相对于原点 O，点 A 和 B 的位置向量分别为 $\mathbf{a}$ and $\mathbf{b}$，其中 $\mathbf{a}$ 和 $\mathbf{b}$ 是非零且不平行的向量。点 M 是 OA 的中点，点 N 在 OB 上且满足 $ON:NB$ 为 $1:2$。位置向量为 $\mathbf{c}$ 的点 C 是直线 AN 和 BM 的交点。 点 O、A 和 C 在一个圆上，其中 OA 是该圆的直径。已知 $\mathbf{a}\cdot\mathbf{b}=\tfrac{1}{2}|\mathbf{a}|^2$。

1. Find the position vectors of $M$ and $N$ relative to the origin.

Since $M$ is the midpoint of $OA$,
$\vec{OM} = \frac{1}{2}\mathbf{a}$
Since $N$ lies on $OB$ such that $ON:NB = 1:2$, the point $N$ is $\frac{1}{3}$ of the way along $OB$. Thus,
$\vec{ON} = \frac{1}{3}\mathbf{b}$

2. Express $\mathbf{c}$ in two different ways using the collinear points $A, C, N$ and $B, C, M$.

Since $C$ lies on the line $AN$, $\vec{AC} = \lambda \vec{AN}$ for some scalar $\lambda$.
$\mathbf{c} - \mathbf{a} = \lambda (\vec{ON} - \vec{OA})$
$\mathbf{c} = \mathbf{a} + \lambda \left(\frac{1}{3}\mathbf{b} - \mathbf{a}\right) = (1-\lambda)\mathbf{a} + \frac{\lambda}{3}\mathbf{b}$
Since $C$ also lies on the line $BM$, $\vec{BC} = \mu \vec{BM}$ for some scalar $\mu$.
$\mathbf{c} - \mathbf{b} = \mu (\vec{OM} - \vec{OB})$
$\mathbf{c} = \mathbf{b} + \mu \left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right) = \frac{\mu}{2}\mathbf{a} + (1-\mu)\mathbf{b}$

3. Compare coefficients of $\mathbf{a}$ and $\mathbf{b}$ to solve for the parameters.

Since vectors $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel, we can equate their coefficients from the two expressions for $\mathbf{c}$:
$1-\lambda = \frac{\mu}{2} \implies \mu = 2 - 2\lambda$
$\frac{\lambda}{3} = 1-\mu$
Substitute the expression for $\mu$ into the second equation:
$\frac{\lambda}{3} = 1 - (2 - 2\lambda)$
$\frac{\lambda}{3} = -1 + 2\lambda$
$1 = 2\lambda - \frac{\lambda}{3}$
$1 = \frac{5\lambda}{3} \implies \lambda = \frac{3}{5}$
Substitute $\lambda = \frac{3}{5}$ back into the first expression for $\mathbf{c}$:
$\mathbf{c} = \left(1-\frac{3}{5}\right)\mathbf{a} + \frac{1}{3}\left(\frac{3}{5}\right)\mathbf{b}$
$\mathbf{c} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b} \text{ (shown)}$

1. State the standard geometrical interpretation of the dot product with a unit vector

The expression $|\mathbf{a}\cdot\hat{\mathbf{c}}|$ mathematically calculates the magnitude of the projection of vector $\mathbf{a}$ onto the direction given by $\hat{\mathbf{c}}$.

Geometrically, this represents the length of the projection of vector $\mathbf{a}$ onto vector $\mathbf{c}$ (or the length of the projection of line segment $OA$ onto line $OC$).

1. Relate vectors $\mathbf{a}$ and $\mathbf{c}$ using the property of the circle.

Since $OA$ is the diameter of the circle containing $O$, $A$, and $C$, the angle in the semicircle is a right angle, so $\angle OCA = 90^\circ$.
Therefore, the vectors $\vec{OC}$ and $\vec{CA}$ are perpendicular.
$\vec{OC} \cdot \vec{CA} = 0$
$\mathbf{c} \cdot (\mathbf{a} - \mathbf{c}) = 0$
$\mathbf{a} \cdot \mathbf{c} - |\mathbf{c}|^2 = 0$
$\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}|^2$

2. Evaluate the dot product $\mathbf{a} \cdot \mathbf{c}$ using the given expressions.

Substitute $\mathbf{c} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}$ into the scalar product:
$\mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot \left(\frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}\right)$
$\mathbf{a} \cdot \mathbf{c} = \frac{2}{5}|\mathbf{a}|^2 + \frac{1}{5}(\mathbf{a} \cdot \mathbf{b})$
Given that $\mathbf{a} \cdot \mathbf{b} = \frac{1}{2}|\mathbf{a}|^2$, we substitute this into the equation:
$\mathbf{a} \cdot \mathbf{c} = \frac{2}{5}|\mathbf{a}|^2 + \frac{1}{5}\left(\frac{1}{2}|\mathbf{a}|^2\right)$
$\mathbf{a} \cdot \mathbf{c} = \frac{2}{5}|\mathbf{a}|^2 + \frac{1}{10}|\mathbf{a}|^2$
$\mathbf{a} \cdot \mathbf{c} = \frac{1}{2}|\mathbf{a}|^2$

3. Combine the results to deduce the ratio $|\mathbf{a}|:|\mathbf{c}|$.

From Step 1, we know $|\mathbf{c}|^2 = \mathbf{a} \cdot \mathbf{c}$. Substituting the result from Step 2:
$|\mathbf{c}|^2 = \frac{1}{2}|\mathbf{a}|^2$
$\frac{|\mathbf{a}|^2}{|\mathbf{c}|^2} = 2$
Taking the positive square root (since magnitudes are positive):
$\frac{|\mathbf{a}|}{|\mathbf{c}|} = \sqrt{2}$
Therefore, the ratio $|\mathbf{a}| : |\mathbf{c}|$ is $\sqrt{2} : 1$.

1. Derive the magnitude of $\mathbf{c}$ in terms of $|\mathbf{a}|$ using the condition that $OA$ is the diameter of the circle (required result from part iii).

Since $OA$ is the diameter of the circle passing through $O, A$, and $C$, the angle in the semicircle $\angle OCA = 90^\circ$. This implies $\vec{OC}$ is perpendicular to $\vec{CA}$:
$\vec{OC} \cdot \vec{CA} = 0$
$\mathbf{c} \cdot (\mathbf{a} - \mathbf{c}) = 0 \implies |\mathbf{c}|^2 = \mathbf{c} \cdot \mathbf{a}$
Substitute $\mathbf{c} = \frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}$ and the given $\mathbf{a} \cdot \mathbf{b} = \frac{1}{2}|\mathbf{a}|^2$:
$|\mathbf{c}|^2 = \left(\frac{2}{5}\mathbf{a} + \frac{1}{5}\mathbf{b}\right) \cdot \mathbf{a} = \frac{2}{5}|\mathbf{a}|^2 + \frac{1}{5}(\mathbf{a} \cdot \mathbf{b})$
$|\mathbf{c}|^2 = \frac{2}{5}|\mathbf{a}|^2 + \frac{1}{5}\left(\frac{1}{2}|\mathbf{a}|^2\right) = \frac{1}{2}|\mathbf{a}|^2$
$|\mathbf{c}| = \frac{1}{\sqrt{2}}|\mathbf{a}|$

2. Calculate the area of the right-angled triangle $OAC$.

Because $\triangle OAC$ is a right-angled triangle at $C$, its area can be found using $\frac{1}{2} \times \text{base} \times \text{height}$.
Using Pythagoras' theorem to find the length of $AC$:
$|\vec{CA}|^2 = |\vec{OA}|^2 - |\vec{OC}|^2 = |\mathbf{a}|^2 - \frac{1}{2}|\mathbf{a}|^2 = \frac{1}{2}|\mathbf{a}|^2$
$|\vec{CA}| = \frac{1}{\sqrt{2}}|\mathbf{a}|$
Therefore, the area of $\triangle OAC$ is:
$\text{Area} = \frac{1}{2} |\vec{OC}| |\vec{CA}| = \frac{1}{2} \left( \frac{1}{\sqrt{2}}|\mathbf{a}| \right) \left( \frac{1}{\sqrt{2}}|\mathbf{a}| \right) = \frac{1}{4}|\mathbf{a}|^2$