We are given a parallelogram $ABCD$, where:

• $AB = 3\sqrt{2}$
• $AD = 8$
• $\angle BAD = 45^\circ$
• $M$ is a point on $CD$ such that $CM:MD = 1:2$
• $N$ is a point on $AD$ such that $AN:ND = 3:1$

We need to find:

a) The length of $AM$
b) The length of $BN$
c) The acute angle between $AM$ and $BN$

Step 1: Set up a coordinate system

Let’s place point $A$ at the origin $(0, 0)$, point $B$ along the $x$-axis, and point $D$ along the $y$-axis. From the given information:

• $A(0, 0)$
• $B(3\sqrt{2}, 0)$
• $D(0, 8)$

Using the property of a parallelogram, $C$, the opposite vertex of $A$, can be found by adding the vectors $\overrightarrow{AD}$ and $\overrightarrow{AB}$:

$C = B + D = (3\sqrt{2}, 8)$

Step 2: Find coordinates of points M and N

• For point $M$ on $CD$:
  The ratio $CM:MD = 1:2$ implies that $M$ divides the line segment $CD$ in the ratio 1:2. Using the section formula, the coordinates of $M$ are:

  $M = \left( \frac{2(0) + 1(3\sqrt{2})}{3}, \frac{2(8) + 1(0)}{3} \right) = \left( \frac{3\sqrt{2}}{3}, \frac{16}{3} \right) = \left( \sqrt{2}, \frac{16}{3} \right)$

• For point $N$ on $AD$:
  The ratio $AN:ND = 3:1$ implies that $N$ divides the line segment $AD$ in the ratio 3:1. Using the section formula, the coordinates of $N$ are:

  $N = \left( \frac{1(0) + 3(0)}{4}, \frac{1(8) + 3(0)}{4} \right) = (0, 2)$

Step 3: Calculate the length of $AM$

The length of $AM$ can be calculated using the distance formula:

$AM = \sqrt{(x_M - x_A)^2 + (y_M - y_A)^2} = \sqrt{(\sqrt{2} - 0)^2 + \left( \frac{16}{3} - 0 \right)^2}$

$AM = \sqrt{2 + \left( \frac{16}{3} \right)^2} = \sqrt{2 + \frac{256}{9}} = \sqrt{\frac{18}{9} + \frac{256}{9}} = \sqrt{\frac{274}{9}} = \frac{\sqrt{274}}{3}$

Step 4: Calculate the length of $BN$

The length of $BN$ can be calculated using the distance formula:

$BN = \sqrt{(x_B - x_N)^2 + (y_B - y_N)^2} = \sqrt{(3\sqrt{2} - 0)^2 + (0 - 2)^2}$

$BN = \sqrt{(3\sqrt{2})^2 + (-2)^2} = \sqrt{18 + 4} = \sqrt{22}$

Step 5: Calculate the acute angle between $AM$ and $BN$

To find the acute angle $\theta$ between $AM$ and $BN$, we use the formula for the dot product:

$\cos \theta = \frac{\vec{AM} \cdot \vec{BN}}{|\vec{AM}| |\vec{BN}|}$

First, compute the vectors $\vec{AM}$ and $\vec{BN}$:

• $\vec{AM} = (\sqrt{2} - 0, \frac{16}{3} - 0) = (\sqrt{2}, \frac{16}{3})$
• $\vec{BN} = (0 - 3\sqrt{2}, 2 - 0) = (-3\sqrt{2}, 2)$

The dot product $\vec{AM} \cdot \vec{BN}$ is:

[ \vec{AM} \cdot \vec{BN} = (\sqrt{2})(-3\sqrt{2}) + \left(\frac{16}{3}\right)(2) = -6 + \frac{32}{3} = \frac{-18