AMC 10 · 2023 · #18
Grade 8 geometry-2dProblem
A rhombic dodecahedron is a solid with 12 congruent rhombus faces. At every vertex, 3 or 4 edges meet, depending on the vertex. How many vertices have exactly 3 edges meet?
Pick an answer.
AMC 10 2023 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.
Toolkit + CCSS Solution
Understand
Restated: A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At each vertex either $3$ or $4$ edges meet. How many vertices have exactly $3$ edges meeting?
Givens: $F = 12$ faces, each a rhombus (so $4$ edges per face); Every vertex has degree $3$ or degree $4$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Unknowns: $x$ = number of degree-$3$ vertices
Understand
Restated: A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At each vertex either $3$ or $4$ edges meet. How many vertices have exactly $3$ edges meeting?
Givens: $F = 12$ faces, each a rhombus (so $4$ edges per face); Every vertex has degree $3$ or degree $4$; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $8$, (E) $9$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #13 Convert to Algebra
The path is a clean three-step climb: (a) edges $E$ from face/edge double-counting, (b) total vertices $V$ from Euler's formula, (c) split $V$ into degree-$3$ and degree-$4$ vertices using the handshake total $3x + 4y = 2E$. Tool #7 (Identify Subproblems) names each rung. The final step is a $2 \times 2$ linear system in $x$ and $y$, which is Tool #13 (Convert to Algebra). No clever symmetry argument is needed — every quantity falls out of two counting identities.
Execute — Answer: D
5.G.B.3 Step 1 - Subproblem A: count edges $E$.
- Each of the $12$ rhombic faces has $4$ edges, and every edge is shared by exactly two faces, so the face-edge incidences total $12 \cdot 4 = 48$ but each edge is counted twice.
- Divide by $2$.
💡 Counting (face, edge) pairs in two ways is the universal polyhedron trick — each edge sits on exactly two faces, so divide by $2$.
5.OA.A.1 Step 2 - Subproblem B: count vertices $V$ using Euler's formula.
- Plug $F = 12$ and $E = 24$ into $V - E + F = 2$.
💡 Euler's $V - E + F = 2$ is the bookkeeping identity for every convex polyhedron — once $E$ and $F$ are known, $V$ is forced.
8.EE.C.8 Step 3 - Subproblem C: set up the two-equation system.
- Let $x$ = number of degree-$3$ vertices and $y$ = number of degree-$4$ vertices.
- Total vertices give $x + y = 14$.
- The handshake lemma — sum of degrees $= 2E$ — gives $3x + 4y = 2 \cdot 24 = 48$.
💡 Each edge has two endpoints, so adding up "how many edges does each vertex see?" double-counts every edge — turning a vertex split into a linear equation.
8.EE.C.8 Step 4 - Solve the system by elimination.
- Multiply the first equation by $4$: $4x + 4y = 56$.
- Subtract the second: $(4x + 4y) - (3x + 4y) = 56 - 48$, giving $x = 8$.
- Then $y = 14 - 8 = 6$.
💡 Subtracting one linear equation from a scaled copy of another removes one variable on the spot.
5.G.B.3 Subproblem A: count edges $E$. Each of the $12$ rhombic faces has $4$ edges, and 5.OA.A.1 Subproblem B: count vertices $V$ using Euler's formula. Plug $F = 12$ and $E = 2 8.EE.C.8 Subproblem C: set up the two-equation system. Let $x$ = number of degree-$3$ ver 8.EE.C.8 Solve the system by elimination. Multiply the first equation by $4$: $4x + 4y = Review
Reasonableness: Cross-check the handshake: $3 \cdot 8 + 4 \cdot 6 = 24 + 24 = 48 = 2 \cdot 24$ — matches $2E$ exactly. Cross-check Euler: $V - E + F = 14 - 24 + 12 = 2$ — matches. Cross-check totals: $x + y = 8 + 6 = 14 = V$. Geometrically, the rhombic dodecahedron is known to have $8$ degree-$3$ vertices (the "acute" corners, sitting at the cube's $8$ corners) and $6$ degree-$4$ vertices (sitting at the cube's $6$ face centers), confirming $\textbf{(D) }8$.
Alternative: Tool #10 (Create a Physical Representation) plus Tool #5 (Look for a Pattern): build (or visualize) a rhombic dodecahedron by gluing a square pyramid onto each face of a cube. The cube's $8$ corners become degree-$3$ vertices (three rhombi meet there) and the apexes above the cube's $6$ faces become degree-$4$ vertices. Count: $8$ degree-$3$ vertices, $6$ degree-$4$ vertices — same answer (D).
CCSS standards used (min grade 8)
5.G.B.3Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category (Reading "$12$ rhombic faces, $4$ edges per rhombus" off the figure's description and double-counting (face, edge) incidences to get $E = 24$.)5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate expressions with these symbols (Substituting into Euler's identity $V - E + F = 2$ to solve $V = 2 + 24 - 12 = 14$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Setting up and solving $\{x + y = 14,\ 3x + 4y = 48\}$ — total vertices and the handshake lemma — to extract $x = 8$.)
⭐ Two counting identities crack this solid: each edge sits on two faces (so $E = \tfrac{12 \cdot 4}{2} = 24$), and Euler's $V - E + F = 2$ gives $V = 14$. Splitting the $14$ vertices into degree-$3$ and degree-$4$ piles using "sum of degrees $= 2E$" yields a $2 \times 2$ system whose answer is $\textbf{(D) }8$ degree-$3$ vertices.
⭐ Two counting identities crack this solid: each edge sits on two faces (so $E = \tfrac{12 \cdot 4}{2} = 24$), and Euler's $V - E + F = 2$ gives $V = 14$. Splitting the $14$ vertices into degree-$3$ and degree-$4$ piles using "sum of degrees $= 2E$" yields a $2 \times 2$ system whose answer is $\textbf{(D) }8$ degree-$3$ vertices.