KINEMATICS O22

We can represent the positions of the ants in the complex plane as $z_A, z_B,$ and $z_C$.
Since the triangle $ABC$ is equilateral (let’s assume counter-clockwise ordering $A \to B \to C$), the position of $C$ can be found by taking the vector $AB$ ($z_B – z_A$) and rotating it by $+60^\circ$ ($e^{i\pi/3}$) relative to $A$.

Step 1: Position Relation
$$ z_C = z_A + (z_B – z_A)e^{i\pi/3} $$

Step 2: Differentiate with respect to time
$$ \dot{z}_C = \dot{z}_A + (\dot{z}_B – \dot{z}_A)e^{i\pi/3} $$ $$ v_C = v_A (1 – e^{i\pi/3}) + v_B e^{i\pi/3} $$ Using the identity $1 – e^{i\pi/3} = 1 – (\frac{1}{2} + i\frac{\sqrt{3}}{2}) = \frac{1}{2} – i\frac{\sqrt{3}}{2} = e^{-i\pi/3}$:

$$ \vec{v}_C = \vec{v}_A e^{-i\pi/3} + \vec{v}_B e^{i\pi/3} $$

Step 3: Modulus and Inequality
The velocity of C is the sum of velocity A rotated by $-60^\circ$ and velocity B rotated by $+60^\circ$. Taking the modulus on both sides and using the Triangle Inequality ($|z_1 + z_2| \leq |z_1| + |z_2|$):

$$ |v_C| = |v_A e^{-i\pi/3} + v_B e^{i\pi/3}| $$ $$ |v_C| \leq |v_A e^{-i\pi/3}| + |v_B e^{i\pi/3}| $$ Since $|e^{i\theta}| = 1$: $$ |v_C| \leq |v_A| + |v_B| $$ $$ v_C \leq v_A + v_B $$

Using the position vectors $\vec{r}_A, \vec{r}_B, \vec{r}_C$. Let $M$ be the midpoint of $AB$.

Step 1: Define Position of C
$$ \vec{r}_C = \vec{r}_M + \vec{HM} $$ where $\vec{r}_M = \frac{\vec{r}_A + \vec{r}_B}{2}$ and $\vec{HM}$ is the altitude vector. The altitude vector can be obtained by rotating the vector $\vec{BA} = (\vec{r}_A – \vec{r}_B)$ by $90^\circ$ and scaling by $\frac{\sqrt{3}}{2}$. $$ \vec{r}_C = \frac{\vec{r}_A + \vec{r}_B}{2} + \frac{\sqrt{3}}{2} (\vec{r}_A – \vec{r}_B) \times \hat{k} $$ (Assuming specific orientation for the cross product to point towards C).

Step 2: Differentiate for Velocity
$$ \vec{v}_C = \frac{\vec{v}_A + \vec{v}_B}{2} + \frac{\sqrt{3}}{2} (\vec{v}_A – \vec{v}_B) \times \hat{k} $$ Grouping the terms by $\vec{v}_A$ and $\vec{v}_B$: $$ \vec{v}_C = \vec{v}_A \left[ \frac{1}{2} + \frac{\sqrt{3}}{2}(\dots \times \hat{k}) \right] + \vec{v}_B \left[ \frac{1}{2} – \frac{\sqrt{3}}{2}(\dots \times \hat{k}) \right] $$

Step 3: Interpretation
The terms in the brackets are rotation operators. The magnitude of a vector does not change upon rotation or cross product with a unit normal vector ($\hat{k}$). Essentially, $\vec{v}_C$ is composed of a rotated version of $\vec{v}_A$ plus a rotated version of $\vec{v}_B$. $$ \vec{v}_C = \vec{v}_A’ + \vec{v}_B’ $$ where $|\vec{v}_A’| = |\vec{v}_A|$ and $|\vec{v}_B’| = |\vec{v}_B|$. Applying the triangle inequality for vectors ($|\vec{a}+\vec{b}| \leq |\vec{a}| + |\vec{b}|$): $$ |\vec{v}_C| \leq |\vec{v}_A’| + |\vec{v}_B’| $$ $$ v_C \leq v_A + v_B $$