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I've found all the possible configurations by case-by-case analysis. I don't know whether my approach would be considered elegant by the author, but it wasn't e
by romwell 8y ago
I've found all the possible configurations by case-by-case analysis. I don't know whether my approach would be considered elegant by the author, but it wasn't exhausting (while being exhaustive).
My cases (or, rather, considerations) were:
1)Three points can form either an equilateral triangle, or an isosceles one;
2)The fourth point must lie on the perpendicular bisector of one of the edges of that triangle;
3)The fourth point may on one of the two sides of that edge.
Given the constraints, that doesn't leave too many cases to consider.
But I, too, wonder what other approaches to the problem there are.
- jobigoud 8y agoHaven't looked for solutions but 3 points can also form a line. Maybe there is a configuration where 3 or even 4 points are colinear?
- romwell 8y agoThree collinear points is just a degenerate obtuse triangle, so this falls under one of the cases above.
- ColinWright 8y agoHow many did you find?
- romwell 8y agoI did find all configurations that way.
- ColinWright 8y agoHow do you know you've found them all? More than one person has told me that they've found them all, only later to discover that they'd missed (at least) one.