Mixing
Sets of 3 integers adding to 180
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If we constrain angles to whole numbers, how many triangles can we make? This is also equivalent to: how many sets of 3 integers add to 180?
combinations
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
$endgroup$
add a comment |
$begingroup$
If we constrain angles to whole numbers, how many triangles can we make? This is also equivalent to: how many sets of 3 integers add to 180?
combinations
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
$endgroup$
4
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08
add a comment |
$begingroup$
If we constrain angles to whole numbers, how many triangles can we make? This is also equivalent to: how many sets of 3 integers add to 180?
combinations
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
$endgroup$
If we constrain angles to whole numbers, how many triangles can we make? This is also equivalent to: how many sets of 3 integers add to 180?
combinations
combinations
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
asked Dec 4 '15 at 15:00
D. McIntyreD. McIntyre
61
61
4
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08
add a comment |
4
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08
4
4
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Assuming every angle has degree > 0. First we need to calculate the number of ways to distribute 177 degrees over 3 angles:
$$binom{177+3-1}{177}=binom{179}{177} = 15931.$$
If you consider symmetrical triangles the same, you have to make sure you subtract the number of permutations. The number of triangles with identical angles is 1 ( 60,60,60). The number of triangles with only two identical angles $b$ and one angle $a(aneq b$) is $3*88$(89 even integers between 2 and 178 minus one for $a=60$, times $3$ to account for permutations). That leaves
$$15931 - 1 - 3*88 = 15666$$
triangles with three different angles. Those triangles have $3!=6$ permutations, thus
$$15666/6 + 88 + 1 = 2700$$
is the number of triangles if you count symmetrical triangles as one.
answered Dec 4 '15 at 17:21
danwindanwin
807
$endgroup$
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
add a comment |
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$begingroup$
Assuming every angle has degree > 0. First we need to calculate the number of ways to distribute 177 degrees over 3 angles:
$$binom{177+3-1}{177}=binom{179}{177} = 15931.$$
If you consider symmetrical triangles the same, you have to make sure you subtract the number of permutations. The number of triangles with identical angles is 1 ( 60,60,60). The number of triangles with only two identical angles $b$ and one angle $a(aneq b$) is $3*88$(89 even integers between 2 and 178 minus one for $a=60$, times $3$ to account for permutations). That leaves
$$15931 - 1 - 3*88 = 15666$$
triangles with three different angles. Those triangles have $3!=6$ permutations, thus
$$15666/6 + 88 + 1 = 2700$$
is the number of triangles if you count symmetrical triangles as one.
answered Dec 4 '15 at 17:21
danwindanwin
807
$endgroup$
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
add a comment |
$begingroup$
Assuming every angle has degree > 0. First we need to calculate the number of ways to distribute 177 degrees over 3 angles:
$$binom{177+3-1}{177}=binom{179}{177} = 15931.$$
If you consider symmetrical triangles the same, you have to make sure you subtract the number of permutations. The number of triangles with identical angles is 1 ( 60,60,60). The number of triangles with only two identical angles $b$ and one angle $a(aneq b$) is $3*88$(89 even integers between 2 and 178 minus one for $a=60$, times $3$ to account for permutations). That leaves
$$15931 - 1 - 3*88 = 15666$$
triangles with three different angles. Those triangles have $3!=6$ permutations, thus
$$15666/6 + 88 + 1 = 2700$$
is the number of triangles if you count symmetrical triangles as one.
answered Dec 4 '15 at 17:21
danwindanwin
807
$endgroup$
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
add a comment |
$begingroup$
Assuming every angle has degree > 0. First we need to calculate the number of ways to distribute 177 degrees over 3 angles:
$$binom{177+3-1}{177}=binom{179}{177} = 15931.$$
If you consider symmetrical triangles the same, you have to make sure you subtract the number of permutations. The number of triangles with identical angles is 1 ( 60,60,60). The number of triangles with only two identical angles $b$ and one angle $a(aneq b$) is $3*88$(89 even integers between 2 and 178 minus one for $a=60$, times $3$ to account for permutations). That leaves
$$15931 - 1 - 3*88 = 15666$$
triangles with three different angles. Those triangles have $3!=6$ permutations, thus
$$15666/6 + 88 + 1 = 2700$$
is the number of triangles if you count symmetrical triangles as one.
answered Dec 4 '15 at 17:21
danwindanwin
807
$endgroup$
Assuming every angle has degree > 0. First we need to calculate the number of ways to distribute 177 degrees over 3 angles:
$$binom{177+3-1}{177}=binom{179}{177} = 15931.$$
If you consider symmetrical triangles the same, you have to make sure you subtract the number of permutations. The number of triangles with identical angles is 1 ( 60,60,60). The number of triangles with only two identical angles $b$ and one angle $a(aneq b$) is $3*88$(89 even integers between 2 and 178 minus one for $a=60$, times $3$ to account for permutations). That leaves
$$15931 - 1 - 3*88 = 15666$$
triangles with three different angles. Those triangles have $3!=6$ permutations, thus
$$15666/6 + 88 + 1 = 2700$$
is the number of triangles if you count symmetrical triangles as one.
answered Dec 4 '15 at 17:21
danwindanwin
807
answered Dec 4 '15 at 17:21
danwindanwin
807
answered Dec 4 '15 at 17:21
danwindanwin
807
answered Dec 4 '15 at 17:21
danwindanwin
807
807
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
add a comment |
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
1
1
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
$begingroup$
I used more of brute force method - examining a case where 1 is an angle and going through the 89 cases with it, then setting 2 as an angle (but subtracting 1 combination since it had been done in the 1 case). Doing this yields a pattern where the even cases are linearly related and the odds are also linearly related. Treating each as an arithmetic series and adding them together yielded 2700 as well. Not quite as elegant, but it worked!
$endgroup$
– D. McIntyre
Dec 5 '15 at 23:40
add a comment |
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4
$begingroup$
Do you consider symmetrical triangles the same? Or order of angles may be important?
$endgroup$
– Abstraction
Dec 4 '15 at 15:03
$begingroup$
This is a partition-problem. And the integers must be positive.
$endgroup$
– Peter
Dec 4 '15 at 15:04
$begingroup$
Three positive integers, maybe? Otherwise, the number is infinite.
$endgroup$
– Thomas Andrews
Dec 4 '15 at 15:08