Sum of angles $mangle (x, y)(0, 0)(0, 1)$ for $0 le x, y le 5$

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The Problem:

Denote by $V_(x, y)$ the vertex at ordered pair $(x, y)$ in the Cartesian coordinate system. Denote by $A_(x, y)$ the measure of angle $angle V_(x, y)V_(0, 0)V_(1, 0)$. Let
$$theta = sum_0 le i, j le 5 (i, j) neq (0, 0)A_(i, j)$$
Find $theta pmod2pi$.

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum
$$sum_0 le i, j le 5 (i, j) neq (0, 0)arctanfrac ij$$
But I don't know how to evaluate this. $arctan(x+y)$ formula fails. I think there is a different approach I am not aware of.

Is there a nice way to generalize to
$$sum_0 le i, j le k (i, j) neq (0, 0)A_(i, j)$$

trigonometry analytic-geometry angle

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edited Jul 20 at 21:38

asked Jul 20 at 20:49

Shrey Joshi

1389

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
  – Poon Levi
  Jul 20 at 20:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
  – WW1
  Jul 20 at 21:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @above, fixed. $A_(0, 0)$ is not part of the sum.
  – Shrey Joshi
  Jul 21 at 0:43
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

The Problem:

Denote by $V_(x, y)$ the vertex at ordered pair $(x, y)$ in the Cartesian coordinate system. Denote by $A_(x, y)$ the measure of angle $angle V_(x, y)V_(0, 0)V_(1, 0)$. Let
$$theta = sum_0 le i, j le 5 (i, j) neq (0, 0)A_(i, j)$$
Find $theta pmod2pi$.

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum
$$sum_0 le i, j le 5 (i, j) neq (0, 0)arctanfrac ij$$
But I don't know how to evaluate this. $arctan(x+y)$ formula fails. I think there is a different approach I am not aware of.

Is there a nice way to generalize to
$$sum_0 le i, j le k (i, j) neq (0, 0)A_(i, j)$$

trigonometry analytic-geometry angle

share|cite|improve this question

edited Jul 20 at 21:38

asked Jul 20 at 20:49

Shrey Joshi

1389

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
  – Poon Levi
  Jul 20 at 20:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
  – WW1
  Jul 20 at 21:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @above, fixed. $A_(0, 0)$ is not part of the sum.
  – Shrey Joshi
  Jul 21 at 0:43
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The Problem:

Denote by $V_(x, y)$ the vertex at ordered pair $(x, y)$ in the Cartesian coordinate system. Denote by $A_(x, y)$ the measure of angle $angle V_(x, y)V_(0, 0)V_(1, 0)$. Let
$$theta = sum_0 le i, j le 5 (i, j) neq (0, 0)A_(i, j)$$
Find $theta pmod2pi$.

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum
$$sum_0 le i, j le 5 (i, j) neq (0, 0)arctanfrac ij$$
But I don't know how to evaluate this. $arctan(x+y)$ formula fails. I think there is a different approach I am not aware of.

Is there a nice way to generalize to
$$sum_0 le i, j le k (i, j) neq (0, 0)A_(i, j)$$

trigonometry analytic-geometry angle

share|cite|improve this question

edited Jul 20 at 21:38

asked Jul 20 at 20:49

Shrey Joshi

1389

The Problem:

Denote by $V_(x, y)$ the vertex at ordered pair $(x, y)$ in the Cartesian coordinate system. Denote by $A_(x, y)$ the measure of angle $angle V_(x, y)V_(0, 0)V_(1, 0)$. Let
$$theta = sum_0 le i, j le 5 (i, j) neq (0, 0)A_(i, j)$$
Find $theta pmod2pi$.

I understand the question and everything, but I am slightly overwhelmed about how to find an organized approach to computing the sum of angles. I got the value down to the sum
$$sum_0 le i, j le 5 (i, j) neq (0, 0)arctanfrac ij$$
But I don't know how to evaluate this. $arctan(x+y)$ formula fails. I think there is a different approach I am not aware of.

Is there a nice way to generalize to
$$sum_0 le i, j le k (i, j) neq (0, 0)A_(i, j)$$

trigonometry analytic-geometry angle

share|cite|improve this question

edited Jul 20 at 21:38

asked Jul 20 at 20:49

Shrey Joshi

1389

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 21:38

edited Jul 20 at 21:38

edited Jul 20 at 21:38

asked Jul 20 at 20:49

Shrey Joshi

1389

asked Jul 20 at 20:49

Shrey Joshi

1389

asked Jul 20 at 20:49

Shrey Joshi

1389

1389

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
  – Poon Levi
  Jul 20 at 20:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
  – WW1
  Jul 20 at 21:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @above, fixed. $A_(0, 0)$ is not part of the sum.
  – Shrey Joshi
  Jul 21 at 0:43
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
  – Poon Levi
  Jul 20 at 20:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
  – WW1
  Jul 20 at 21:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @above, fixed. $A_(0, 0)$ is not part of the sum.
  – Shrey Joshi
  Jul 21 at 0:43
  
  

  
  
  
  

1

1

Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
– Poon Levi
Jul 20 at 20:54

Hint: find a relation between $A_(i,j)$ and $A_(j,i)$
– Poon Levi
Jul 20 at 20:54

1

1

$A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
– WW1
Jul 20 at 21:10

$A(0,0)$ doesn't seem to be well defined but appears to be part of the sum
– WW1
Jul 20 at 21:10

@above, fixed. $A_(0, 0)$ is not part of the sum.
– Shrey Joshi
Jul 21 at 0:43

@above, fixed. $A_(0, 0)$ is not part of the sum.
– Shrey Joshi
Jul 21 at 0:43

add a comment | 

1 Answer
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I got it now, Note that if we represent $A_m, n+A_n, m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $kpi/4$. $$(k(k+1)/2)(pi/2)+kpi/4=frack(k+2)pi4$$ Which is the general solution. Substituting $k=5$ gives $35pi/4 equiv 3pi/4 pmod2pi$

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

add a comment | 

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1 Answer
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I got it now, Note that if we represent $A_m, n+A_n, m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $kpi/4$. $$(k(k+1)/2)(pi/2)+kpi/4=frack(k+2)pi4$$ Which is the general solution. Substituting $k=5$ gives $35pi/4 equiv 3pi/4 pmod2pi$

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

add a comment | 

up vote
0
down vote

I got it now, Note that if we represent $A_m, n+A_n, m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $kpi/4$. $$(k(k+1)/2)(pi/2)+kpi/4=frack(k+2)pi4$$ Which is the general solution. Substituting $k=5$ gives $35pi/4 equiv 3pi/4 pmod2pi$

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

add a comment | 

up vote
0
down vote

up vote
0
down vote

I got it now, Note that if we represent $A_m, n+A_n, m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $kpi/4$. $$(k(k+1)/2)(pi/2)+kpi/4=frack(k+2)pi4$$ Which is the general solution. Substituting $k=5$ gives $35pi/4 equiv 3pi/4 pmod2pi$

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

I got it now, Note that if we represent $A_m, n+A_n, m$ in terms of complex numbers we get $arg(m+ni)(n+mi)=pi/2$. There are $k(k+1)$ of these numbers (ignnoring diagonals so we get $(k(k+1)/2)(pi/2)$. Adding back the diagonal (ignoring $(0, 0)$) we get $kpi/4$. $$(k(k+1)/2)(pi/2)+kpi/4=frack(k+2)pi4$$ Which is the general solution. Substituting $k=5$ gives $35pi/4 equiv 3pi/4 pmod2pi$

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 17:36

Shrey Joshi

1389

answered Jul 22 at 17:36

Shrey Joshi

1389

answered Jul 22 at 17:36

Shrey Joshi

1389

1389

add a comment | 

add a comment | 

 

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