Mixing
Number of isosceles triangle.
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Prove that the number of isosceles triangles with integer sides, no side exceeding n, is $ frac{1}{4}(1+3n^2)$ or $frac{3}{4}(n^2)$ according to whether n is odd or even.
I am able to count the number of isosceles triangles which have the length of equal sides greater than or equal to length of base. I cannot count the remaining triangles.
combinatorics
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
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add a comment |
$begingroup$
Prove that the number of isosceles triangles with integer sides, no side exceeding n, is $ frac{1}{4}(1+3n^2)$ or $frac{3}{4}(n^2)$ according to whether n is odd or even.
I am able to count the number of isosceles triangles which have the length of equal sides greater than or equal to length of base. I cannot count the remaining triangles.
combinatorics
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
$endgroup$
$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
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– Arthur
Oct 28 '15 at 14:53
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57
add a comment |
$begingroup$
Prove that the number of isosceles triangles with integer sides, no side exceeding n, is $ frac{1}{4}(1+3n^2)$ or $frac{3}{4}(n^2)$ according to whether n is odd or even.
I am able to count the number of isosceles triangles which have the length of equal sides greater than or equal to length of base. I cannot count the remaining triangles.
combinatorics
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
$endgroup$
Prove that the number of isosceles triangles with integer sides, no side exceeding n, is $ frac{1}{4}(1+3n^2)$ or $frac{3}{4}(n^2)$ according to whether n is odd or even.
I am able to count the number of isosceles triangles which have the length of equal sides greater than or equal to length of base. I cannot count the remaining triangles.
combinatorics
combinatorics
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
edited Oct 28 '15 at 14:51
user940
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
asked Oct 28 '15 at 14:43
doubts_94doubts_94
204
204
$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
$endgroup$
– Arthur
Oct 28 '15 at 14:53
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57
add a comment |
$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
$endgroup$
– Arthur
Oct 28 '15 at 14:53
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57
$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
$endgroup$
– Arthur
Oct 28 '15 at 14:53
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
$endgroup$
– Arthur
Oct 28 '15 at 14:53
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If the base is $2i-1$ for some $igeq 1$ (i.e. odd), then there are $n-i + 1$ isosceles (non-degenerate) triangles. If the base is $2j$ (i.e. even) for some $j geq 1$, there are $n-j$ isosceles triangles. Also note that there are $lceil n/2rceil$ odd numbers and $lfloor n/2rfloor$ even numbers that are $leq n$ (and positive). That means that we sum up
$$
overbrace{sum_{i = 1}^{lceil n/2rceil}(n-i + 1)}^{text{Odd bases}} + overbrace{sum_{j = 1}^{lfloor n/2rfloor} (n - j)}^{text{Even bases}}
$$
It should be quite easy from here to divide up into $n$ even or odd and do the direct calculations from there using the sum of arithmetic series.
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
$endgroup$
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If the base is $2i-1$ for some $igeq 1$ (i.e. odd), then there are $n-i + 1$ isosceles (non-degenerate) triangles. If the base is $2j$ (i.e. even) for some $j geq 1$, there are $n-j$ isosceles triangles. Also note that there are $lceil n/2rceil$ odd numbers and $lfloor n/2rfloor$ even numbers that are $leq n$ (and positive). That means that we sum up
$$
overbrace{sum_{i = 1}^{lceil n/2rceil}(n-i + 1)}^{text{Odd bases}} + overbrace{sum_{j = 1}^{lfloor n/2rfloor} (n - j)}^{text{Even bases}}
$$
It should be quite easy from here to divide up into $n$ even or odd and do the direct calculations from there using the sum of arithmetic series.
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
$endgroup$
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
add a comment |
$begingroup$
If the base is $2i-1$ for some $igeq 1$ (i.e. odd), then there are $n-i + 1$ isosceles (non-degenerate) triangles. If the base is $2j$ (i.e. even) for some $j geq 1$, there are $n-j$ isosceles triangles. Also note that there are $lceil n/2rceil$ odd numbers and $lfloor n/2rfloor$ even numbers that are $leq n$ (and positive). That means that we sum up
$$
overbrace{sum_{i = 1}^{lceil n/2rceil}(n-i + 1)}^{text{Odd bases}} + overbrace{sum_{j = 1}^{lfloor n/2rfloor} (n - j)}^{text{Even bases}}
$$
It should be quite easy from here to divide up into $n$ even or odd and do the direct calculations from there using the sum of arithmetic series.
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
$endgroup$
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
add a comment |
$begingroup$
If the base is $2i-1$ for some $igeq 1$ (i.e. odd), then there are $n-i + 1$ isosceles (non-degenerate) triangles. If the base is $2j$ (i.e. even) for some $j geq 1$, there are $n-j$ isosceles triangles. Also note that there are $lceil n/2rceil$ odd numbers and $lfloor n/2rfloor$ even numbers that are $leq n$ (and positive). That means that we sum up
$$
overbrace{sum_{i = 1}^{lceil n/2rceil}(n-i + 1)}^{text{Odd bases}} + overbrace{sum_{j = 1}^{lfloor n/2rfloor} (n - j)}^{text{Even bases}}
$$
It should be quite easy from here to divide up into $n$ even or odd and do the direct calculations from there using the sum of arithmetic series.
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
$endgroup$
If the base is $2i-1$ for some $igeq 1$ (i.e. odd), then there are $n-i + 1$ isosceles (non-degenerate) triangles. If the base is $2j$ (i.e. even) for some $j geq 1$, there are $n-j$ isosceles triangles. Also note that there are $lceil n/2rceil$ odd numbers and $lfloor n/2rfloor$ even numbers that are $leq n$ (and positive). That means that we sum up
$$
overbrace{sum_{i = 1}^{lceil n/2rceil}(n-i + 1)}^{text{Odd bases}} + overbrace{sum_{j = 1}^{lfloor n/2rfloor} (n - j)}^{text{Even bases}}
$$
It should be quite easy from here to divide up into $n$ even or odd and do the direct calculations from there using the sum of arithmetic series.
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
answered Oct 28 '15 at 15:08
ArthurArthur
117k7116200
117k7116200
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
add a comment |
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
$begingroup$
That does do the trick. Elegant approach. Cheers.
$endgroup$
– doubts_94
Oct 28 '15 at 15:19
add a comment |
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$begingroup$
Will keep in mind.
$endgroup$
– doubts_94
Oct 28 '15 at 14:50
$begingroup$
Do you count degenerate triangles (i.e. with sides $2, 1$ and $1$)? Also, rather than "longer or shorter base", I think it's easier to divide into "even or odd base".
$endgroup$
– Arthur
Oct 28 '15 at 14:53
$begingroup$
The question does not explicitly mention them. I am not very convinced they should be considered. I' ll try the even/odd approach.
$endgroup$
– doubts_94
Oct 28 '15 at 14:57