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Mathematics-Online problems:

Problem of the week

The hanging mobile in the figure below is constructed of four rods - the top rod has length 2 and all the other rods have length 1 each.

\includegraphics[clip,width=10cm]{Mobile_Bild1}

What is the width $ b$ of the mobile, if only the masses of the spheres as indicated have to be considered? How many different hanging mobiles, that cannot be transferred into each other by turning the rods (e.g. the arrangements 3-1-1-1-2 and 1-3-1-1-2 are not different), can be constructed by permutation of the weights? Which arrangement of the weights has the maximum width $ b_{max}$?

Note:
\includegraphics[bb=135 545 340 620,clip,width=.42\linewidth]{Mobile_Bild3bb}
$ l_1\,:\,l_2=m_2\,:\,m_1$


Answer:

$ b$ $ =$ $ /$
Number of mobiles $ =$
Arrangement $ =$ ; ; ; ;
$ b_{max}$ $ =$ $ /$

(Reduce the fractions to their lowest terms.)


   

[solution to the problem of the previous week]