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\begin{center}
\begin{tikzpicture}

% Define parameters
\def\r{1.2}      % radius
\def\h{2.1}      % height of cone

% Draw the hemisphere bottom curve (solid line)
\draw (-\r,0) arc (180:360:\r);

% Draw the dashed line for the opening/junction
% \draw[dashed] (-\r,0) -- (\r,0);

% Draw the elliptical curve to indicate 3D shape on the hemisphere
\draw[dashed] plot[domain=0:180,samples=50] ({\r*cos(\x)},{0.3*\r*sin(\x)});
\draw plot[domain=180:360,samples=50] ({\r*cos(\x)},{0.3*\r*sin(\x)});

% Draw the cone
\draw (-\r,0) -- (0,\h) -- (\r,0);

% Mark the center point
% \node at (0,0) {$O$};
\fill (0,0) circle (1.5pt);

% Add the radius label with an arrow
\draw[->] (0,0) -- (\r,0) node[midway, below] {$a$};

% Add the height label with an arrow
\draw[->] (0,0) -- (0,\h) node[midway, right] {$\lambda a$};

\end{tikzpicture}
\end{center}

By symmetry the centre of mass will lie on the main axis. Taking the plane faces as $x = 0$ we have the following centers of mass:

\begin{align*}
    && \text{COM} && \text{Mass} \\
    \text{Hemisphere} && -\frac{3a}{8} && \frac{2\pi a^3}{3} \\
    \text{Cone} && \frac{\lambda a}{4} && \frac{\lambda \pi a^3}{3} \\
    \text{Toy} && \bar{x} && \frac{(\lambda + 2)\pi a^3}{3} \\
\end{align*}

Therefore,

\begin{align*}
    && \frac{(\lambda + 2)\pi a^3}{3} \cdot \bar{x} &= -\frac{3a}{8} \cdot \frac{2\pi a^3}{3} + \frac{\lambda a}{4} \cdot \frac{\lambda \pi a^3}{3} \\
    \Rightarrow && (\lambda + 2) \bar{x} &= \frac{(\lambda^2 -3)a}{4}  
\end{align*}

Therefore the centre of mass will be inside the hemisphere (and it will always move to an upright position) iff $\bar{x} < 0 \Leftrightarrow \lambda < \sqrt{3}$.

\begin{center}
\begin{tikzpicture}

\def\r{1.2}      % radius
    \def\h{2.1}      % height of cone

\draw (0, 0) -- (\h, 0) -- (\h/2, {sqrt(3)*\h/2}) -- cycle;

\coordinate (base) at ({\h/2*cos(150) + \h/4},{\h/2*sin(150) + sqrt(3)*\h/4}); 
    \coordinate (centr) at (\h/4, {sqrt(3)*\h/4});
    \coordinate (top) at (\h,0);
    \coordinate (O) at (0,0);
    \draw plot[domain=60:240,samples=50] ({\h/2*cos(\x) + \h/4},{\h/2*sin(\x) + sqrt(3)*\h/4});

\draw[dashed] (\h, 0) -- (base);
    \draw[dashed] (0, \h) -- (O);

\fill ($(base)!0.8!(centr)$) circle (1.5pt);

\pic [draw, angle radius=0.8cm, "$\theta$"] {angle = base--top--O};
    
\end{tikzpicture}
\end{center}

For the toy to topple from this position, $\bar{x}$ must be longer than it would need to be to form a right-angled triangle with the vertical at the plane face. The angle at this point will be $\theta$, so we need:

$\bar{x} > a\tan \theta = a \frac{a}{\lambda a} = \frac{a}{\lambda}$

\begin{align*}
    && \bar{x} &> \frac{a}{\lambda} \\
    \Leftrightarrow && \frac{(3-\lambda^2)a}{4(\lambda + 2)} &> \frac{a}{\lambda} \\
    \Leftrightarrow && {(3-\lambda^2)\lambda} &> {4(\lambda + 2)} \\
    \Leftrightarrow && -\lambda^3-\lambda -8 &> 0 \\
\end{align*}

Contradiction! Therefore it can never topple when laid on its side.

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