Are there such numbers?

Maybe better to ask for help on this question here: Are there eight numbers $0< x_{1},\ldots,x_{4},y_{1},\ldots,y_{4}<1$, such that $$\frac{\max\left(\left(x_{i}-x_{j}\right){}^{2}+\left(y_{i}-y_{j}\right){}^{2},\left(x_{j}-x_{k}\right){}^{2}+\left(y_{j}-y_{k}\right){}^{2},\left(x_{k}-x_{i}\right){}^{2}+\left(y_{k}-y_{i}\right){}^{2}\right)}{\left(x_{k}\left(y_{i}-y_{j}\right)+x_{i}\left(y_{j}-y_{k}\right)+x_{j}\left(y_{k}-y_{i}\right)\right)^{2}}$$ less than $5/8$ for all distinct $i$, $j$ and $k$, $1\le i,j,k\le4$? Thanks.

-
Can you explain why would you think there exist such numbers (or not?) or give us a hint about where this comes from? If you asked this elsewhere (probable Math.stackexchange.com) please add a link to the other question (and do the same there) – Mariano Suárez-Alvarez Nov 22 '12 at 7:47
(In fact, it is rather best not to ask the question in both places with such little difference of time; but that ship has already sailed) – Mariano Suárez-Alvarez Nov 22 '12 at 7:48
I think Mariano is asking: where does this question arise? What is special about 5/8? – Yemon Choi Nov 22 '12 at 8:17
Why 8 numbers? If it worked for $\{i,j,k\} = \{1,2,3\}$ then where would $x_4$ and $y_4$ come in to it? – Gordon Royle Nov 22 '12 at 8:19

The quantity at the numerator is the length of the largest edge of the triangle with vertices $(x_i,y_i), (x_j,y_j), (x_k,y_k)$, squared. The denominator is its area, squared, so the fraction is four times the reciprocal squared of the minimal altitude of that triangle. The whole question may be therefore rephrased more geometrically: Are there $4$ points in the unit square such that the $12$ altitudes of the $4$ triangles they form are all larger than $2\sqrt{8/5}$? No, because no triangle inside the unit square can have an altitude larger than $\sqrt 2 < 2\sqrt{8/5}$.

 Since we are still here, we may ask what is the optimal configuration for four points in the square, that is, the one forming four triangles with larger minimal altitude $h$. Four points on a square either form a convex quadrilateral (A), or a triangle with an interior point (B). It is convenient to consider separately the two cases.

(A) Among all choices of four points that form a convex quadrilateral, by a symmetrization argument, the optimal configuration can be reached on rectangles (any quadrilateral contains a rectangle with the diagonals on the same lines, and with same minimal altitude $h$). Moreover, a rectangle can be enlarged and rotated until all its vertices touch the edges of the unit square, and this will not decrease $h$. This way it follows that the optimal configuration for convex quadrilaterals is the unit square, giving the optimal $h_A=\sqrt{2}/2$.

(B) Given a triangle inside the unit square, the choice of a fourth point inside the triangle, giving the best $h$ is the incenter of the triangle (all three interior altitudes are equal). In this case, the problem may be stated: find the largest radius $h$ of circles contained in triangles contained in the unit square. This is a computation that you may like to complete (one vertex of the triangle must coincide with a vertex of the unit square, by an analogous argument as before); in any, case it will give an optimal $h_B < 1/2$, worse than the above $h_A$.

-
1) In a triangle, the altitude of minimum length (among the three) joins a vertex and a point of the opposite side (not in the prolongation of the side, otherwise it is not the minimum one). 2) Hence, in a convex quadrilateral, the minimum altitude (among the 12) joins a vertex and a point on a diagonal (not on an edge, because there is a shorter altitude to a diagonal). 3) Therefore, in a convex quadrilateral, the minimum altitude is the one from the closest vertex V to the intersection point P of the diagonals, to the other diagonal. – Pietro Majer Apr 17 '13 at 8:19
4) If we shrink the convex quadrilateral, leaving P and V fixed, and moving the other vertices towards P, to the same distance as V, we obtain a rectangle included in the quadrilateral, whose minimum altitude is the same (by the above characterization). – Pietro Majer Apr 17 '13 at 8:43
@Pietro: Thank you for your answer, but why is it among the 12? – Op_Bgh Apr 20 '13 at 16:39
sorry, would you clarify the question? – Pietro Majer Apr 21 '13 at 16:28