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Mathematics-Online problems:

Solution to the problem of the (previous) week


The figure shows a regular hexagon with side length 1, whose vertices are connected with each other.


Determine the lengths of the line segments and the areas of the triangles and quadrilaterals generated by connecting all the vertices.


Length of the line segments:
  $ a = $
  $ b = $
  $ c = $
  $ d = $


  $ E = $
  $ F = $
  $ G = $

(The results should be correct to three decimal places.)



The rhombus shown above is dissected into two equal parts by the horizontal line. Thus we obtain the two angles shown in the figure. Hence $ a=b$ and $ a+b=1,$ which leads to $ a=1/2$ and $ b=1/2.$

$ (d+c)$ is the altitude in the equilateral triangle with side length $ 1,$ that is $ d+c=\sqrt{1-(1/2)^2}=\sqrt{3}/2.$ Then we have $ \tan(30^\circ)=c/(1/2).$ Due to this and $ \tan(30^\circ)=1/\sqrt{3},$ we at once obtain $ c=1/(2\sqrt{3}) \approx 0.289.$ Finally, we get $ d=(d+c)-c$ and hence $ d=\sqrt{3}/2-1/(2\sqrt{3})=1/\sqrt{3} \approx 0.577.$


By symmetry, the three areas $ E, F, G$ can also be identified as shown here.
When we calculated the lengths of the line segments, we obtained $ \sqrt{3}/2$ for the altitude of the equilateral triangle. Hence, its area is $ \sqrt{3}/4.$ Consequently, we have $ \sqrt{3}/4=E+F+2G.$


If, in addition, we draw the third median, we find that by symmetry all the resulting areas are equal and hence $ F=2G=E.$ Considering also that $ \sqrt{3}/4=E+F+2G,$ we obtain $ E=1/(4\sqrt{3}) \approx 0.144,$ $ F=1/(4\sqrt{3}) \approx 0.144,$ $ G=1/(8\sqrt{3}) \approx 0.072.$

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