Four persons K, L, M, N are initially at the four corners of a square of side d. Each person now mov

Q: Wait, do the particles travel in straight lines to the center?

No, they travel in curved paths called logarithmic spirals. As they move, the square they form rotates, so the direction they are heading constantly changes. However, their *distance* to the center decreases at a constant rate.

Q: Why can we assume the radial velocity is constant?

Because of perfect symmetry, the particles always form a square. The angle between the side of a square and its diagonal to the center is always $45^\circ$. Since the velocity is always along the side, its component towards the center is always $v \cos 45^\circ$.

Q: Can we just use relative velocity between two adjacent particles to solve this?

Yes! This is a brilliant shortcut. The velocity of approach between $K$ and $L$ is $v - v \cos 90^\circ = v$. Since the initial distance is $d$, the time is simply $\frac{d}{v}$.

Q: What if there were 3 particles forming an equilateral triangle?

The same logic applies! They would meet at the centroid. The angle between the side and the line to the center would be $30^\circ$. The radial velocity would be $v \cos 30^\circ$, and you could solve it similarly.

Q: Does the size of the particles matter?

In this idealized kinematics problem, we treat the persons as point masses (particles) with zero size. If they had physical dimensions, they would collide before reaching the exact mathematical center.

Question

Four persons $K, L, M, N$ are initially at the four corners of a square of side $d$. Each person now moves with a uniform speed $v$ in such a way that $K$ always moves directly towards $L$, $L$ directly towards $M$, $M$ directly towards $N$ and $N$ directly towards $K$. The four persons will meet at a time ......

Sigma Percentile · Accepted Answer

### The Deceptive Complexity of the Pursuit

Imagine four people standing at the four corners of a square. Suddenly, they all start running. But they aren't running in a straight line to the center; instead, each person is relentlessly chasing the person to their immediate right.

At first glance, this sounds like a terrifying calculus problem. Because the target is always moving, the pursuer must constantly change direction. This results in a beautiful, spiraling path known as a logarithmic spiral.

However, we don't need heavy calculus to solve this. By leveraging the physical geometry of the situation, we can bypass the complex curves entirely and arrive at an elegant solution.

### The Power of Symmetry

The secret weapon in this problem is **symmetry**. Because all four people are moving at the exact same speed $v$ and are chasing each other in a perfectly symmetric loop, the shape they form at any given instant will *always* be a square.

As they run, this square will continuously shrink and rotate. But no matter how small it gets, it remains a square.

The only point in space that remains completely invariant during this symmetric collapse is the geometric center of the original square. Therefore, logic dictates that all four people must eventually collide exactly at the center point, $O$.

### The Radial Approach: Cutting Through the Math

Since we know they will meet at the center, let's stop worrying about the complex spiral path and instead focus on the straight-line distance from a corner to the center. Let's track just one person, say $K$.

The initial distance from the corner of a square of side $d$ to its center is half of the diagonal. We can calculate this initial radial distance $r$ as:

$$r = \frac{d\sqrt{2}}{2} = \frac{d}{\sqrt{2}}$$

Now, person $K$ is running with a velocity $v$ directed along the edge of the square. But we only care about how fast they are closing the distance to the center. We need the component of their velocity along the radial line.

Because the shape is always a square, the angle between the edge (their velocity vector) and the diagonal (the line to the center) is always exactly $45^\circ$. This is the crucial insight! The radial velocity $v_r$ is constant throughout the entire motion:

$$v_r = v \cos 45^\circ = \frac{v}{\sqrt{2}}$$

Since both the distance to be covered and the speed at which it is being covered are known, we can simply use the basic kinematics formula for time:

$$t = \frac{r}{v_r} = \frac{\frac{d}{\sqrt{2}}}{\frac{v}{\sqrt{2}}} = \frac{d}{v}$$

### The Relative Velocity Shortcut: A Stroke of Genius

If you thought the radial approach was elegant, there is an even faster way to solve this using the concept of relative velocity of approach.

Consider person $K$ chasing person $L$. The initial distance between them is $d$. Person $K$ is running directly towards $L$ with speed $v$.

What is $L$ doing? Person $L$ is running towards $M$, which is exactly perpendicular to the line connecting $K$ and $L$. Therefore, the component of $L$'s velocity along the line connecting them is $v \cos 90^\circ = 0$.

The relative velocity at which the gap between them is closing is simply:

$$v_{app} = v - 0 = v$$

Because they always form a square, this perpendicular relationship holds true at every single instant. The gap between them is always closing at a constant rate of $v$.

To close an initial gap of $d$ at a constant rate of $v$, the time taken is simply:

$$t = \frac{d}{v_{app}} = \frac{d}{v}$$

### Conclusion

Whether you use the radial component method or the relative velocity shortcut, the answer is beautifully simple. What initially appears to be a nightmare of spiraling calculus collapses into a single line of algebra when viewed through the lens of symmetry.

Visualized Solution

$I ni t ia l S e t u p$

$S y mm e t r y an d t h e M ee t in g P o in t$

$T h e R a d ia l P a t h$

$I ni t ia l R a d ia l D i s t an ce$

$R eso l v in g V e l oc i t y$

$C a l c u l a t in g R a d ia l V e l oc i t y$

$T im e t o M ee t$

The Sigma Insight: Relative Velocity

The Deceptive Complexity of the Pursuit

The Power of Symmetry

The Radial Approach

The Relative Velocity Shortcut

Conclusion

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Visualized Solution

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