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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

On a pyramid having a square base and edge length $ a$ a path ascends from one of the bottom vertices to the apex at a slope of $ 25\, \%$.


\includegraphics{pyramide}


Determine the height $ h$ of the pyramid, the length $ c$ of the first section of the path, the lengths $ b$ and $ a',$ as well as the total length $ L$ of the path, all of them in terms of $ a.$

Hint: Let the slope here be the ratio of the difference in altitude to the distance covered (i.e. $ f:g$ in the drawing below).


\begin{picture}(2217,708)(664,-523)\thicklines \put(676,-511){\line( 1, 0){20... ...1){\makebox(0,0)[lb]{$f$}} \put(1581, -60){\makebox(0,0)[lb]{$g$}} \end{picture}


Answer:

$ h$ =   $ a$
$ c$ =   $ a$
$ b$ =   $ a$
$ a'$ =   $ a$
$ L$ =   $ a$

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

Solution:

The height of the pyramid is

$\displaystyle h = \sqrt{a^2 - \left(\frac{\sqrt{2}a}{2}\right)^2} = \frac{a}{\sqrt{2}} \approx 0.707 a\,. $

Let $ h_1$ denote the distance between the end point of the line segment $ c$ and the pyramid's base. Since $ c$ ascends by $ 25\, \%$ (compare the definition of the slope in the problem), we have

$\displaystyle h_1 = \frac{c}{4}. $

The edge of the pyramid with $ b$ being part of it forms an angle of $ 45^\circ$ with the base, for the height of the pyramid is equal to half the length of the diagonal. Therefore, we obtain the relation

$\displaystyle b = \sqrt{2}\,h_1 = \sqrt{2}\,\frac{c}{4}. $

The application of the law of cosine yields

$\displaystyle c^2 = a^2 + b^2 - 2ab\cos{60^\circ} = a^2 + b^2 - ab. $

Substituting the value for $ b$ into the previous equation, we obtain

$\displaystyle \frac{7}{8}\,c^2 + \frac{\sqrt{2}}{4}\,ac - a^2 = 0, $

and hence
$\displaystyle c$ $\displaystyle = \displaystyle \frac{\sqrt{2}}{7} (\sqrt{29} - 1)a$ $\displaystyle \approx 0.886\,a,$  
$\displaystyle b$ $\displaystyle = \displaystyle \frac{1}{14}(\sqrt{29} - 1)a$ $\displaystyle \approx 0.313\,a,$  
$\displaystyle a'=a-b$ $\displaystyle = \displaystyle \frac{1}{14}(15 - \sqrt{29})a$ $\displaystyle \approx 0.687\,a.$  

Finally, the total length $ L$ is obtained as a geometric series:

$\displaystyle L = c + c\,\frac{a'}{a} + c\left(\frac{a'}{a}\right)^2 + \cdots = c\cdot\frac{1}{1-\displaystyle\frac{a'}{a}} \approx 2.828\,. $

[problem of the week]