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Consider an equilateral triangle with side length 10, which is divided perfectly into little triangles with unit side length. Find the number of the little rhombuses with side length at least 2.

(Deleted incorrect attempted work)

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    Do the rhombuses have to be aligned with the triangle's grid? (And your answer is too low either way!) – 2017-01-13
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    Yes. Please point out if there is any mistake in my work. – 2017-01-13
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    Actually I think you're right, but I have a simpler way of arriving at 150 that should be more intuitive. Would you mind me posting it as an answer? – 2017-01-13
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    I see my mistake already. Please post your solution. Thank you. – 2017-01-13
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    I mean, you got the right answer, but in a more roundabout fashion... – 2017-01-13

1 Answers 1

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Consider the rhombuses directly:

^ * * ` ` ` ^ * * ` `
 * ` * ` ` ` * ` * `
  * * * ` ` ` * * *
   ` ` ` ` ` ` ` `
    ` ` ` ` ` ` `
     ` ` ` ` ` `
      ^ * * ` `
       * ` * `
        * * *
         ` `
          `

There are three ways to orient a rhombus in the triangular grid. For a given orientation there are $T_7=28$ rhombuses of side length 2 in the given equilateral triangle, as evidenced by the carets in the diagram above ($T_n=\frac{n(n+1)}2$ is the $n$th triangular number). Similarly we can find $T_5=15$ rhombuses of side 3, $T_3=6$ of side 4 and $T_1=1$ of side 5 with a given orientation. Adding these counts up and multiplying by three gives the answer as $$3(28+15+6+1)=150$$

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    Thank you very much, Parcly Taxel. – 2017-01-14