Geometry & Figures

Problem

Count all the rhombuses in a figure of small triangles

An elongated hexagon is tiled by 14 identical small equilateral triangles in two rows of 7, one above the other, split by a horizontal centerline from the left tip to the right tip. In each row the triangles alternate point-up and point-down. I must count every rhombus, where each rhombus is exactly one point-up triangle joined to a neighboring point-down triangle (2 small triangles), in all of its possible orientations.
Geometry
Your answer
How to solve
Strategy Make a Systematic List — Label the 14 small triangles (top row t1..t7 left to right, bottom row b1..b7 left to right) and list rhombuses by orientation so none is missed or double-counted. Each rhombus is a fixed up+down pair sharing an edge; the shared edge can be horizontal (a tall vertical diamond) or one of two slanted edges (a left-leaning or a right-leaning diamond). Count each orientation as a subproblem, then add.
1STEP 1

Label the 14 small triangles

Number the top row t1-t7 and bottom row b1-b7. Every rhombus is one up-triangle glued to a neighboring down-triangle along a shared edge.

top: t₁,…,t₇; bottom: b₁,…,b₇
2STEP 2

Count the tall vertical rhombuses (shared edge on the centerline)

Up+down pairs sharing a base on the centerline: t1+b1, t3+b3, t5+b5, t7+b7. That is 4 vertical rhombuses.

t₁+b₁, t₃+b₃, t₅+b₅, t₇+b₇ → 4
3STEP 3

Count the left-leaning rhombuses

Pairs sharing a backslash-tilted edge: top t1+t2, t3+t4, t5+t6; bottom b2+b3, b4+b5, b6+b7. That is 3 + 3 = 6 left-leaning rhombuses.

t₁+t₂, t₃+t₄, t₅+t₆, b₂+b₃, b₄+b₅, b₆+b₇ → 6
4STEP 4

Count the right-leaning rhombuses

Pairs sharing a slash-tilted edge: top t2+t3, t4+t5, t6+t7; bottom b1+b2, b3+b4, b5+b6. That is 3 + 3 = 6 right-leaning rhombuses.

t₂+t₃, t₄+t₅, t₆+t₇, b₁+b₂, b₃+b₄, b₅+b₆ → 6
5STEP 5

Add the three orientations

Each rhombus has exactly one orientation, so the three lists don't overlap. Add the vertical, left-leaning, and right-leaning counts.

4 + 6 + 6 = 16
Answer
16
4 + 6 + 6 = 16
List every up+down pair explicitly: 4 vertical + 6 left-leaning + 6 right-leaning = 16. Cross-check by up-triangle — t1:2, t3:3, t5:3, t7:2, b2:2, b4:2, b6:2 — also totals 16, with no pair counted twice and none missed.
Takeaway

Sort the diamonds by how they lean: 4 stand upright, 6 lean one way, 6 lean the other. Organized counting gives 4 + 6 + 6 = 16!

  • Label the 14 small triangles
  • Count the tall vertical rhombuses (shared edge on the centerline)
  • Count the left-leaning rhombuses
  • Count the right-leaning rhombuses
  • Add the three orientations
Where next?
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