One can state the original conjecture in terms of loci as follows: fix the first five vertices and denote the sixth by $P=(x,y)$. Consider the two loci

1. those $P$ for which the concurrency holds;

2. those $P$ for which the sine condition holds.

The conjecture that these two loci coincide is, as shown above, false--the first locus is a proper subset of the second. One can make this result more precise as follows: the first locus is a straight line (the one through $A_5$ and the intersection of $A_1A_4$ and $A_2A_5$). The second is the union of this line and a conic section.

The proof is computational. One calculates the area of the triangle carved out by the three main diagonals (it is a linear form in $x,y$) and the difference of the two sine products (after simplification, this is a quartic). The latter quartic is the product of the square of the above linear form and a quadratic form. The required conic is the zero set of the latter.

One can state the original conjecture in terms of loci as follows: fix the first five vertices and denote the sixth by $P=(x,y)$. Consider the two loci

1. those $P$ for which the concurrency holds;

2. those $P$ for which the sine condition holds.

The conjecture that these two loci coincide is, as shown above, false--the first locus is a proper subset of the second. One can make this result more precise as follows: the first locus is a straight line (the one through $A_3$ and the intersection of $A_1A_4$ and $A_2A_5$). The second is the union of this line and a conic section.

The proof is computational. One calculates the area of the triangle carved out by the three main diagonals (it is a linear form in $x,y$) and the difference of the two sine products (after simplification, this is a quartic). The latter quartic is the product of the square of the above linear form and a quadratic form. The required conic is the zero set of the latter.