AMC 8 · 2024 · #20
Easy mode Grade 3Problem
Picture a cube. Label its 8 corners P,Q,R,S,T,U,V,W as shown in the figure below.
Pick any 3 of these 8 corners and connect them with straight lines. You get a triangle. (For example, picking P, Q, and R gives an isosceles triangle △PQR.)
We are only interested in triangles that are equilateral (all three sides the same length) AND have P as one of their three corners.
How many such triangles are there?
Pick an answer.
AMC 8 2024 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: Pick any three of the eight vertices of cube $PQRSTUVW$ to form a triangle. The question asks: **how many of these triangles are equilateral and contain vertex $P$**?
Givens: A cube has $8$ vertices, $12$ equal-length edges, and $6$ faces that are congruent squares; $P$ has exactly $3$ edge-neighbor vertices: $Q, S, W$; The vertex directly opposite to $P$ across the cube (the other end of the space diagonal) is $U$; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 6
Unknowns: The number of equilateral triangles formed by cube vertices that include $P$
Understand
Restated: Pick any three of the eight vertices of cube $PQRSTUVW$ to form a triangle. The question asks: **how many of these triangles are equilateral and contain vertex $P$**?
Givens: A cube has $8$ vertices, $12$ equal-length edges, and $6$ faces that are congruent squares; $P$ has exactly $3$ edge-neighbor vertices: $Q, S, W$; The vertex directly opposite to $P$ across the cube (the other end of the space diagonal) is $U$; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 6
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List, #3 Eliminate Possibilities
A cube is hard to keep straight in your head, so start with Tool #10 — grab a **real cube** (a die or a small box), label one corner $P$, and feel the situation in your hand. The other $7$ vertices then naturally sort into just three "distance types" from $P$. Next, instead of hunting for the vertices that **do** form an equilateral triangle with $P$, use Tool #16 (change focus) to **rule out** the distance types that **can't** work — that shrinks the candidate set fast. Once the candidates are few, Tool #2 (systematic list) enumerates the triangles, and Tool #3 matches the count to the answer choices.
Execute — Answer: D
K.G.B.4 Step 1 - Hold a real cube (a die or paper box), mark one vertex $P$, and sort the other $7$ vertices by how they relate to $P$.
- There are exactly three buckets: ① **edge-neighbors** of $P$ (joined to $P$ by an edge): $Q, S, W$ — that's $3$ vertices, ② vertices that sit **across a face from** $P$ (the other end of a face diagonal): $R, T, V$ — another $3$ vertices, and ③ the **opposite corner of the cube** from $P$ (the other end of the space diagonal): just $U$.
- Total: $3 + 3 + 1 = 7$, accounting for every non-$P$ vertex.
💡 Turning a real cube and grouping its corners into "next-door, across a face, all the way opposite" is exactly the Kindergarten skill of analyzing and comparing 3D shapes.
3.G.A.1 Step 2 - All six faces of a cube are **congruent squares**, so all face diagonals must be equal in length.
- By the same reasoning all edges are equal, and inside any square the diagonal is longer than a side, so edges and face diagonals are different lengths.
- The space diagonal cuts through the whole cube, so it is longer still.
- So the three distance types from $P$ go in the strict order **edge < face diagonal < space diagonal** — three different lengths, no ties between types.
💡 "All six faces are the same square, so all their diagonals match" is Grade 3 reasoning about shapes in the same category sharing attributes.
K.G.B.4 Step 3 - An equilateral triangle needs three equal sides, so any equilateral triangle with vertex $P$ needs the **other two vertices to be the same distance type from $P$**.
- **Rule out** the bad types first (Tool #16): the space-diagonal type contains only $U$, so you can't pick a second vertex at that distance — no equilateral triangle uses the space-diagonal length.
- The edge type forces the other two vertices to be edge-neighbors of $P$ (chosen from $Q, S, W$); but any two of $Q, S, W$ sit across a face from each other (a face diagonal apart), so the third side is too long — no equilateral triangle uses the edge length either.
- The only surviving distance type is the **face diagonal**, with candidate vertices $R, T, V$.
💡 "Throw away the buckets that can't work first" is the same sort-and-eliminate move children practice in Kindergarten when classifying shapes.
K.OA.A.3 Step 4 - Pick any two of the three candidates $R, T, V$ to pair with $P$, listed in alphabetical order: ① $\{R, T\}$ → $\triangle PRT$, ② $\{R, V\}$ → $\triangle PRV$, ③ $\{T, V\}$ → $\triangle PVT$.
- That's $3$ possible triangles to check.
💡 Listing all the ways to split three letters $R, T, V$ into a pair is at the same level as Kindergarten "break 3 into two groups" practice.
2.G.A.1 Step 5 - Check that all three triangles really are equilateral by feeling the cube: is the third side ($RT$, $RV$, or $VT$) also a face diagonal?
- ① $R$ and $T$ are opposite corners of the back face $RSTU$ — face diagonal.
- ② $R$ and $V$ are opposite corners of the right face $QRUV$ — face diagonal.
- ③ $T$ and $V$ are opposite corners of the bottom face $TUVW$ — face diagonal.
- Since all six faces are congruent squares (from Step 2), all three diagonals have the same length, so each triangle has three equal sides — all three are equilateral.
💡 Confirming "all three sides are face diagonals, so they're equal" matches Grade 2 work on recognizing shapes by attributes like "three equal sides."
K.OA.A.5 Step 6 - Match the count $3$ to the answer choices.
- (A) $0$, (B) $1$, (C) $2$ are too small because we already exhibited three working triangles $\triangle PRT, \triangle PRV, \triangle PVT$.
- (E) $6$ is impossible because there are only three candidate vertices $R, T, V$, so the total number of triangles using $P$ and a pair of them is at most $\binom{3}{2} = 3$.
- The only choice left is (D) $3$.
💡 Counting up to $5$ and picking the matching number is exactly the Kindergarten fluency within $5$.
K.G.B.4 Hold a real cube (a die or paper box), mark one vertex $P$, and sort the other $ 3.G.A.1 All six faces of a cube are **congruent squares**, so all face diagonals must be K.G.B.4 An equilateral triangle needs three equal sides, so any equilateral triangle wit K.OA.A.3 Pick any two of the three candidates $R, T, V$ to pair with $P$, listed in alpha 2.G.A.1 Check that all three triangles really are equilateral by feeling the cube: is th K.OA.A.5 Match the count $3$ to the answer choices. (A) $0$, (B) $1$, (C) $2$ are too sma Review
Reasonableness: For an equilateral triangle with $P$ to exist, at least two more vertices must sit at the **same distance type** from $P$. The space-diagonal type has only one member ($U$), and the edge type fails because $P$'s three edge-neighbors are pairwise a face diagonal apart. So everything has to come from the face-diagonal type — and crucially $R, T, V$ are also a face diagonal apart from each other. Getting exactly $3$ triangles also matches the symmetry of the cube: every vertex sits in exactly three such equilateral triangles, which is a nice sanity check.
Alternative: An alternative is Tool #17 (visualize spatially): notice that $P, R, T, V$ form a **regular tetrahedron** hidden inside the cube. All $4$ faces of a regular tetrahedron are equilateral triangles, and exactly $3$ of those $4$ faces contain $P$ — namely $\triangle PRT, \triangle PRV, \triangle PVT$. Same answer $3$, but it requires holding a tetrahedron in your head, so the physical-cube + list approach used here is safer for younger learners.
CCSS standards used (min grade 3)
K.OA.A.3Decompose numbers less than or equal to 10 into pairs (Listing all ways to choose a pair from the three candidates $\{R, T, V\}$, yielding $\{R,T\}, \{R,V\}, \{T,V\}$.)K.OA.A.5Fluently add and subtract within 5 (Counting the resulting triangles ($3$) and matching to answer choice (D).)K.G.B.4Analyze and compare two- and three-dimensional shapes (Sorting the cube's $7$ non-$P$ vertices into edge-neighbors, face-diagonal partners, and the opposite corner, and eliminating distance types that can't yield an equilateral triangle.)2.G.A.1Recognize and draw shapes having specified attributes (Confirming that the three triangles $\triangle PRT, \triangle PRV, \triangle PVT$ each have three equal sides (the equilateral attribute).)3.G.A.1Understand that shapes in different categories share attributes (Deducing that because all six faces of a cube are congruent squares, all face diagonals share a single common length.)
⭐ This AMC 8 problem only needs Grade 3 understanding that shapes in the same category share attributes (every face of a cube is the same square!) you already know!
⭐ This AMC 8 problem only needs Grade 3 understanding that shapes in the same category share attributes (every face of a cube is the same square!) you already know!
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