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Problem Text
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} so \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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