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January 25, 2015

January 25, 2015

Posted by **Mark** on Monday, July 15, 2013 at 4:57pm.

- Math -
**MathMate**, Monday, July 15, 2013 at 6:07pmShe cannot subdivide the equilateral triangle into 2 isosceles triangles. This is the smallest N which is impossible.

She can subdivide it into 3 congruent isosceles triangles by joining the centroid with each of the vertices. Each triangle is then 30-30-120 as angles.

Each of the 30-30-120 triangles can in turn be subdivided into 3 isosceles triangles with angles 60-60-60, and 30-30-120, thus adding two more triangles.

So the next possible N are then 5, 7, 9.

Since each of the smaller 30-30-120 triangles can again be subdivided, indefinitely, we conclude that the original equilateral triangle can be subdivided into odd N above 1. Even N are therefore not possible. Thus there is no maximum number N, since N->∞, as long as N=2k, where k∈integers.

- Math -
**MysteriousPerson**, Thursday, July 18, 2013 at 5:37pmMathMate, I would disagree with you because 4 triangles are possible. We can break up the equilateral triangle to 4 different equilateral triangles.

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