On 9/17/11, William Rex Marshall wrote:
One possible definition of "most scalene": The triangle which is worst at covering its own mirror image.
Let A(T) be the maximum possible area of intersection of a triangle T of area 1 and its mirror image T' (also of area 1) in the Euclidean plane. For which T is A(T) minimised?
Though I haven't attempted to prove this, it seems plausible to assume that the maximum overlap occurs when either (1) The longest edges of T,T' and both pairs of their vertices are superposed, with triangular overlap region. Suppose T has vertices (-c,0), (+c,0), (+a,b), and T' has (-c,0), (+c,0), (-a,b), where we could normalise to c = 1; squared edge-lengths in ascending order are b^2 + (a-c)^2, b^2 + (a+c)^2, (c+c)^2; the overlap ratio comes out to be c/(a+c). (2) Both pairs of longer edges of T,T', and their common vertices are superposed, with kite-shaped overlap region. Suppose T has vertices (0,v), (+u,0), (-s,t), and T' has (0,v), (-u,0), (+s,t), with s/u + t/v = 1; squared edge-lengths in ascending order are t^2 + (u+s)^2, s^2 + (v-t)^2, u^2+v^2; the overlap ratio comes out to be 2s/(u+s). For fixed b, as a increases towards more isosceles T, case(1) becomes inferior to case(2): solving equations yields the crossover at b^2 = 4/(1+2*a)^2 - (1+a)^2, giving complex b when a > (sqrt(17) - 3)/2. At the latter value the overlap ratio A(T) = 1/(a+1) is approximately 0.6403882031 which I conjecture is minimal; unfortunately, as happened also in Dan's analysis, since b = 0 here the corresponding "triangle" is again linear. Fred lunnon On 9/17/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Following up various suggestions for scalene test triangles T, I computed the conic determinant (normalised for scale invariance) when the "centre" S is the centroid and the 6 subsidiary points are incentres: the result quantifies relatively how far the hexad departs from "conicity" (as it were) for T with vertices shown.
0 {[0, 0], [0, 1], [1, 0]} right isosceles 1.6e-8 {[3, 4], [-12, 5], [9, -9]} WFL old test 9.7e-8 {[0, 0], [0, 12], [15, 20]} WFL sides {12, 17, 25} -1.4e-7 {[0, 0], [0, 1], [3^(1/2), 0]} DA angles {1/6, 1/3, 1/2} -6.6e-8 {[0, 0], [0, 1], [0.256001685, 0.124419097]} KM sides ~{263, 697, 924}
Precision is 50 decimals; notice how small all the discrepancies are.
Convincing win for Dan's remarkably simple suggestion then --- which I had also considered, but discarded as a result of somehow mistakenly concluding that the conjecture succeeds for right-angled cases, instead of isosceles as explained earlier by PJCM . [WRM's ingenious suggestion --- minimising maximal overlap with the mirror-image triangle --- must await more determined analysis.]
Incidentally, the rational {12, 17, 25}-sided triangle turns out to have a rational incentre at [10/3, 10] --- I haven't looked at other special points.
Fred Lunnon