Find number of triangles with integral sides and side lengths ≤ 2n
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
add a comment |
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
$begingroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
$endgroup$
Find number of triangles with integral sides and side lengths less than or equal to $2n$.
I approached this method by recursion.
Say $A_{2n} $is the number of triangles with integral sides and side lengths less than or equal to $2n$.
$A_{2n-1} $ is the number of triangles with integral sides and side lengths less than or equal to $2n-1$ .
So $A_{2n}=A_{2n-1}+left( text{number of triangles having at least one side equal to} 2n right) .$
How to count the number of triangles having at least one side equal to $2n$?
Also, is there any other better method to this other than generalization as well?
geometry combinations recursion
geometry combinations recursion
edited Dec 17 '18 at 14:43
Word Shallow
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
edited Dec 17 '18 at 14:43
Word Shallow
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
edited Dec 17 '18 at 14:43
Word Shallow
768518
edited Dec 17 '18 at 14:43
Word Shallow
768518
edited Dec 17 '18 at 14:43
Word Shallow
768518
768518
asked Dec 17 '18 at 12:49
tanyatanya
105
asked Dec 17 '18 at 12:49
tanyatanya
105
asked Dec 17 '18 at 12:49
tanyatanya
105
105
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
$endgroup$
add a comment |
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
$endgroup$
add a comment |
$begingroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
$endgroup$
Number of triangles having at least one side equal to $2n$ is the number of couple $(a,b)$ with $age b$ and $a+b > 2n$. That is,
$$begin{aligned}
sum^{2n}_{b=1}sum^{2n}_{a = max(b,2n + 1-b)} 1 &= sum^{2n}_{b=1} (2n+1 - max(b,2n+1-b))\
&=sum^{2n}_{b=n+1} ((2n+1) - b) + sum^{n}_{b=1} (2n+1 - (2n+1-b))\
&=sum^{1}_{c=n}c + sum^{n}_{b=1} bhspace{5em}(text{where } c = 2n+1-b)\
&= n(n+1)
end{aligned}$$
In order to argue recursively, you would need to compute $A_{2n-1}$ in similar fashion as well.
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
answered Dec 17 '18 at 15:55
Quang HoangQuang Hoang
12.9k1132
12.9k1132
add a comment |
add a comment |
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$begingroup$
Perhaps it would be easier to count up instead of counting down...
$endgroup$
– abiessu
Dec 17 '18 at 12:54
$begingroup$
Welcome to Math.SE! Please use MathJax.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 17 '18 at 12:55