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A336328
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Primitive triples for integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.
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11
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57, 65, 73, 73, 88, 95, 43, 147, 152, 127, 168, 205, 97, 185, 208, 111, 221, 280, 49, 285, 296, 95, 312, 343, 296, 315, 361, 152, 343, 387, 323, 392, 407, 147, 377, 437, 285, 464, 469, 255, 343, 473, 247, 408, 485, 469, 589, 624, 403, 725, 728, 871, 901, 931
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OFFSET
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1,1
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COMMENTS
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Inspired by Project Euler, Problem 143 (see link).
The triples are displayed in increasing order of largest side c, and if largest sides coincide then by increasing order of the middle side b; so, each triple (a, b, c) is in increasing order.
If one angle of the triangle, for example C, is >= 2*Pi/3 then the Fermat point F is this vertex C, so, FA + FB + FC becomes CA + CB, while when all angles are < 2*Pi/3, then the Fermat point is inside the triangle (see link Fermat points), this last condition means that c^2 < a^2 + a*b + b^2.
As a < b < c, then FA > FB > FC.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d: 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2 (see Martin Gardner).
Equivalently: if a point M is inside an equilateral triangle A'B'C' and integer distances to vertices are MA' = a = A072054(n), MB' = b = A072053(n), MC' = c = A072052(n), then the side of this equilateral triangle A'B'C' is equal to d = FA + FB + FC = A061281(n) where F is the Fermat point of the triangle ABC with sides (a,b,c) (see Martin Gardner).
+-----+-----+-----+-----------+-----------+-----------+-----+-------+
| a | b | c | FA | FB | FC | d | a+b+c |
+-----------+-----+-----------+-----------+-----------+-----+-------+
| 57 | 65 | 73 | 325/7 | 264/7 | 195/7 | 112 | 195 |
| 73 | 88 | 95 | 440/7 | 325/7 | 264/7 | 147 | 256 |
| 43 | 147 | 152 | 5016/37 | 1064/37 | 765/37 | 185 | 342 |
| 127 | 168 | 205 | 39360/283 | 27265/283 | 13464/283 | 283 | 500 |
| 97 | 185 | 208 | 14800/91 | 6528/91 | 3515/91 | 273 | 490 |
| 111 | 221 | 280 | 70720/331 | 34200/331 | 4641/331 | 331 | 612 |
| 49 | 285 | 296 | 91200/331 | 12376/331 | 5985/331 | 331 | 630 |
| 95 | 312 | 343 | 3864/13 | 1015/13 | 360/13 | 403 | 750 |
| 296 | 315 | 361 | 9405/43 | 8512/43 | 6120/43 | 559 | 972 |
| 152 | 343 | 387 | 30429/97 | 11520/97 | 5096/97 | 485 | 882 |
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From the previous table, we observe that every FA, FB, FC is a fraction while FA + FB + FC = d is an integer (A336329). Jinyuan Wang has found that the 37th triple is the first for which the common denominator of these fractions is 1 (A351477).
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REFERENCES
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Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
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LINKS
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FORMULA
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If FA + FB + FC = d, then
d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))), or,
d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.
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EXAMPLE
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The table begins:
57, 65, 73;
73, 88, 95;
43, 147, 152;
127, 168, 205;
97, 185, 208;
111, 221, 280;
49, 285, 296;
.............
For first triple (57, 65, 73) and corresponding d = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, relation gives: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.
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CROSSREFS
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Cf. A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), A351802 (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other FA + FB + FC).
Cf. A333391 (with isogonic center).
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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