[math-fun] Update on IMO/RKG-inspired conics thru' 6 points of triangle
Notation: initial triangle T with corners P,Q,R; side lengths a,b,c; angles A,B,C; centre point S = (B_1 P + B_2 Q + B_3 R)/(B_1+B_2+B_3), where homogeneous barycentric coordinates B are some function of T; 3 cevian points PS.QR = U,V,W; 6 subsidiary triangles SPU etc. The 6 subsidiary centres appear to lie on a conic, for (all) triangle centres lying at the (1) centroid, B = [1,1,1]; (2) incentre, B = [a,b,c] or [sin A, ... ]; (3) e-centres, B = [-a,b,c], etc --- reliant on vertex ordering!; (4) circumcentre, B = [a^2(b^2+c^2-a^2), ... ] or [sin 2A, ... ]; (5) orthocentre B = [1/(b^2+c^2-a^2), ... ] or [tan A, ... ]; (6) first isogenic point X13, B = [a sec(A - pi/6), ... ]; In general (non-isosceles) they fail to lie on a conic for (7) symmedian point B = [a^2,b^2,c^2 or 1 - cos 2A, ... ]; (8) second isogenic point X14, B = [a sec(A + pi/6), ... ]; (9) fixed weight with B general constant vector (correcting earlier claim). The conjectures are based on evaluating the 6x6 determinant of quadratic monomials in the (projective) coordinate components. Fermat points (6),(7) cause particular difficulty, with the determinant evaluated only approximately: for example WFL's {12,17,25}-sided test triangle required at least 70 decimal-digit precision to confirm case (6), and case (7) was rejected with determinant ~ -2.538 E-13 . Fred Lunnon
Hi Fred, Please clarify a few points. Are you using the same kind of center for the initial triangle and for the six subsidiary centers? Or are the centers (1) ... (9) computed with respect to the centroid of the initial triangle? Did you do the calculation algebraically or numerically? Exactly what is it you are conjecturing, and what is quite certainly proved? Best wishes, Branko On Sep 24, 2011, at 8:48 PM, Fred lunnon wrote:
Notation: initial triangle T with corners P,Q,R; side lengths a,b,c; angles A,B,C; centre point S = (B_1 P + B_2 Q + B_3 R)/(B_1+B_2+B_3), where homogeneous barycentric coordinates B are some function of T; 3 cevian points PS.QR = U,V,W; 6 subsidiary triangles SPU etc.
The 6 subsidiary centres appear to lie on a conic, for (all) triangle centres lying at the (1) centroid, B = [1,1,1]; (2) incentre, B = [a,b,c] or [sin A, ... ]; (3) e-centres, B = [-a,b,c], etc --- reliant on vertex ordering!; (4) circumcentre, B = [a^2(b^2+c^2-a^2), ... ] or [sin 2A, ... ]; (5) orthocentre B = [1/(b^2+c^2-a^2), ... ] or [tan A, ... ]; (6) first isogenic point X13, B = [a sec(A - pi/6), ... ];
In general (non-isosceles) they fail to lie on a conic for (7) symmedian point B = [a^2,b^2,c^2 or 1 - cos 2A, ... ]; (8) second isogenic point X14, B = [a sec(A + pi/6), ... ]; (9) fixed weight with B general constant vector (correcting earlier claim).
The conjectures are based on evaluating the 6x6 determinant of quadratic monomials in the (projective) coordinate components. Fermat points (6),(7) cause particular difficulty, with the determinant evaluated only approximately: for example WFL's {12,17,25}-sided test triangle required at least 70 decimal-digit precision to confirm case (6), and case (7) was rejected with determinant ~ -2.538 E-13 .
Fred Lunnon
Dear Branko, Sorry about the delay --- I took a week's vacation on impulse. "Are you using the same kind of center for the initial triangle and for the six subsidiary centers? Or are the centers (1) ... (9) computed with respect to the centroid of the initial triangle?" The former: in my current investigation, all centres are of the same type. "Did you do the calculation algebraically or numerically?" All computations were exact, with the exception of Fermat / isogenic points. "Exactly what is it you are conjecturing, and what is quite certainly proved?" The negative results (no conic) have been proved via counterexample. The positive result for centroid was proved by Gene Salamin via a simple affine transformation argument. The other positive results are conjectural, supported only by a handful of sample computations. Yours, Fred On 9/25/11, Branko Grünbaum <grunbaum@math.washington.edu> wrote:
Hi Fred,
Please clarify a few points.
Are you using the same kind of center for the initial triangle and for the six subsidiary centers? Or are the centers (1) ... (9) computed with respect to the centroid of the initial triangle?
Did you do the calculation algebraically or numerically?
Exactly what is it you are conjecturing, and what is quite certainly proved?
Best wishes,
Branko
On Sep 24, 2011, at 8:48 PM, Fred lunnon wrote:
Notation: initial triangle T with corners P,Q,R; side lengths a,b,c; angles A,B,C; centre point S = (B_1 P + B_2 Q + B_3 R)/(B_1+B_2+B_3), where homogeneous barycentric coordinates B are some function of T; 3 cevian points PS.QR = U,V,W; 6 subsidiary triangles SPU etc.
The 6 subsidiary centres appear to lie on a conic, for (all) triangle centres lying at the (1) centroid, B = [1,1,1]; (2) incentre, B = [a,b,c] or [sin A, ... ]; (3) e-centres, B = [-a,b,c], etc --- reliant on vertex ordering!; (4) circumcentre, B = [a^2(b^2+c^2-a^2), ... ] or [sin 2A, ... ]; (5) orthocentre B = [1/(b^2+c^2-a^2), ... ] or [tan A, ... ]; (6) first isogenic point X13, B = [a sec(A - pi/6), ... ];
In general (non-isosceles) they fail to lie on a conic for (7) symmedian point B = [a^2,b^2,c^2 or 1 - cos 2A, ... ]; (8) second isogenic point X14, B = [a sec(A + pi/6), ... ]; (9) fixed weight with B general constant vector (correcting earlier claim).
The conjectures are based on evaluating the 6x6 determinant of quadratic monomials in the (projective) coordinate components. Fermat points (6),(7) cause particular difficulty, with the determinant evaluated only approximately: for example WFL's {12,17,25}-sided test triangle required at least 70 decimal-digit precision to confirm case (6), and case (7) was rejected with determinant ~ -2.538 E-13 .
Fred Lunnon
participants (2)
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Branko Grünbaum -
Fred lunnon