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In any triangle ABC, two points are marked on each side, in order to divide the side into three equal parts. When tracing the cevians to each of the six marked points there will be a hexagon in the center of the figure. Give the ratio between the area of ​​the hexagon and the area of ​​the triangle ABC.

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Tassandro Cavalcante
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    Could you please, at least, include a sketch of that? – Hussain-Alqatari Mar 21 '20 at 21:50
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    My hint is: the ratio is the same as it would be if the triangle were equilateral. (There exists an affine transformation that maps any triangle into an equilateral triangle. Affine maps preserve area ratios.) – Batominovski Mar 21 '20 at 22:52
  • I rolled the edit back to the original version (with one minor change--fixing a typo). The revision by @MicahWindsor seems to be different from what the OP intended to say. I however decided not to clarify what I think the OP means, because I am not too certain that my interpretation is correct either. The OP should attach a drawing to make sure people understand correctly, or reword the question. – Batominovski Mar 21 '20 at 23:03
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    Here's a similar question: https://math.stackexchange.com/questions/3304882/morley-theorem-variant-trisecting-a-triangles-sides-instead-of-its-angles – also, http://www.gogeometry.com/problem/p122_marion_walter_theorem_proof_area.htm – Gerry Myerson Mar 22 '20 at 11:42
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    Here's a nice reference: https://girlsangle.wordpress.com/2014/07/15/marion-walters-theorem-via-mass-points/ – also, http://www2.edc.org/makingmath/mathprojects/marionwalter/links/marionwalter_lnk_1.asp and https://mathworld.wolfram.com/MarionsTheorem.html and probably others. – Gerry Myerson Mar 22 '20 at 11:53
  • Had a look at any of those links, Tass? – Gerry Myerson Mar 23 '20 at 11:51
  • Yes! Thanks!!!! – Tassandro Cavalcante Mar 23 '20 at 16:33

1 Answers1

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Let [] denote areas. For the trisecting points D, E and F on the respective sides of the triangle ABC, it is well-known that

$$[XYZ] = \frac17[ABC]$$

It can also be shown that

$$\frac{XU}{XZ} = \frac14,\>\>\>\>\>\frac{XV}{XY} = \frac25$$

As a result,

$$[XVU] = \frac14\cdot\frac25 [XYZ] = \frac1{10}[XYZ]$$

Similarly, $$[YTS] = [ZWH] =\frac1{10}[XYZ]$$

Thus, the area of the hexagon is

$$[UVSTHW] = [XYZ] - [XVU] - [YTS] - [ZWH] $$ $$= (1-\frac3{10})[XYZ]=\frac7{10}[XYZ]=\frac1{10}[ABC]$$

Thus, the are ratio between ​​the hexagon and the triangle ABC is $\frac1{10}$.

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