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The diagram shows a circle, of radius $r$ and centre $I$, touching the three sides of a triangle $ABC$. We write $a$ for the length of $BC$ and $\alpha$ for the angle $\angle BAC$ and so on. 
Let $s=\frac{1}{2}\left(a+b+c\right)$ and let $\triangle$ be the area of the triangle. 
\begin{center}
    \begin{tikzpicture}
    % 1. Define Vertices
    \coordinate (C) at (0,0);
    \coordinate (B) at (6,0);
    \coordinate (A) at (3.5, 5.5);
    % 2. Find the Incenter (I)
    % We construct the bisectors geometrically to find I.
    
    % --- Bisector of Angle A ---
    % Find two points 1cm away from A on both sides
    \coordinate (A_on_B) at ($(A)!1cm!(B)$);
    \coordinate (A_on_C) at ($(A)!1cm!(C)$);
    % The midpoint of these two defines the direction of the bisector
    \coordinate (A_bisector_point) at ($(A_on_B)!0.5!(A_on_C)$);
    % --- Bisector of Angle B ---
    \coordinate (B_on_A) at ($(B)!1cm!(A)$);
    \coordinate (B_on_C) at ($(B)!1cm!(C)$);
    \coordinate (B_bisector_point) at ($(B_on_A)!0.5!(B_on_C)$);
    % --- Intersection ---
    % Use the core 'intersection of' syntax (no library needed)
    % This finds where the line A--A_bis intersects B--B_bis
    \coordinate (I) at (intersection of A--A_bisector_point and B--B_bisector_point);

% 3. Calculate Tangent Points (Projections)
    % Syntax: ($(Start)!(ProjectionTarget)!(End)$)
    \coordinate (a) at ($(C)!(I)!(B)$);
    \coordinate (b) at ($(C)!(I)!(A)$);
    \coordinate (c) at ($(A)!(I)!(B)$);
    % 4. Draw Triangle
    \draw (A) -- (B) -- (C) -- cycle;
    % 5. Draw Incircle
    % Calculate radius 'r' distance using the 'let' syntax
    \draw let \p1 = ($(I)-(a)$), \n{r} = {veclen(\x1,\y1)} 
          in (I) circle (\n{r});
    % 6. Draw internal segments
    \draw (I) -- (a) node[midway, right] {$r$};
    \draw (I) -- (b);
    \draw (I) -- (c);
    % 7. Draw Angle Alpha
    \pic [draw, angle radius=0.8cm, "$\alpha$", angle eccentricity=1.3] {angle = C--A--B};
    % 8. Add Labels
    \node[above] at (A) {$A$};
    \node[below right] at (B) {$B$};
    \node[below left] at (C) {$C$};
    \node[above, yshift=2pt] at (I) {$I$};
    \node[below] at (a) {$a$};
    \node[left] at (b) {$b$};
    \node[right] at (c) {$c$};
\end{tikzpicture}
\end{center}
\begin{questionparts}
\item By considering the area of the triangles $AIB,$ $BIC$ and $CIA$,
or otherwise, show that $\Delta=rs$. 
\item By using the formula $\Delta=\frac{1}{2}bc\sin\alpha$, show that
\[
\Delta^{2}=\tfrac{1}{16}[4b^{2}c^{2}-\left(2bc\cos\alpha\right)^{2}].
\]
Now use the formula $a^{2}=b^{2}+c^{2}-2bc\cos\alpha$ to show that
\[
\Delta^{2}=\tfrac{1}{16}[(a^{2}-\left(b-c\right)^{2})(\left(b+c\right)^{2}-a^{2})]
\]
and deduce that 
\[
\Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.
\]
\item A hole in the shape of the triangle $ABC$ is cut in the top of a level table. A sphere of radius $R$ rests in the hole. Find the height of the centre of the sphere above the level of the table top, expressing your answer in terms of $a,b,c,s$ and $R$. 
\end{questionparts}

Solution (Optional)

\begin{questionparts}
\item $[AIB] = \frac12br$, $[BIC] = \frac12ar$, $[CIA] = \frac12 rc$, therefore $\Delta = [AIB] +[BIC] + [CIA] = \frac12r(a+b+c) = sr$

\item $\,$
\begin{align*}
&& \Delta &= \frac12 bc \sin \alpha \\
\Rightarrow &&  \Delta^2  &= \frac14 b^2c^2 \sin^2 \alpha \\
&&&= \frac14 \left (b^2c^2 - b^2c^2\cos^2 \alpha \right) \\
&&&=  \frac1{16} \left (4b^2c^2 - (2bc\cos \alpha )^2\right) \\
\\
\Rightarrow && \Delta^2 &= \frac1{16} \left (4b^2c^2 - (b^2+c^2-a^2 )^2\right) \\
&&&= \frac1{16} (2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2) \\
&&&= \frac{1}{16}(a^2-(b-c)^2)((b+c)^2-a^2) \\
&&&= \frac1{16}(a-b+c)(a+b-c)(b+c-a)(b+c+a) \\
&&&= (s - b)(s-c)(s-a)s \\
\Rightarrow && \Delta &= \sqrt{s(s-a)(s-b)(s-c)}
\end{align*}

\item We have the setting like this,

\begin{center}
    \begin{tikzpicture}
    % 1. Define Variables
    \def\R{3}       % Radius of the sphere
    \def\h{1.5}     % Height (distance from center to cross-section)
    \def\ang{15}    % Perspective angle for the ellipse flatness
    
    % Calculate the radius of the cross-section (r) using Pythagoras
    % r = sqrt(R^2 - h^2)
    \pgfmathsetmacro{\r}{sqrt(\R*\R - \h*\h)}
    
    % 2. Define Coordinates
    \coordinate (O) at (0,0);           % Center of sphere
    \coordinate (C) at (0,-\h);         % Center of cross-section
    \coordinate (P) at (\r,-\h);        % Point on the edge of cross-section
    
    % 3. Draw the Sphere
    \draw[thick] (O) circle (\R);
    
    % 4. Draw the Cross-Section (Ellipse)
    % We draw it in two parts: dashed for the back, solid for the front
    % The vertical radius of the ellipse is set to 0.6 for perspective
    \draw[dashed, color=gray] (\r,-\h) arc (0:180:{\r} and {0.6});
    \draw[thick] (\r,-\h) arc (0:-180:{\r} and {0.6});
    
    % 5. Draw the Triangle connecting R, r, and h
    \draw[dashed] (O) -- (C) node[midway, left] {$h$};
    \draw[thick] (C) -- (P) node[midway, below] {$r$};
    \draw[thick] (O) -- (P) node[midway, above right] {$R$};
    
    % 6. Mark the Right Angle
    \draw ($(C) + (0, 0.3)$) -| ($(C) + (0.3, 0)$);
    
    % 7. Add Center Point
    \fill (O) circle (1.5pt);
    
    % Optional: Add a faint equator for better 3D effect
    %\draw[gray!30] (\R,0) arc (0:-180:{\R} and {0.3});
    %\draw[gray!30, dashed] (\R,0) arc (0:180:{\R} and {0.3});

\end{tikzpicture}
\end{center}

\begin{align*}
&& h & = \sqrt{R^2-r^2} \\
&&&= \sqrt{R^2-\frac{\Delta^2}{s^2}} \\
&&&= \sqrt{R^2 - \frac{(s-a)(s-b)(s-c)}{s}}
\end{align*}
\end{questionparts}

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