Problem source

The cuboid $ABCDEFGH$ is such  $AE$, $BF$, $CG$, $DH$ are perpendicular  to the opposite faces $ABCD$ and $EFGH$, and $AB =2, BC=1, AE={\lambda}$. Show that if $\alpha$ is the acute angle between the diagonals $AG$ and $BH$ then 
$$\cos {\alpha} 
=  |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$
Let $R$ be the ratio of the volume of the cuboid to its surface area. Show that
$R<\frac{1}{3}$ for all possible
values of $\lambda$.
Prove that, if $R\ge \frac{1}{4}$, then $\alpha \le \arccos \frac{1}{9}$.

Solution source

\begin{center}
    \begin{tikzpicture}[
    x={(-0.6cm,-0.4cm)}, y={(1cm,0cm)}, z={(0cm,1cm)},
]

    % --- Parameters ---
    % Mathematical dimensions: AB=2, BC=1, AE=lambda
    % We scale them for visual clarity in the diagram
    \def\cuboidW{2}   % Width (AB)
    \def\cuboidD{1}   % Depth (BC) - represented by x-vector direction here effectively
    \def\cuboidH{2.5} % Height (AE) - lambda

    % --- Coordinates ---
    % Orientation: A is origin. 
    % A(0,0,0), B(0,2,0) ... wait, let's map strictly to the vector definitions above.
    % Vectors defined: x is going back-left, y is going right, z is going up.
    % Let's adjust coordinates to match standard A front-left logic with the defined vectors.
    
    % Let's use a simpler coordinate mapping manually:
    % x axis -> Right (AB)
    % y axis -> Up (AE)
    % z axis -> Depth (AD)
    
    \begin{scope}[x={(1cm,0cm)}, y={(0cm,1cm)}, z={(0.5cm,0.3cm)}]
        
        % Vertices
        \coordinate (A) at (0,0,0);
        \coordinate (B) at (\cuboidW,0,0);
        \coordinate (C) at (\cuboidW,0,\cuboidD);
        \coordinate (D) at (0,0,\cuboidD);
        
        \coordinate (E) at (0,\cuboidH,0);
        \coordinate (F) at (\cuboidW,\cuboidH,0);
        \coordinate (G) at (\cuboidW,\cuboidH,\cuboidD);
        \coordinate (H) at (0,\cuboidH,\cuboidD);

        % --- Drawing the Cuboid ---
        
        % Hidden Edges (Back and Bottom-Left)
        \draw[dashed, thin, gray] (A) -- (D);
        \draw[dashed, thin, gray] (D) -- (C);
        \draw[dashed, thin, gray] (D) -- (H);

        % Visible Edges
        \draw[thick] (A) -- (B) -- (C) -- (G) -- (H) -- (E) -- cycle; % Outer perimeter
        \draw[thick] (A) -- (E); % Front left vertical
        \draw[thick] (B) -- (F); % Front right vertical
        \draw[thick] (F) -- (E); % Front top horizontal
        \draw[thick] (F) -- (G); % Right top horizontal

        % --- Diagonals AG and BH ---
        % M is the intersection (center of cuboid)
        \coordinate (M) at ($(A)!0.5!(G)$);

        \draw[blue, thick] (A) -- (G) node[pos=1, above right, black] {};
        \draw[red, thick]  (B) -- (H) node[pos=1, above left, black] {};

        % --- Marking the Angle alpha ---
        % We mark the angle at the intersection M.
        % We choose the angle facing the top (between MG and MH) or side depending on lambda.
        % To look good, we draw an arc between points on the segments MG and MH.
        
        \coordinate (P_G) at ($(M)!0.25!(G)$); % Point on MG
        \coordinate (P_H) at ($(M)!0.25!(H)$); % Point on MH
        
        % Draw the arc in the plane formed by the diagonals
        \draw[fill=yellow!30, opacity=0.8] (M) -- (P_G) to[bend right=20] (P_H) -- cycle;
        \node at ($(M)!0.4!($(P_G)!0.5!(P_H)$)$) [above] {$\alpha$};

        % --- Labels ---
        
        % Vertices
        \node[below left]  at (A) {$A$};
        \node[below right] at (B) {$B$};
        \node[right]       at (C) {$C$};
        \node[left, gray]  at (D) {$D$};
        \node[left]        at (E) {$E$};
        \node[above]       at (F) {$F$};
        \node[above right] at (G) {$G$};
        \node[above left]  at (H) {$H$};

        % Dimensions
        \node[below] at ($(A)!0.5!(B)$) {2};
        \node[below right] at ($(B)!0.5!(C)$) {1};
        \node[left]  at ($(A)!0.5!(E)$) {$\lambda$};
        
        % Right Angle Markers (optional, to show perpendicularity)
        % Angle at A (EA perp AB)
        \draw[thin] ($(A)+(0,0.2,0)$) -- ($(A)+(0.2,0.2,0)$) -- ($(A)+(0.2,0,0)$);
        
    \end{scope}
\end{tikzpicture}
\end{center}


Set $A$ to be the origin, then $B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle$, in particular

\begin{align*}
&& AG&= \langle 2, 1, \lambda \rangle \\
&& BH &= \langle -2, 1, \lambda \rangle \\
\Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\
&&&= |\frac{-3+\lambda^2}{5+\lambda^2}|
\end{align*}

\begin{align*}
&& \text{Volume} &= 2\lambda \\
&& \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\
\Rightarrow &&  R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\
&& \frac14 &\leq R \\
\Rightarrow && 3\lambda +2 &\leq 4\lambda \\
\Rightarrow &&2 & \leq  \lambda 
\end{align*}

Then $\frac{\lambda^2-3}{5+\lambda^2}$ is increasing as $\lambda$ increases, in particularly the smallest value is $\frac{1}{9}$.