edge length be $s = 2$ (so midpoints l

[2] #7 8.G.B.8 Find the four side lengths of $EJCI$ using the 3D distance formula (Pythagoras i

[3] #7 6.G.A.1 All four sides equal $\sqrt{5}$, so $EJCI$ is a rhombus. The area of a rhombus i

[4] #17 8.G.B.8 Compute $EC$ — the cube's space diagonal — from $C=(0,0,0)$ to $E=(2,2,2)$: $EC

[5] #7 8.EE.A.2 Plug the two diagonals into the rhombus area formula, then divide by the face ar

Review

Reasonableness: $R^2 = \tfrac{3}{2}$ means the cross-section has area $\tfrac{3}{2}$ times a face — about $1.22$ times larger linearly ($R = \sqrt{6}/2 \approx 1.225$). 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) $\tfrac{9}{4}$ would mean the cross-section is $2.25\times$ a face, which is too big for a region bounded by edge-midpoints; choice (A) $\tfrac{5}{4}$ would mean barely larger than a face, which under-counts the stretch from the space-diagonal direction. $\tfrac{3}{2}$ sits in the right window.

Alternative: Tool #1 (Draw a Diagram) gives a coordinate-free version: notice $\overline{FBDH}$ is a rectangle inside the cube with sides $s$ (the vertical edges $\overline{FB}, \overline{HD}$) and $s\sqrt{2}$ (the face diagonals $\overline{BD}, \overline{FH}$). Since $J$ and $I$ are the midpoints of the two short sides, segment $\overline{JI}$ is parallel to $\overline{BD}$ and has the same length $s\sqrt{2}$. The other diagonal $\overline{EC}$ is the space diagonal $s\sqrt{3}$. Rhombus area $= \tfrac{1}{2}(s\sqrt{3})(s\sqrt{2}) = \tfrac{s^2\sqrt{6}}{2}$, so $R = \tfrac{\sqrt{6}}{2}$ and $R^2 = \tfrac{3}{2}$.

CCSS standards used (min grade 8)

5.G.A.2 Represent real-world and mathematical problems by graphing points (Placing the cube in a 3D coordinate system with $C$ at the origin and $E$ at $(2,2,2)$ so each vertex and midpoint has clean integer coordinates.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Applying the rhombus area formula $\text{Area} = \tfrac{1}{2} d_1 d_2$ once $EJCI$ is identified as a rhombus.)
8.G.B.8 Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing all four side lengths of $EJCI$ and both diagonals $EC$ and $JI$ from coordinate differences using the 3D distance formula.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Simplifying radicals ($\sqrt{12} = 2\sqrt{3}$, $\sqrt{8} = 2\sqrt{2}$) and squaring $R = \tfrac{\sqrt{6}}{2}$ to get $R^2 = \tfrac{3}{2}$.)

⭐ This AMC 8 problem only needs Grade 8 distance-by-Pythagoras-in-coordinates that you already know — even in 3D, distance is just $\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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