Question 201375
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If you join the mid-points of two sides of a triangle then you form a smaller similar triangle the sides of which are each one-half the measure of the sides of the original triangle, hence the perimeter of the smaller triangle is also one-half the perimeter of the original.  If the perimeter of the first triangle is *[tex \Large P], then the perimeter of the smaller triangle is *[tex \Large \frac{P}{2}].  The next one would be *[tex \Large \frac{P}{4}], and so on.


Therefore the sum of the perimeters would be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P + P\sum_{k=1}^\infty\,\frac{1}{2^k}]


But since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{n\rightarrow\infty}\ \sum_{k=1}^n\,\frac{1}{2^k} = 1]


The sum of the perimeters becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P + P\cdot 1 = 2P]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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