Vector Representation and Rectangle Properties

The position vectors $\vec{a}$ and $\vec{b}$ represent adjacent sides $\vec{OA}$ and $\vec{OB}$ of rectangle $OACB$. Since it's a rectangle, the sides $\vec{a}$ and $\vec{b}$ are perpendicular, meaning their dot product is zero: $\vec{a} \cdot \vec{b} = 0$. The diagonals of the rectangle are represented by the vectors $\vec{OC} = \vec{a} + \vec{b}$ and $\vec{AB} = \vec{b} - \vec{a}$.

Magnitude Ratio from Given Condition

The given condition is $16|\vec{a} \times \vec{b}| = 3(|\vec{a}| + |\vec{b}|)^2$. Because $\vec{a}$ and $\vec{b}$ are perpendicular, the magnitude of their cross product is $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|$. Let $a = |\vec{a}|$ and $b = |\vec{b}|$. The condition simplifies to $16ab = 3(a+b)^2$. This relationship implies that the ratio of the side lengths is either $|\vec{a}|/|\vec{b}| = \sqrt{3}$ or $|\vec{b}|/|\vec{a}| = \sqrt{3}$.

Angle Between Diagonals Calculation

Let $\theta$ be the acute angle between the diagonals $\vec{OC}$ and $\vec{AB}$. The cosine of this angle can be calculated using the dot product formula:

$ \cos \theta = \frac{(\vec{a} + \vec{b}) \cdot (\vec{b} - \vec{a})}{|\vec{a} + \vec{b}| |\vec{b} - \vec{a}|} $

Expanding the dot product and magnitudes:

$ \cos \theta = \frac{\vec{a}\cdot\vec{b} - |\vec{a}|^2 + |\vec{b}|^2 - \vec{b}\cdot\vec{a}}{|\vec{a}|^2 + |\vec{b}|^2} = \frac{|\vec{b}|^2 - |\vec{a}|^2}{|\vec{a}|^2 + |\vec{b}|^2} $

Using the ratio derived from the condition, where $|\vec{b}|^2 = 3|\vec{a}|^2$ or $|\vec{a}|^2 = 3|\vec{b}|^2$:

Case 1: $|\vec{b}|^2 = 3|\vec{a}|^2$ $ \cos \theta = \frac{3|\vec{a}|^2 - |\vec{a}|^2}{|\vec{a}|^2 + 3|\vec{a}|^2} = \frac{2|\vec{a}|^2}{4|\vec{a}|^2} = \frac{1}{2} $

Case 2: $|\vec{a}|^2 = 3|\vec{b}|^2$ $ \cos \theta = \frac{|\vec{b}|^2 - 3|\vec{b}|^2}{3|\vec{b}|^2 + |\vec{b}|^2} = \frac{-2|\vec{b}|^2}{4|\vec{b}|^2} = -\frac{1}{2} $

Since $\theta$ is defined as the acute angle, we consider $\cos \theta = 1/2$, which means $\theta = 60^\circ$.

Finding $\tan(\theta/2)$

We use the half-angle identity for tangent:

$ \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \cos \theta}{1 + \cos \theta} $

Substitute the value $\cos \theta = 1/2$:

$ \tan^2\left(\frac{\theta}{2}\right) = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} $

Since $\theta$ is an acute angle ($0^\circ < \theta < 90^\circ$), $\theta/2$ is also acute ($0^\circ < \theta/2 < 45^\circ$), ensuring $\tan(\theta/2)$ is positive.

$ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} $