$OABC$ 是一个菱形。点 $A$ 和 $C$ 满足 $\overrightarrow{OA}=\mathbf{a}$ 且 $\overrightarrow{OC}=\mathbf{c}$。点 $M$ 是该菱形的中心，且角 $AOC$ 为 $60^{\circ}$。

1. Express the vector $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

Since $OABC$ is a rhombus, $\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{c}$.
The centre $M$ is the midpoint of the diagonal $OB$, so:
$\overrightarrow{OM} = \frac{1}{2}\overrightarrow{OB} = \frac{1}{2}(\mathbf{a} + \mathbf{c})$

2. State the formula for the length of projection of $\overrightarrow{OM}$ on $\overrightarrow{OC}$.

The length of projection of $\overrightarrow{OM}$ on $\overrightarrow{OC}$ is given by:
$\text{Length of projection} = |\overrightarrow{OM} \cdot \hat{\mathbf{c}}| = \frac{\overrightarrow{OM} \cdot \mathbf{c}}{|\mathbf{c}|}$

3. Evaluate the scalar product $\mathbf{a} \cdot \mathbf{c}$ using the properties of the rhombus.

Since $OABC$ is a rhombus, the side lengths are equal, so $|\mathbf{a}| = |\mathbf{c}|$.
The angle between $\mathbf{a}$ and $\mathbf{c}$ is $\angle AOC = 60^{\circ}$.
$\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos 60^{\circ}$
$\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}| |\mathbf{c}| \left(\frac{1}{2}\right) = \frac{1}{2}|\mathbf{c}|^2$

4. Calculate $\overrightarrow{OM} \cdot \mathbf{c}$ and find the final length of projection.

Substitute $\overrightarrow{OM}$ into the scalar product:
$\overrightarrow{OM} \cdot \mathbf{c} = \frac{1}{2}(\mathbf{a} + \mathbf{c}) \cdot \mathbf{c}$
$\overrightarrow{OM} \cdot \mathbf{c} = \frac{1}{2}(\mathbf{a} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{c})$
Since $\mathbf{c} \cdot \mathbf{c} = |\mathbf{c}|^2$:
$\overrightarrow{OM} \cdot \mathbf{c} = \frac{1}{2}\left(\frac{1}{2}|\mathbf{c}|^2 + |\mathbf{c}|^2\right) = \frac{1}{2}\left(\frac{3}{2}|\mathbf{c}|^2\right) = \frac{3}{4}|\mathbf{c}|^2$
Therefore, the length of projection is:
$\frac{\frac{3}{4}|\mathbf{c}|^2}{|\mathbf{c}|} = \frac{3}{4}|\mathbf{c}| \text{ (Shown)}$

1. Express the position vector of M and state the area of triangle OMC in terms of a vector product.

Since $M$ is the centre of the rhombus, it is the midpoint of the diagonal $AC$ (or $OB$).
$M$ can be found using the diagonals bisecting each other:
$\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OC}) = \frac{1}{2}(\mathbf{a} + \mathbf{c})$

The area of $\triangle OMC = \frac{1}{2} |\overrightarrow{OM} \times \overrightarrow{OC}|$

2. Simplify the vector product using the distributive property.

$\overrightarrow{OM} \times \overrightarrow{OC} = \frac{1}{2}(\mathbf{a} + \mathbf{c}) \times \mathbf{c}$
$= \frac{1}{2}(\mathbf{a} \times \mathbf{c}) + \frac{1}{2}(\mathbf{c} \times \mathbf{c})$

Since the vector product of any vector with itself is the zero vector ($\mathbf{c} \times \mathbf{c} = \mathbf{0}$):
$\overrightarrow{OM} \times \overrightarrow{OC} = \frac{1}{2}(\mathbf{a} \times \mathbf{c})$

3. Calculate the magnitude of the simplified vector product to find the area and the constant $k$.

$\text{Area} = \frac{1}{2} \left| \frac{1}{2}(\mathbf{a} \times \mathbf{c}) \right| = \frac{1}{4} |\mathbf{a} \times \mathbf{c}|$

By the definition of the magnitude of a vector product:
$|\mathbf{a} \times \mathbf{c}| = |\mathbf{a}| \,|\mathbf{c}| \sin(60^{\circ})$

Since $OABC$ is a rhombus, all sides are equal in length, meaning $|\mathbf{a}| = |\mathbf{c}|$.
$|\mathbf{a} \times \mathbf{c}| = |\mathbf{c}| \,|\mathbf{c}| \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}|\mathbf{c}|^2$

$\text{Area} = \frac{1}{4} \left( \frac{\sqrt{3}}{2}|\mathbf{c}|^2 \right) = \frac{\sqrt{3}}{8}|\mathbf{c}|^2$
$\therefore k = \frac{\sqrt{3}}{8}$

1. Find the position vector of D, $\overrightarrow{OD}$, in terms of $\mathbf{c}$.

Since $D$ lies on $OC$ produced and $OC:CD = 2:3$, the length of $OD$ is $5$ parts compared to $OC$'s $2$ parts.
$\overrightarrow{OD} = \frac{2+3}{2}\overrightarrow{OC} = \frac{5}{2}\mathbf{c}$

2. Formulate an expression for the shortest distance from $D$ to the line $OA$.

The shortest distance $d$ from point $D$ to the line $OA$ can be found using the right-angled triangle formed by the perpendicular from $D$ to $OA$.
$d = |\overrightarrow{OD}| \sin(\text{angle between } \overrightarrow{OD} \text{ and } \overrightarrow{OA})$
Since $D$ lies on the line passing through $O$ and $C$, the angle between $\overrightarrow{OD}$ and $\overrightarrow{OA}$ is the same as the angle $AOC$, which is $60^{\circ}$.
$d = \left| \frac{5}{2}\mathbf{c} \right| \sin 60^{\circ}$

3. Calculate the exact value of the shortest distance to find $t$.

$d = \frac{5}{2}|\mathbf{c}| \left( \frac{\sqrt{3}}{2} \right)$
$d = \frac{5\sqrt{3}}{4}|\mathbf{c}|$
Comparing this to the given form $t|\mathbf{c}|$, we have:
$t = \frac{5\sqrt{3}}{4}$