AMC 8 · 2024 · #20

• 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.

• 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.

• 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.

• 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.

• 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."

• 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 $5$.