AMC 10 · 2005 · #25

Grade 12 geometry-3dcounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

spatial-visualizationequilateral-triangledot-productlaw-of-cosines systematic-enumerationcaseworksymmetry-argument ↑ Prerequisites: spatial-visualizationequilateral-trianglecoordinate-geometry

📏 Long solution 💡 4 insights

Problem

A three by three by three block of lattice points holds 27 points. Count the equilateral triangles whose three corners all lie among them.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Visualize Spatial Relationships

Checking triangles one at a time is hopeless, since 27 points give 2925 triples. Instead classify by side length. Writing the two sides at a vertex as vectors turns "equilateral" into a single equation about the dot product, and that equation rules out most of the few available lengths straight away. Only three lengths survive, and each becomes its own small counting subproblem that can be done exactly rather than by inspection.

1STEP 1

Picture the grid

The block is small enough to organise but far too big to check one by one.

2STEP 2

Turn the triangle into two vectors

Two vectors from one corner describe the whole triangle.

u = B - A, v = C - A, C - B = v - u

3STEP 3

Find the rule the side length must obey

Equal sides force a relation making the squared length even.

|v-u|² = |u|² + |v|² - 2(u · v) ⟹ u · v = 1/2|u|²

4STEP 4

List the squared lengths available

Only five squared lengths survive that test.

|u|² ∈ {1, 2, 3, 4, 5, 6, 8, 9, 12} → {2, 4, 6, 8, 12}

5STEP 5

Eliminate two more lengths

Checking the relation directly removes two more, leaving three.

|u|² = 4 → u · v ∈ {0, ± 4} ≠ 2, |u|² = 12 → u · v ∈ {± 4, ± 12} ≠ 6

6STEP 6

Set up a counting frame that cannot double count

A counting frame reaches each triangle six times.

#triangles = (#{(A, u, v)})/(3 · 2)

7STEP 7

Count where a given pair of arrows fits

Cube symmetry means one representative settles each family's placements.

#{A} = Π_i=1³(3 - (max(0, u_i, v_i) - min(0, u_i, v_i)))

8STEP 8

Count each family

The three families give 64, 8 and 8.

(12 · 4 · 8)/6 = 64, (24 · 2 · 1)/6 = 8, (12 · 4 · 1)/6 = 8

9STEP 9

Add the three families

Adding gives 80, choice (C).

64 + 8 + 8 = 80

Answer

Each family has a picture that independently confirms its size. The 64 triangles of side √(2) are the corner triangles of the eight unit cubes sitting inside S: each unit cube splits into two regular tetrahedra with 4 faces apiece, so 8 × 8 = 64, and the smallest box containing such a triangle is exactly its unit cube, so no triangle is counted for two different cubes. The 8 triangles of side √(8) = 2√(2) are the same construction applied to the single big 2 × 2 × 2 cube. The 8 triangles of side √(6) join midpoints of three edges of the big cube; each such triangle is perpendicular to a space diagonal, and each of the 4 space diagonals carries exactly 2 of them, so 4 × 2 = 8. Both routes give 64 + 8 + 8 = 80. The trap answer 72 is what appears if the √(6) family is never noticed, which is why the parity argument in step 3 matters: it proves in advance that exactly three side lengths are possible, so the search has a stopping point.

💡Key takeaway

Before hunting for shapes, work out which side lengths are even possible; that turns an endless search into three short lists.

Picture the grid
Turn the triangle into two vectors
Find the rule the side length must obey
List the squared lengths available
Eliminate two more lengths
Set up a counting frame that cannot double count
Count where a given pair of arrows fits
Count each family
Add the three families