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Here is my question:

Let P be a regular polygon with 24 sides. Consider all the triangles whose vertices are also vertices of P. Find the number of such triangles that are neither isosceles nor equilateral.

And here is my attempt:

If we can calculate the number of isosceles and equilateral triangles that can be formed then we will get the answer.

To calculate the number of such isosceles triangles, Notice that, each sides of polygon are equal. Now if we fix a point and choose two other points by bypassing equal number of vertices both sides of our fixed point then we get a isosceles triangle. For better understanding suppose we name the vertices 1,2,..,24. Now we take the fixed point as 1. Then other two points can be (2,24),(3,23),(4,22),...,(13,13) we can choose those two points in 12 ways. And we can choose the fixed point in 24 ways. Now in this way we count each point three times. So total number of isosceles triangle is (24×12)/3 = 96

The number of equilateral triangle is zero. As using the same logic now the dictance between all there points must be equal. And there are no such point. So the final answer is 24C3 - 96.

Is it correct, or can there be any improvement upon this?

FD_bfa
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Albert
  • 117

2 Answers2

2
  • The total number of triangles is ${24 \choose 3}=2024$ which you wrote as 24C3.
  • Your isosceles triangle method would suggest $24 \times 11=264$ triangles including equilateral triangles and most of these are distinct so you should not be dividing by $3$; you said $\times 12$ but that included the non-triangle $(1,13,13)$.
  • There are $\frac{24}3=8$ equilateral triangles (here you do divide by $3$), such as $(1,9,17)$ and $(8,16,24)$, and each of these would have been counted three times among those isosceles triangles while you want them counted only once, so the actual number of isosceles triangles including equilateral triangles is $264 - 2 \times 8 = 248$.
  • So the number of non-isosceles and non-equilateral triangles is $2024-248 = 1776$.
Henry
  • 169,616
1

The question asks for the number of scalene triangles that share vertices with a regular $24$-gon.

The number of scalene triangles that share vertices with a regular $n$-gon in general is given by $Q_\text{scalene}$ in the following formula from my answer to "How many non-congruent triangles can be formed by the vertices of a regular polygon of $n$ sides" under the heading "Scalene triangles": $$ Q_\text{scalene} = \begin{cases} \dfrac16 n (n - 2) (n - 4) & n \equiv \pm 2 \pmod 6, \\ \dfrac16 n (n - 1) (n - 5) & n \equiv \pm 1 \pmod 6, \\ \dfrac16 n (n^2 - 6 n + 12) & n \equiv 0 \pmod 6, \\ \dfrac16 n(n - 3)^2 & n \equiv 3 \pmod 6. \end{cases} $$

For $n = 24,$ the applicable case is $n \equiv 0 \pmod 6,$ and therefore $$ \begin{align} Q_\text{scalene} &= \dfrac16 n (n^2 - 6 n + 12) \\ &= \dfrac16 \cdot 24(24^2 - 6\cdot 24 + 12) \\ &= 4\cdot 37\cdot 12\\ &= 1776. \end{align} $$

David K
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