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Solution to the problem of the (previous) week

Problem:

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:

Width $ b$ of the hanging mobile:
$ l_1 = \dfrac{m_2}{m_1+m_2},\ \ $ $ l_2 = \dfrac{m_3}{m_1+m_2+m_3},\ \ $ $ l_3 = \dfrac{m_4}{m_4+m_5}.$

\includegraphics[clip,width=.6\linewidth]{Mobile_Bild2}

If $ m_1=3,$ $ m_2=1,$ $ m_3=1,$ $ m_4=1,$ and $ m_5=2,$ we obtain

$\displaystyle b = l_1+l_2+2+l_3 = \frac{1}{4}+\frac{1}{5}+2+\frac{1}{3}= \frac{167}{60}. $

Number of the different hanging mobiles:

Placing the sphere with the mass 3 in the positions 1, 3, and 4, results in different hanging mobiles. For these positions, there are 3, 2, and 3 possibilities for the sphere whose mass is 2. Altogether, we obtain

$\displaystyle 3+2+3=8$

different possible arrangements.

Arrangement with maximum width:

The length of the top rod is always 2, no matter how the masses are distributed.
Regard $ l_1=\dfrac{m_2}{m_1+m_2}$: If $ m_1<m_2,$ $ l_1$ will reach its maximum. This means that the higher weight must be located towards the center. Hence, $ m_1=1$ and $ m_5=1$.
Regard $ l_2=\dfrac{m_3}{m_1+m_2+m_3}$: If $ m_3$ has the greatest value of the three masses, $ l_2$ will reach its maximum. Then, three possibilities are remaining:

$ m_1$ $ m_2$ $ m_3$ $ m_4$ $ m_5$ $ l_1+l_2+l_3$
1 1 2 3 1 $ \frac{7}{4}$
1 1 3 2 1 $ \frac{53}{30}$
1 2 3 1 1 $ \frac{5}{3}$

Consequently, the second possibility yields the maximum width

$\displaystyle b_{max}=\frac{53}{30}+2=\frac{113}{30}.$

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