Symmetry Function

Geometry Level 5

"There exists a 2D figure with exactly $m$ line(s) of symmetry and rotational symmetry of order $n$ ."

Suppose the above statement is called $T(m, n)$ . Let the function $f : \mathbb{Z_{\geq 0}} \times \mathbb{Z_{> 0}} \rightarrow \mathbb{Z}$ be defined as:

$f(m ,n) = \begin{cases} m + n & \text{if } T(m, n) \text{ is true} \\ 0 & \text{if } T(m, n) \text{ is false} \\ \end{cases}$

If $g : \mathbb{Z_{>0}} \rightarrow \mathbb{Z}$ is defined as $g(x) = \displaystyle \sum _{m = 0} ^{x} \sum _{n = 1} ^{x} f(m, n)$ ,
find $\displaystyle \lim_{x \to \infty} \frac{g(x)}{x^{2}}$

The answer is 1.5.

1 solution

Pranshu Gaba
Sep 3, 2015

Theorem If a figure has $m = 0$ lines of symmetry, then it can have rotational symmetry of order $n = 1, 2, 3, \ldots$ . If a figure has $m > 0$ lines of symmetry, then it will have rotational symmetry of order $n = m$ .

Proof (Coming Soon)

Therefore $f(0, n) = n$ for all positive integers $n$ , $f(m, n) = m + n$ if $m = n$ and $f(m, n) = 0$ if $m \neq n$ .

We have $g(x) = \displaystyle \sum_{n = 1} ^{x} f(0, n) + \sum_{n = 1} ^{x} f(n, n) = \displaystyle \sum_{n = 1} ^{x} n + \sum_{n = 1} ^{x} (n + n) = \displaystyle \sum_{n = 1} ^{x} 3n = \frac{3x(x + 1)}{2}$

We obtain $\dfrac{g(x)}{x^{2}} = \dfrac{3}{2} \cdot \dfrac{x^2 + x}{x^2}$

Therefore $\displaystyle \lim_{x \to \infty} \frac{g(x)}{x^{2}} = \frac{3}{2} \times 1 = \boxed{1.5}$ $_\square$

Moderator note:

Yes, this boils down to proving that claim. Is there a simple way of showing it?

Symmetry Function

The answer is 1.5.

1 solution

Moderator note:

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