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Solution to the problem of the (previous) week

Problem:

#./interaufg90_en.tex#The figure shows a regular hexagon with side length 1, whose vertices are connected with each other.

\includegraphics[width=0.6\linewidth]{6eck.eps}

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


Answer:

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


Areas:
  $ E = $
  $ F = $
  $ G = $

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

Solution:

\includegraphics[width=0.6\linewidth]{l_00_2_bild1}

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.$

\includegraphics[width=0.2\linewidth]{l_00_2_bild2}

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.$

\includegraphics[width=0.2\linewidth]{l_00_2_bild3}

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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