special interior angles of regular $n$-gon
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Let $K$ be a regular $n$-gon in the plane. Assume the following:
$$ninBbb N,quad ngeq3\I={0,1,...,n-1}\ forall iin I, quad M(i):=operatorname{mod}(i,n)$$
And we define $P_i$ as the $i$-th vertex of $K$ in the counter-clockwise direction for all $iin I$.
Then we define the angles which are the topic of my investigation:
$$forall iin I,quad alpha_i:=angle P_0P_{M(i)}P_{M(i+1)}$$
I am trying to find a closed form expression for $alpha_i$.
My Attempts So Far:
Since $K$ is a regular polygon,
$$forall iin I,quad angle P_{M(i-1)}P_{M(i)}P_{M(i+1)}=frac{n-2}npi$$
And trivially, $$alpha_0=frac{n-2}npi$$
If we construct the segment $P_0P_2$, the do a little trigonometry we can show that
$$alpha_1=frac{n-2}2pi$$
Also we know that $$alpha_{n-1}=0$$
and $$sum_{iin I}alpha_ileq (n-2)pi$$
But other than that I do not know how to find a closed form for $alpha_i$.
Could anyone help?
geometry modular-arithmetic euclidean-geometry closed-form pattern-recognition
asked Nov 15 at 18:06
clathratus
2,074220
add a comment |
up vote
0
down vote
favorite
Let $K$ be a regular $n$-gon in the plane. Assume the following:
$$ninBbb N,quad ngeq3\I={0,1,...,n-1}\ forall iin I, quad M(i):=operatorname{mod}(i,n)$$
And we define $P_i$ as the $i$-th vertex of $K$ in the counter-clockwise direction for all $iin I$.
Then we define the angles which are the topic of my investigation:
$$forall iin I,quad alpha_i:=angle P_0P_{M(i)}P_{M(i+1)}$$
I am trying to find a closed form expression for $alpha_i$.
My Attempts So Far:
Since $K$ is a regular polygon,
$$forall iin I,quad angle P_{M(i-1)}P_{M(i)}P_{M(i+1)}=frac{n-2}npi$$
And trivially, $$alpha_0=frac{n-2}npi$$
If we construct the segment $P_0P_2$, the do a little trigonometry we can show that
$$alpha_1=frac{n-2}2pi$$
Also we know that $$alpha_{n-1}=0$$
and $$sum_{iin I}alpha_ileq (n-2)pi$$
But other than that I do not know how to find a closed form for $alpha_i$.
Could anyone help?
geometry modular-arithmetic euclidean-geometry closed-form pattern-recognition
asked Nov 15 at 18:06
clathratus
2,074220
You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $K$ be a regular $n$-gon in the plane. Assume the following:
$$ninBbb N,quad ngeq3\I={0,1,...,n-1}\ forall iin I, quad M(i):=operatorname{mod}(i,n)$$
And we define $P_i$ as the $i$-th vertex of $K$ in the counter-clockwise direction for all $iin I$.
Then we define the angles which are the topic of my investigation:
$$forall iin I,quad alpha_i:=angle P_0P_{M(i)}P_{M(i+1)}$$
I am trying to find a closed form expression for $alpha_i$.
My Attempts So Far:
Since $K$ is a regular polygon,
$$forall iin I,quad angle P_{M(i-1)}P_{M(i)}P_{M(i+1)}=frac{n-2}npi$$
And trivially, $$alpha_0=frac{n-2}npi$$
If we construct the segment $P_0P_2$, the do a little trigonometry we can show that
$$alpha_1=frac{n-2}2pi$$
Also we know that $$alpha_{n-1}=0$$
and $$sum_{iin I}alpha_ileq (n-2)pi$$
But other than that I do not know how to find a closed form for $alpha_i$.
Could anyone help?
geometry modular-arithmetic euclidean-geometry closed-form pattern-recognition
asked Nov 15 at 18:06
clathratus
2,074220
Let $K$ be a regular $n$-gon in the plane. Assume the following:
$$ninBbb N,quad ngeq3\I={0,1,...,n-1}\ forall iin I, quad M(i):=operatorname{mod}(i,n)$$
And we define $P_i$ as the $i$-th vertex of $K$ in the counter-clockwise direction for all $iin I$.
Then we define the angles which are the topic of my investigation:
$$forall iin I,quad alpha_i:=angle P_0P_{M(i)}P_{M(i+1)}$$
I am trying to find a closed form expression for $alpha_i$.
My Attempts So Far:
Since $K$ is a regular polygon,
$$forall iin I,quad angle P_{M(i-1)}P_{M(i)}P_{M(i+1)}=frac{n-2}npi$$
And trivially, $$alpha_0=frac{n-2}npi$$
If we construct the segment $P_0P_2$, the do a little trigonometry we can show that
$$alpha_1=frac{n-2}2pi$$
Also we know that $$alpha_{n-1}=0$$
and $$sum_{iin I}alpha_ileq (n-2)pi$$
But other than that I do not know how to find a closed form for $alpha_i$.
Could anyone help?
geometry modular-arithmetic euclidean-geometry closed-form pattern-recognition
geometry modular-arithmetic euclidean-geometry closed-form pattern-recognition
asked Nov 15 at 18:06
clathratus
2,074220
asked Nov 15 at 18:06
clathratus
2,074220
asked Nov 15 at 18:06
clathratus
2,074220
asked Nov 15 at 18:06
clathratus
2,074220
asked Nov 15 at 18:06
clathratus
2,074220
2,074220
You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13
add a comment |
You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13
You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13
You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13
add a comment |
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You don't really need the $M(i)$ notation; it just confuses things (at least for me). All vertices have indices less than $n$, and the angle $angle P_0P_{n-1}P_n = angle P_0P_{n-1}P_0 = 0$ trivially. Also, I think you mean $alpha_1 = frac{n-2}{n}pi$ (not $alpha_0$). Finally, I think the next angle, which you have labeled $alpha_1$, is actually $alpha_2 = frac{n-3}{n}$.
– rogerl
Nov 15 at 20:13