Dissect an equilateral triangle into 9 similar triangles with angles of 45, 60 and 75 degrees; one at each corner and the other six meeting at a point and forming a hexagon.

Preliminary.

Consider the central angles. Either there will be at least one set of 3 adjacent angles all different, creating some line parallel to a side (the 'base case', e.g. VOY in Brian's diagram) in which case the problem is equivalent to fitting 4 of the similar triangles into a smaller equilateral triangle, or the central angles are {45º,45º,b,b,c,c} where b and c are 60º and 75º or vice versa.

In such case, there must be a 45º angle in the centre opposite a corner, and another 45º angle in the centre in the triangle next to it. But if we create such a second triangle in one direction, the line at that angle through the centre is perpendicular to a side, whereas if we create it in the other, all angles are fixed and the construction is easily shown not to work (the last two triangles have angles of {≈37º,≈83º,60º} and {≈98º,≈22º,60º}

So we need only consider the 'base case' as it relates to fitting 4 similar triangles into a smaller equilateral triangle.

Proof of impossibility.

Start at corner A with the angle of the first similar triangle nearest C=45º and the third angle = 75º. Then the similar triangle including C must have angles of 75º, closest to the 45º angle already mentioned, and 45º nearest B. But the triangle on B must also have angles of 75º nearest C, and 45º nearest A, leaving a central triangle with angles 60º,60º,60º. So the desired dissection is impossible.

This comment is dedicated to my former IT manager, who, whenever asked to do anything, invariably replied 'Can't be done'.

Edited on July 20, 2012, 7:56 am

Posted by broll on 2012-07-20 07:02:34