点 A 和 B 相对于原点 O 的位置向量分别为 $\mathbf{a}$ 和 $\mathbf{b}$。点 C 满足 OACB 为平行四边形。AC 上的点 P 满足 AP : PC = 1 : 2，且 BC 上的点 Q 满足 BQ : QC = 1 : 2。

1. Find the vectors $\overrightarrow{QC}$ and $\overrightarrow{CP}$ by using the properties of the parallelogram and given ratios.

Since $OACB$ is a parallelogram, $\overrightarrow{BC} = \overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}$.
Given $BQ : QC = 1 : 2$, the length of $QC$ is $\frac{2}{3}$ of $BC$. Thus, $\overrightarrow{QC} = \frac{2}{3}\overrightarrow{BC} = \frac{2}{3}\mathbf{a}$.
Given $AP : PC = 1 : 2$, the length of $PC$ is $\frac{2}{3}$ of $AC$. Thus, $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{AC} = \frac{2}{3}\mathbf{b}$, which gives $\overrightarrow{CP} = -\frac{2}{3}\mathbf{b}$.

2. Determine $\overrightarrow{QP}$ using vector addition.

$\overrightarrow{QP} = \overrightarrow{QC} + \overrightarrow{CP} = \frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}$

1. Determine $\overrightarrow{OC}$ and state $\overrightarrow{QP}$ from part (i).

Since $OACB$ is a parallelogram,
$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB} = \mathbf{a} + \mathbf{b}$
From part (i), we have found:
$\overrightarrow{QP} = \frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}$

2. Set up the scalar product $\overrightarrow{OC} \cdot \overrightarrow{QP}$ and expand.

$\overrightarrow{OC} \cdot \overrightarrow{QP} = (\mathbf{a} + \mathbf{b}) \cdot \left( \frac{2}{3}\mathbf{a} - \frac{2}{3}\mathbf{b} \right)$
$= \frac{2}{3}(\mathbf{a} \cdot \mathbf{a}) - \frac{2}{3}(\mathbf{a} \cdot \mathbf{b}) + \frac{2}{3}(\mathbf{b} \cdot \mathbf{a}) - \frac{2}{3}(\mathbf{b} \cdot \mathbf{b})$

3. Simplify using properties of the scalar product to find $\lambda$ and $\mu$.

Since the scalar product is commutative ($\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$) and $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$:
$- \frac{2}{3}(\mathbf{a} \cdot \mathbf{b}) + \frac{2}{3}(\mathbf{b} \cdot \mathbf{a}) = 0$
$\overrightarrow{OC} \cdot \overrightarrow{QP} = \frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2$
Comparing this to $\lambda|\mathbf{a}|^2+\mu|\mathbf{b}|^2$, we find:
$\lambda = \frac{2}{3}$
$\mu = -\frac{2}{3}$

1. Concisely evaluate $\overrightarrow{OC} \cdot \overrightarrow{QP}$ using results from parts (i) and (ii).

From part (i), $\overrightarrow{QP} = \frac{2}{3}(\mathbf{a} - \mathbf{b})$.
Since OACB is a parallelogram, $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \mathbf{a} + \mathbf{b}$.
Evaluating the dot product as required in part (ii):
$\overrightarrow{OC} \cdot \overrightarrow{QP} = (\mathbf{a} + \mathbf{b}) \cdot \frac{2}{3}(\mathbf{a} - \mathbf{b})$
$= \frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2$

2. Set the scalar product to zero to find the relationship between the lengths of $\mathbf{a}$ and $\mathbf{b}$.

Given that $\overrightarrow{OC} \cdot \overrightarrow{QP} = 0$, we substitute the expression from part (ii):
$\frac{2}{3}|\mathbf{a}|^2 - \frac{2}{3}|\mathbf{b}|^2 = 0$
$|\mathbf{a}|^2 = |\mathbf{b}|^2$
$|\mathbf{a}| = |\mathbf{b}|$
This implies that the magnitudes of $\mathbf{a}$ and $\mathbf{b}$ are the same, meaning adjacent sides $OA$ and $OB$ are equal in length.

3. Identify the specific shape of the parallelogram and justify it.

A parallelogram with adjacent sides of equal length is a rhombus.
Therefore, OACB is a rhombus.