[math-fun] Rohan notices that integer 60º triangles come in pairs:
On 2018-07-12 05:57, Fred Lunnon wrote:
"Two sides and a non-included angle" --- the one case when a triangle is not uniquely specified by three of its six angles and sides.
Knew that would prove useful some day ...
WFL
The pons ASSinorum.
On 7/12/18, Allan Wechsler <acwacw@gmail.com> wrote:
This makes perfect sense. Notice that if you take the sum of the varying values, you get the longest side: 3+5 = 8, 7+8 = 15, 5+16 = 21.
In other words, the "average length" of the varying side is half the long side.
This all becomes completely clear if you imagine putting a compass at the vertex between the two constant sides, and opening it to the length of
the
"hypotenuse" (which is, counterintuitively, the side of median length, not the longest side, which is a leg). When you sweep the compass, it cuts the opposite side in two places, producing the two lengths of the variable side.
The Pythagoroid Formula for the sixty-degree case is A^2 + B^2 = C^2 + AB. So, 64 + 9 = 49 + 24 for the first triple, and 64 + 25 = 49 + 40 for the second.
But no parametric solution, like with true Pythagorean? --rwg
On Thu, Jul 12, 2018 at 7:40 AM, Bill Gosper <billgosper@gmail.com>
wrote:
{7 | 8 | 3, 7 | 8 | 5, 13 | 15 | 7, 13 | 15 | 8, 19 | 21 | 5, 19 | 21 | 16, 31 | 35 | 11, 31 | 35 | 24, . . ., 859 | 989 | 429, 859 | 989 | 560, 889 | 999 | 295, 889 | 999 | 704,. . . Sometimes very close pairs: 181 | 209 | 104, 181 | 209 | 105, . . ., 2521 | 2911 | 1455, 2521 | 2911 | 1456, . . . --rwg
But no parametric solution, like with true Pythagorean?
Tom Rokicki had already found one! In[373]:= Block[{a = x^2 + 2 x y, b = y^2 + 2 x y, c = x^2 + y^2 + x y}, c^2 == a^2 + b^2 - a b // Simplify] Out[373]= True How about c^2 == a^2 + b^2 + a b? In[376]:= Block[{a = -x^2 - 2 x y, b = y^2 + 2 x y, c = x^2 + y^2 + x y}, c^2 == a^2 + b^2 + a b // Simplify] Out[376]= True Hmm. --rwg On Thu, Jul 12, 2018 at 11:40 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2018-07-12 05:57, Fred Lunnon wrote:
"Two sides and a non-included angle" --- the one case when a triangle is not uniquely specified by three of its six angles and sides.
Knew that would prove useful some day ...
WFL
The pons ASSinorum.
On 7/12/18, Allan Wechsler <acwacw@gmail.com> wrote:
This makes perfect sense. Notice that if you take the sum of the varying values, you get the longest side: 3+5 = 8, 7+8 = 15, 5+16 = 21.
In other words, the "average length" of the varying side is half the
long
side.
This all becomes completely clear if you imagine putting a compass at the vertex between the two constant sides, and opening it to the length of the "hypotenuse" (which is, counterintuitively, the side of median length, not the longest side, which is a leg). When you sweep the compass, it cuts the opposite side in two places, producing the two lengths of the variable side.
The Pythagoroid Formula for the sixty-degree case is A^2 + B^2 = C^2 + AB. So, 64 + 9 = 49 + 24 for the first triple, and 64 + 25 = 49 + 40 for the second.
But no parametric solution, like with true Pythagorean? --rwg
On Thu, Jul 12, 2018 at 7:40 AM, Bill Gosper <billgosper@gmail.com>
wrote:
{7 | 8 | 3, 7 | 8 | 5, 13 | 15 | 7, 13 | 15 | 8, 19 | 21 | 5, 19 | 21 | 16, 31 | 35 | 11, 31 | 35 | 24, . . ., 859 | 989 | 429, 859 | 989 | 560, 889 | 999 | 295, 889 | 999 | 704,. .
.
Sometimes very close pairs: 181 | 209 | 104, 181 | 209 | 105, . . ., 2521 | 2911 | 1455, 2521 | 2911 | 1456, . . . --rwg
Yeah, the second one you found works for 120 degree integer triangles. There are other parameterizations that may be easier to work with. For 60 degree triangles, the three angles form an arithmetic progression; these are the only triangles for which this is true. On Fri, Jul 13, 2018 at 11:20 AM Bill Gosper <billgosper@gmail.com> wrote:
But no parametric solution, like with true Pythagorean?
Tom Rokicki had already found one! In[373]:= Block[{a = x^2 + 2 x y, b = y^2 + 2 x y, c = x^2 + y^2 + x y}, c^2 == a^2 + b^2 - a b // Simplify]
Out[373]= True
How about c^2 == a^2 + b^2 + a b? In[376]:= Block[{a = -x^2 - 2 x y, b = y^2 + 2 x y, c = x^2 + y^2 + x y}, c^2 == a^2 + b^2 + a b // Simplify]
Out[376]= True
Hmm. --rwg
On Thu, Jul 12, 2018 at 11:40 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2018-07-12 05:57, Fred Lunnon wrote:
"Two sides and a non-included angle" --- the one case when a triangle is not uniquely specified by three of its six angles and sides.
Knew that would prove useful some day ...
WFL
The pons ASSinorum.
On 7/12/18, Allan Wechsler <acwacw@gmail.com> wrote:
This makes perfect sense. Notice that if you take the sum of the
varying
values, you get the longest side: 3+5 = 8, 7+8 = 15, 5+16 = 21.
In other words, the "average length" of the varying side is half the long side.
This all becomes completely clear if you imagine putting a compass at the vertex between the two constant sides, and opening it to the length of the "hypotenuse" (which is, counterintuitively, the side of median length, not the longest side, which is a leg). When you sweep the compass, it cuts the opposite side in two places, producing the two lengths of the variable side.
The Pythagoroid Formula for the sixty-degree case is A^2 + B^2 = C^2 + AB. So, 64 + 9 = 49 + 24 for the first triple, and 64 + 25 = 49 + 40 for the second.
But no parametric solution, like with true Pythagorean? --rwg
On Thu, Jul 12, 2018 at 7:40 AM, Bill Gosper <billgosper@gmail.com>
wrote:
{7 | 8 | 3, 7 | 8 | 5, 13 | 15 | 7, 13 | 15 | 8, 19 | 21 | 5, 19 |
21 |
16, 31 | 35 | 11, 31 | 35 | 24, . . ., 859 | 989 | 429, 859 | 989 | 560, 889 | 999 | 295, 889 | 999 | 704,. . . Sometimes very close pairs: 181 | 209 | 104, 181 | 209 | 105, . . ., 2521 | 2911 | 1455, 2521 | 2911 | 1456, . . . --rwg
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participants (2)
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Bill Gosper -
Tomas Rokicki