Animated Solution for Physics - Kinematics: Three ants A, B and C are crawling on a large horizontal tabletop always occupying vertices of an equilateral triangle, size of which may vary with time. If at an instant, speeds of A and B are $v_A$ and $v_B$, which of the following conclusions can you make for speed $v_C$ of C?

Imagine three ants crawling on a large horizontal tabletop. They are moving in such a perfectly coordinated dance that, at any given instant, they always form the vertices of an equilateral triangle. The size of this triangle might be expanding or shrinking, and it might be rotating, but its shape remains strictly equilateral. We are given the speeds of ants A and B as $v_A$ and $v_B$, and we need to find a constraint on the speed of ant C, $v_C$.

While we could try to solve this using standard 2D vectors and rotation matrices, the algebra gets incredibly messy. Instead, we will use a much more elegant tool: Complex Numbers.

By treating the tabletop as a complex plane, we can represent the positions of the ants as complex numbers $z_A$, $z_B$, and $z_C$. The beauty of complex numbers is that a rotation in 2D space is achieved simply by multiplying by a complex exponential, $e^{i\theta }$.

Because the ants always form an equilateral triangle, the vector pointing from A to C is simply the vector pointing from A to B, rotated by $60^{\circ }$ (which is $\pi /3$ radians). Depending on whether the vertices are arranged clockwise or counter-clockwise, this rotation could be positive or negative.

Mathematically, we write this as:

Let's isolate $z_C$ to see how its position depends on the other two ants. Expanding the bracket and grouping the $z_A$ terms together, we get:

Now, we encounter a term that looks a bit complicated: $1-e^{i\pi /3}$. But watch what happens when we expand it using Euler's formula ($e^{i\theta }=\cos \theta +i\sin \theta$):

Subtracting the real parts, we get:

This is exactly the expansion of $e^{-i\pi /3}$! This beautiful symmetry means that $1-e^{\pm i\pi /3}=e^{\mp i\pi /3}$. Substituting this back into our position equation, we get a wonderfully clean and symmetric relation:

We are asked about speeds, which are the magnitudes of velocities. To find the velocities, we simply differentiate our position equation with respect to time $t$. Since the rotation factors $e^{\pm i\pi /3}$ are constants, they remain unchanged during differentiation:

This equation tells us that the velocity vector of C is a linear combination of the velocity vectors of A and B, each rotated by $60^{\circ }$ in opposite directions.

To find the relationship between their speeds, we must take the modulus (magnitude) of both sides. Here, we apply the Triangle Inequality, which states that the magnitude of a sum of vectors is always less than or equal to the sum of their individual magnitudes ($|\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|$):

The magnitude of any complex exponential $e^{i\theta }$ is exactly $1$. Therefore, the rotation factors disappear when we take the modulus, leaving us with our final, elegant conclusion:

When does equality hold? The speed of C will be exactly equal to the sum of the speeds of A and B only when the two rotated velocity vectors point in the exact same direction. Geometrically, this requires the original velocity vectors of A and B to be at an angle of $120^{\circ }$ to each other.