Return to Answer

One solution thus far: 323,392, and 645 for the distances between points of different triangles (red), 407 and 713 for the equilaterals' edge lengths. The two triangles are rotated relative to each other by arccos(421/1147)

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the Diophantine equations), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter, using one edge for one of the equilateral triangles, and the remaining vertex to locate a corner of the other equilateral. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below.... the rotation angle yields a t of +/- 28/11, which mercifully seems to check out 'by hand'.

Per their suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest? Is there anything else special about this solution, which might suggest where to look for others? I eyeballed the ratio of the larger and smaller triangles to be suspiciously close to a relevant irrational, but looking closer reducing the larger a few steps would be a better approximation.

One solution thus far: 323,392, and 645 for the distances between points of different triangles (red), 407 and 713 for the equilaterals' edge lengths. The two triangles are rotated relative to each other by arccos(421/1147)

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the prospect of solving the Diophantine equations analytically), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter, using one edge for one of the equilateral triangles, and the remaining vertex to locate a corner of the other equilateral. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below.... the rotation angle yields a t of +/- 28/11, which mercifully seems to check out 'by hand'.

Per their suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest? Is there anything else special about this solution, which might suggest where to look for others? I eyeballed the ratio of the larger and smaller triangles to be suspiciously close to a relevant irrational, but looking closer reducing the larger a few steps would be a better approximation.

One solution thus far: 323,392, and 645 for the distances between points of different triangles (red), 407 and 713 for the equilaterals' edge lengths. The two triangles are rotated relative to each other by arccos(421/1147)

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the Diophantine equations), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below.... the rotation angle yields a t of +/- 28/11, which mercifully seems to check out 'by hand'.

Per their suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest? Is there anything else special about this solution, which might suggest where to look for others? I eyeballed the ratio of the larger and smaller triangles to be suspiciously close to a relevant irrational, but looking closer reducing the larger a few steps would be a better approximation.

One solution thus far: 323,392, and 645 for the distances between points of different triangles (red), 407 and 713 for the equilaterals' edge lengths. The two triangles are rotated relative to each other by arccos(421/1147)

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the Diophantine equations), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter, using one edge for one of the equilateral triangles, and the remaining vertex to locate a corner of the other equilateral. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below.... the rotation angle yields a t of +/- 28/11, which mercifully seems to check out 'by hand'.

Per their suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest? Is there anything else special about this solution, which might suggest where to look for others? I eyeballed the ratio of the larger and smaller triangles to be suspiciously close to a relevant irrational, but looking closer reducing the larger a few steps would be a better approximation.

323 392 645 407 713

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the Diophantine equations), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below....

Per his suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest?

One solution thus far: 323,392, and 645 for the distances between points of different triangles (red), 407 and 713 for the equilaterals' edge lengths. The two triangles are rotated relative to each other by arccos(421/1147)

After briefly checking pairs of Eisenstein integers (and even more briefly looking in horror at the Diophantine equations), I decided to force three of the five lengths to rational by iterating over integer triangles of increasing perimeter. Then, using the law of cosines and sagemath's complex numbers, I solved for the locations of the remaining points and the two remaining distances directly. I think I was using a few too many unexpected steps for the symbolic processor to handle (it didn't report back sqrt(49) as a rational) so I used complex double precision...I didn't dare use less as there were some amazingly close misses for such small integers. For example, plugging in 54,146 for the two red sides and 140 for the black equilateral edge gives 186.999999078213 and 198.999999133798 for the other two lengths.

I actually stopped my search just shy of the first success, figuring 1000 as a perimeter of the smallest red-red-black triangle was as good a time to optimize as any. Using the identity recently mentioned below (sum of squares of red edges=sum of squares of triangle lengths) didn't require any non-integer math at all, so I built a lookup table of sums of distinct nonzero squares and switched to forcing the three red edges to integers, culling the overwhelming majority of cases without having to invoke complex math. Also, I recognized that the three triangles formed with two red edges and one of the equilateral legs had an exploitable relationship: of the three angles where two red edges meet, two of them sum to the third. Three acos calls were still a lot faster than solving for all the points, and filtered out even more candidates, so cross-checking my results up through 1000 only took a few minutes now. I finally pushed past 1000 and found the above pentad shortly thereafter. Code available upon request, but it's pretty messy :p Now to use the resulting value of t to make sure it satisfies WhatsUp's Diophantine eq. below.... the rotation angle yields a t of +/- 28/11, which mercifully seems to check out 'by hand'.

Per their suggestion, I really would like to see what the professionals at mathoverflow can do with this. Is there some reason why there seem to be more 'near misses' than a uniform distribution of sums of square roots would suggest? Is there anything else special about this solution, which might suggest where to look for others? I eyeballed the ratio of the larger and smaller triangles to be suspiciously close to a relevant irrational, but looking closer reducing the larger a few steps would be a better approximation.