Is there a theorem for the area of the hexagon formed by connecting the points formed when the sides of a triangle are trisected? It appears that the ratio of the area of the triangle to the area of the hexagon is 4.5.
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4$\begingroup$ For this and many similar statements, whatever value occurs in an equilateral triangle is the same for any triangle because linear transformations preserve area ratios and equal divisions of line segments (but not equal divisions of angles). $\endgroup$– Douglas ZareCommented Jul 21, 2014 at 19:40
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2 Answers
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There seems to be some confusion about the formulation. I understand it in the sense that if the triangle is $ABC$ and $C_1$, $C_2$ trisect $AB$ and so on, then you are concerned with the area of the hexagon $A_1A_2B_1B_2C_1C_2A_1$. Since areas (more precisely, proportions thereof) are invariant under affine transformations, one can reduce to the case where $A=(0,0)$, $B=(1,0)$ and $C=(0,1)$. I leave the simple computations in this case to the OP.
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$\begingroup$ The comment by Douglas Zare was posted while I was typing the above. $\endgroup$ Commented Jul 21, 2014 at 19:50
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$\begingroup$ This is even better than the equilateral triangle because it makes it obvious that the proportion is rational. $\endgroup$ Commented Jul 22, 2014 at 1:16
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$\begingroup$ This question has been closed but it might be of interest to the poser that the general case, i.e., where the three sides are cut into bits of varying sizes (so that one has six parameters) can be solved just as easily and one gets a simple and elegant expression in these quantities $\endgroup$ Commented Jul 22, 2014 at 11:56
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Is this what you have in mind?
http://girlsangle.wordpress.com/2014/07/15/marion-walters-theorem-via-mass-points/
answered Jul 21, 2014 at 18:53
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$\begingroup$ No, but this did spark my investigation. I decided to connect all trisection points. Not with the vertices of the triangle, but just the trisection points. When I did this, there was an internal hexagon formed. And this hexagon was 1/4.5 in area with the triangle. $\endgroup$ Commented Jul 22, 2014 at 17:49