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9 of the same shape (Posted on 2012-07-19) Difficulty: 3 of 5
Determine whether the following construction is possible:

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.

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Hints/Tips Can't be done. | Comment 2 of 3 |

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 {45,45,b,b,c,c} where b and c are 60 and 75 or vice versa.

In such case, there must be a 45 angle in the centre opposite a corner, and another 45 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 {≈37,≈83,60} and {≈98,≈22,60}

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=45 and the third angle = 75. Then the similar triangle including C must have angles of 75, closest to the 45 angle already mentioned, and 45 nearest B. But the triangle on B must also have angles of 75 nearest C, and 45 nearest A, leaving a central triangle with angles 60,60,60. 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

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