Thank you to all those who sent in solutions to October’s Monthly Maths Problem – please see below the workings of Bryoni Pidsley who correctly found five more configurations of 4 points in a plane with only 2 distinct values for distances between pairs of points:
Configuration 1 :
An equilateral triangle with point 4 in the centre. Distance 1, between the three points of the triangle, distance 2, between the centre oint and the other three points.
Configuration 2:
An isosceles trapezium (is trapeziod a more accurate description? – blame google for my indecision) with the top length equal to the two sides (distance 1) and the bottom length distance 2. With two angles of 108 degrees (at the top) and two angles of 72 degrees (at the bottom) to make sure the distance between the top right point and the bottom left point is also distance 2 etc.
Configuration 3:
A rhombus with angles 60, 60, 120 and 120 degrees so that distance 1 is the length of the sides and one of the diagonals and distance 2 is the other diagonal.
Configuration 4:
A kite with angles 150, 75, 75 and 60 degrees so that distance 1 is the length of the two shorter sides and distance 2 is the length of the two longer sides and also the two diagonals.
Configuration 5:
An isosceles triangles with one side length of distance 1 and two lengths of distance 2 with angles 75, 75 and 30 degrees with a fourth point in the middle which is distance 1 from the other three points.
Why don’t you have a go at November’s problem?
