Problem source

A thin circular path with diameter $AB$ is laid on horizontal ground. A vertical flagpole is erected with its base at a point $D$ on the diameter $AB$. The angles of  elevation of the top of the flagpole from $A$ and $B$ are $\alpha$ and $\beta$ respectively (both are acute).  The point $C$ lies on the circular path with $DC$ perpendicular to $AB$ and the angle of elevation of the top of the flagpole from $C$ is $\phi$. Show that $\cot\alpha\cot \beta  = \cot^2\phi$.
Show that, for any $p$ and $q$,
\[
\cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) =  \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)
.\]
Deduce that, if $p$ and $q$ are positive and  $ p+q  \le \tfrac12 \pi$,
then
\[
 \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \,
\]
and hence show that $\phi \le \tfrac12(\alpha+\beta)$ when $ \alpha +\beta \le \tfrac12 \pi\,$.

Solution source

\begin{center}
    \begin{tikzpicture}

    % --- Parameters ---
    \def\R{3.5}        % Radius of the circle in "world" units
    \def\yscale{0.3}   % Vertical compression factor for the ground (0.3 makes it look flat)
    \def\rot{-25}      % Rotation of diameter AB on the ground plane (degrees)
    \def\dOffset{1.0}  % Distance of D from center (along AB)
    \def\h{3.5}        % Height of the flagpole
    
    % --- Coordinate Transformation Function ---
    % We calculate coordinates in a flat "circle world" first, 
    % then rotate them, then squash the y-coordinate to create the perspective.
    
    % Center
    \coordinate (O) at (0,0);
    
    % 1. Define points in Circle World (Unrotated, Unsquashed)
    % A and B are on the x-axis
    % D is at x = \dOffset
    % C is at x = \dOffset, y = -sqrt(R^2 - d^2) (Negative to be "in front")
    \pgfmathsetmacro{\xc}{\dOffset}
    \pgfmathsetmacro{\yc}{-sqrt(\R*\R - \dOffset*\dOffset)}
    
    % 2. Apply Rotation (\rot) and Perspective (\yscale)
    % Formula: 
    % x_final = x*cos(rot) - y*sin(rot)
    % y_final = (x*sin(rot) + y*cos(rot)) * yscale
    
    % Point A (-R, 0)
    \pgfmathsetmacro{\Ax}{-\R*cos(\rot)}
    \pgfmathsetmacro{\Ay}{-\R*sin(\rot)*\yscale}
    \coordinate (A) at (\Ax, \Ay);
    
    % Point B (R, 0)
    \pgfmathsetmacro{\Bx}{\R*cos(\rot)}
    \pgfmathsetmacro{\By}{\R*sin(\rot)*\yscale}
    \coordinate (B) at (\Bx, \By);
    
    % Point D (\dOffset, 0)
    \pgfmathsetmacro{\Dx}{\dOffset*cos(\rot)}
    \pgfmathsetmacro{\Dy}{\dOffset*sin(\rot)*\yscale}
    \coordinate (D) at (\Dx, \Dy);
    
    % Point C (\xc, \yc)
    \pgfmathsetmacro{\Cx}{\xc*cos(\rot) - \yc*sin(\rot)}
    \pgfmathsetmacro{\Cy}{(\xc*sin(\rot) + \yc*cos(\rot))*\yscale}
    \coordinate (C) at (\Cx, \Cy);
    
    % Top of Flagpole (Same x as D, y is D.y + h)
    \coordinate (T) at (\Dx, \Dy + \h);

    % --- Drawing ---

    % 1. Ground Plane (Ellipse)
    % The ellipse is aligned with the page axes to ensure it looks "flat"
    \draw[thick, gray!80!black] (O) ellipse ({\R} and {\R*\yscale});
    % Optional: Add a lighter fill or dashed back line for better depth?
    % Let's just redraw the back half dashed for clarity
    \begin{scope}
        \clip (-\R-1, 0) rectangle (\R+1, \R);
        \draw[dashed, gray] (O) ellipse ({\R} and {\R*\yscale});
    \end{scope}
    
    % 2. Ground Lines
    % Diameter AB
    \draw[thick, black] (A) -- (B);
    
    % Chord DC (Ground)
    % Since DC is perpendicular to AB in reality, we draw the line connecting them.
    \draw[dashed, black] (D) -- (C);

    % 3. Vertical Geometry
    % Flagpole
    \draw[ultra thick, black] (D) -- (T) node[pos=0.5, right] {$h$};
    
    % Sight Lines
    \draw[thick, blue] (A) -- (T);
    \draw[thick, blue] (B) -- (T);
    \draw[thick, blue] (C) -- (T);

    % --- Angles ---
    
    % Angle Alpha at A
    % We approximate the perspective arc
    \pic [draw, red, angle radius=.8cm, angle eccentricity=1.5, "$\alpha$"] {angle = D--A--T};
    \pic [draw, red, angle radius=.8cm, angle eccentricity=1.5, "$\beta$"] {angle = T--B--D};
    \pic [draw, red, angle radius=.8cm, angle eccentricity=1.5, "$\phi$"] {angle = D--C--T};
    

    % --- Right Angle Markers ---
    
    % Right angle at D (Ground)
    % Use coords to create a small parallelogram aligned with AB and DC
    \coordinate (D_arm_AB) at ($(D)!0.25cm!(B)$);
    \coordinate (D_arm_DC) at ($(D)!0.25cm!(C)$);
    \coordinate (D_corner) at ($(D_arm_AB) + (D_arm_DC) - (D)$);
    \draw[thin, black!70] (D_arm_AB) -- (D_corner) -- (D_arm_DC);
    
    % Right angle at D (Pole)
    % A small rectangle upright
    \draw[thin, black!70] ($(D)+(0,0.25)$) -- ++($(D_arm_AB)-(D)$) -- (D_arm_AB);

    % --- Labels ---
    \fill[black] (A) circle (1.5pt) node[left] {$A$};
    \fill[black] (B) circle (1.5pt) node[right] {$B$};
    \fill[black] (C) circle (1.5pt) node[below] {$C$};
    \fill[black] (D) circle (1.5pt) node[below right, xshift=-1pt, yshift=-1pt] {$D$};
    \fill[black] (T) circle (1.5pt) node[above] {Top};

\end{tikzpicture}
\end{center}

\begin{align*}
    && \cot \alpha &= \frac{AD}{h} \\
    && \cot \beta &= \frac{BD}{h} \\
    && \cot \phi &= \frac{DC}h \\
    && CD^2 &= AB \cdot BD  \tag{intersecting chords} \\
    \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta
\end{align*}

\begin{align*}
    && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\
    &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\
    &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\
    &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\
    &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\
    &&&= RHS
\end{align*}

Therefore

\begin{align*}
    \cot p \cot q -\cot^2 \tfrac12 (p+q) &=  \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\
    &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)}
\end{align*}
Since $p,q$ are acute, the denominator is positive. Since $p+q \leq \frac{\pi}{2}$, we have $\cos(p+q) \geq 0$. Also $(1-\cos(p-q)) \geq 0$. Thus, the expression is $\geq 0$.

So we must have

\begin{align*}
    && \cot^2 \phi &= \cot \alpha \cot \beta \\
    &&&\geq \cot^2 \tfrac12(\alpha+\beta)
\end{align*}

Since $\cot$ is decreasing on $(0, \tfrac12 \pi)$ we can deduce $\phi \leq \tfrac12 (\alpha+\beta)$