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Mathematics-Online problems:

Solution to the problem of the (previous) week


Problem:

The logo
\includegraphics[width=0.8\linewidth]{moLogo_en.eps}
is produced by rolling a circle with radius 1 along the letter `` M''. This process is illustrated in the following drawings for a slightly modified ``M'' with vertices $ (\pm a, \pm a)$ and $ (0,0)$.
\includegraphics[width=0.6\linewidth]{moLogo1.eps}                  \includegraphics[width=0.6\linewidth]{moLogo2.eps}

From the drawings to scale read the number $ n$ of the revolutions the arrow accomplishes when the circle rotates once completely along the ``M''.
Determine the length $ x,$ the change of the rotation angle $ \alpha_k$ of the arrow between the positions $ k$ and $ k+1$ depending on length $ a,$ as well as the length $ a$ itself.


Answer:

$ n$ =
$ x $ =
$ \alpha_1 $ = $ a$ $ +$
$ \alpha_4 $ = $ a$ $ +$
$ \alpha_5 $ = $ a$ $ +$
$ \alpha_6 $ = $ a$ $ +$
$ \alpha_7 $ = $ a$ $ +$
$ a$ =
   
(The results should be correct to three decimal places.)


Solution:

From the drawings we can see that

$\displaystyle n=6.$


From the second drawing follows

$\displaystyle x=1+\sqrt{2} \approx 2.414. $


Moreover, we obtain
$\displaystyle \hspace*{1.9cm} \alpha_1 =$ $\displaystyle \sqrt{2}a-1$ $\displaystyle \approx 1.414a-1$  
$\displaystyle \alpha_4 =$ $\displaystyle \pi$ $\displaystyle \approx 3.142$  
$\displaystyle \alpha_5 =$ $\displaystyle 2a-x$ $\displaystyle \approx 2a-2.414$  
$\displaystyle \alpha_6 =$ $\displaystyle \sqrt{2}a-x$ $\displaystyle \approx 1.414a-2.414$  
$\displaystyle \alpha_7 =$ $\displaystyle \frac{\pi}{4}$ $\displaystyle \approx 0.785.$  

The total angle is

$\displaystyle 2 \cdot \sum_{k=1}^7{\alpha_k}=
(4\sqrt{2}+8)a-4x-2+4\pi=(4\sqrt{2}+8)a-6-4\sqrt{2}+4\pi , $

and equating it with $ 6 \cdot 2\pi$ yields
$\displaystyle (4\sqrt{2}+8)a-6-4\sqrt{2}+4\pi$ $\displaystyle =$ $\displaystyle 12\pi$  
$\displaystyle \Longrightarrow\qquad a$ $\displaystyle =$ $\displaystyle \frac{3+2\sqrt{2}+4\pi}{4+2\sqrt{2}} \approx 2.694\hdots$  


[problem of the week]