%
\question
{Lucas Mussnich} % Enter your name
{03/14/11} % Enter mm/dd/yy
{mussnich@ufmg.br} % Enter e-mail address
{1-6} % Enter the number and question.
%question
{
Each carbon atom in a C$_{60}$ molecule has one pentagonal and two hexagonal rings. Calculate the angles (a) between the two hexagonal rings and (b) between the hexagonal ring and the pentagonal ring. \\
(a) See Fig. \ref{fig:Q1-6-1} . If we apply the law of cosines to the triangles ABC and ADC, with respect to the side AC, we obtain
\begin{equation}
\label{eq:cossines}
\begin{array}{l}
d^2 = 2 l^2 (1 - \cos(\beta))\\
d^2 = 2 h^2 (1 - \cos(\alpha))
\end{array}
\end{equation}
% if you want to include figure.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,clip]{Q1-6-1.eps}\\
\begin{minipage}{10cm}
\caption{Auxialiary geometric construction to help determining $\alpha$, the angle between the two hexagon sheets.}
\label{fig:Q1-6-1}
\end{minipage}
\end{center}
\end{figure}
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For the case of fullerene, the parameters in the figure are adapted in the following manner: $\beta$ is the angle of the pentagon, $\beta = 3 \pi / 5$; $l$ is the side of the hexagon/pentagon, and the planes depicted are the hexagon planes. We may, for convenience, define $l = 2R$. Since $h$ is at right angles to the axis that intersect the two hexagon planes, $h$ is the hexagon height, $h = \sqrt{3} l / 2 = \sqrt{3} R$.
Therefore, the Eq. \ref{eq:cossines} yields
\begin{equation}
\label{eq:alpha}
\alpha = \cos^{-1}[3 - 4(1 - \cos(3 \pi / 5)))/3] \approx 139^o.
\end{equation}
(b) See Fig. \ref{fig:Q1-6-2}. The angle we wish to calculate is $\gamma$.
% if you want to include figure.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,clip]{Q1-6-2.eps}\\
\begin{minipage}{10cm}
\caption{$\gamma$ is the angle between the pentagon and the hexagon planes.}
\label{fig:Q1-6-2}
\end{minipage}
\end{center}
\end{figure}
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Let us now consider the ABCD prism in Fig. \ref{fig:Q1-6-3}.
% if you want to include figure.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,clip]{Q1-6-3.eps}\\
\begin{minipage}{10cm}
\caption{Focused view of prism ABCD from Fig.\ref{fig:Q1-6-2} }
\label{fig:Q1-6-3}
\end{minipage}
\end{center}
\end{figure}
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We may construct two perpendicular segments forming an angle $\gamma \prime$: the first one is the height of the triangle ABC with respect to the side AB. Let us call it CE (see Fig. \ref{fig:Q1-6-2}). The other one is constructed in the following manner: trace a segment perpendicular to AB, starting from point E, in the plane ABD, which intersects the straight line AD on the point P. That is segment the segment EP. Fig. \ref{fig:Q1-6-4} shows EP in the plane of the triangle ABD. We shall thus define $\gamma \prime$, by this construction, as the angle between CE and EP.
% if you want to include figure.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,clip]{Q1-6-4.eps}\\
\begin{minipage}{10cm}
\caption{EP segment on the ABD triangle's plane.}
\label{fig:Q1-6-4}
\end{minipage}
\end{center}
\end{figure}
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As it was meant to be, $\gamma \prime = \pi - \gamma$. Let us now consider the triangle EPC as in Fig \ref{fig:Q1-6-5}.
% if you want to include figure.
\begin{figure}[h]
\begin{center}
\includegraphics[width=6cm,clip]{Q1-6-5.eps}\\
\begin{minipage}{10cm}
\caption{Triangle EPC.}
\label{fig:Q1-6-5}
\end{minipage}
\end{center}
\end{figure}
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Let us call u the magnitude of the segment EC and x the magnitude of the segment EP. We note that since the planes of the triangles ABD and ADC are perpendicular, (see Fig. \ref{fig:Q1-6-3}), trigonometry on a rectangle triangle yields
\begin{equation}
\label{eq:gammanotyet}
\cos(\gamma \prime) = - \cos(\gamma) = x / u
\end{equation}
In terms of the known parameters, who are $x$ and $u$? Fig. \ref{fig:Q1-6-2} shows that $u = l \sin(\beta)$. If we define $v = l \cos(\beta)$, the magnitude of the segment BE, then Fig. \ref{fig:Q1-6-4} shows us that $x = (l - v)\tan(\pi / 6)$. Therefore, using that $\beta = 3 \pi / 5$, we have
\begin{equation}
\label{eq:gamma}
\gamma = \cos^{-1}[\sqrt{3}(\cos(\beta) - 1)/3 \sin(\beta)] \approx 143^o.
\end{equation}
} % Do not forget this.