I first discovered this problem in a posting to the NMBRTHRY listserver from Jim Buddenhagen.
Assume that the sides of a triangle are a,b,c with corresponding opposite angles A,B,C.
Then the cosine rule gives
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|
Assume that tan(C/2) = n / d with n,d integers with no common factors. Then
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If we define a / g = j, c / g = j, p = d n j, q = d2 n2 k j, then
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This quartic has an obvious rational point, so is birationally equivalent to an elliptic curve. Using standard methods, and some algebraic manipulation, we find the family of curves
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with p = s / t and ±q = d n ( d2 - n2 ) + 2 t - s2 / t2.
As an example, let n = 2, d = 1. Then s = 3, t = -1
lies on the curve
s2 = t ( t -
8
) ( t + 2 ),
so p = 3, j = 3/2, q =
17,
k = 17/6, giving
There are clearly 4 points of order 2, t = 0, t = d n3, t = -d3 n, and the point at infinity. I can show that there are no points of order 4, but am unable to prove that the torsion subgroup is Z2×Z2 and not Z2 ×Z6, though experimental evidence suggests this is true. Anybody out there help?
The 4 points of order 2 do not lead to a triangle, so, if these are the only torsion points, a triangle only exists if the rank of the curve is > 0.
The Birch and Swinnerton-Dyer conjecture was then
used
to estimate the rank for the cases
(a) 1 <= n <= 99 , d = 1 ,
(b) n = 1 , 1 <= d <= 99,
with the results in the following tables.
Rank of curve for 1 <= n <= 99 , d = 1
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 00+ | - | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 |
| 10+ | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| 20+ | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 30+ | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 40+ | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 50+ | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 60+ | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
| 70+ | 1 | 2 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
| 80+ | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
| 90+ | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 2 | 0 |
Rank of curve for n = 1 , 1 <= d <= 99
| d | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 00+ | - | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 |
| 10+ | 0 | 0 | 0 | 0 | 0 | 2 | 1 | 0 | 1 | 0 |
| 20+ | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 |
| 30+ | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 2 |
| 40+ | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 50+ | 1 | 0 | 0 | 0 | 0 | 3 | 1 | 0 | 1 | 0 |
| 60+ | 0 | 1 | 1 | 1 | 2 | 1 | 0 | 1 | 0 | 1 |
| 70+ | 0 | 2 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 80+ | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 2 |
| 90+ | 0 | 1 | 0 | 1 | 0 | 2 | 0 | 0 | 0 | 1 |
For most of the curves of rank > 0, a rational point has been found, and hence a triangle satisfying the criteria of the problem. Readers interested in these solutions can find the results in the following links (zero values indicate no known solution):
(a) results for 1 <= n <= 99 , d = 1 ,
(b) results for n = 1 , 1
<=
d <= 99.
It should be pointed out that there are 9 unsolved
curves
in the first group, but only 2 in the second. Some of the predicted
heights
of the rational points in the first group are huge. The list of
unsolved
curves is in the following table.
| n | d | Height estimate | Information | |
| 1 | 1 | 87 | 69.1 | |
| 2 |
1 | 93 | 73.2 | |
| 3 |
92 | 1 | 101.5 | |
| 4 |
88 | 1 | 106.7 | |
| 5 |
91 | 1 | 144.0 | |
| 6 |
85 | 1 | 184.3 | |
| 7 |
82 | 1 | 186.6 | |
| 8 |
67 | 1 | 360.6 | |
| 9 |
74 | 1 | 435.2 | |
| 10 | 94 | 1 | 440.5 | |
| 11 | 61 | 1 | 470.6 |