The distance to Abbey Road?

This problem can be solved without calculation:

Due to the symmetry of the problem, the angles between beetles stay constant throughout the entire process. Therefore, the path of each beetle remains at right angles to the path of the neighboring beetles.

Given these perpendicular trajectories, each beetle is always traveling directly toward its target beetle while its target beetle is performing no motion to either increase or decrease the gulf between them (as it is always moving perpendicularly).

Then, each beetle simply traverses the original distance which separated them from the next beetle; the length of the spiral path is equal to the length of a side of the original square, $\SI{2}{\meter}.$

Let s be the distance between John and Paul, and v = 0.01m/s. After a short time dt, the new distance between them is s' = sqrt((s-v dt)^2+(v dt)^2). s' = s+ds, so substituting and squaring both sides gives us (s+ds)^2 = (s-v dt)^2+(v dt)^2 2s ds+(ds)^2 = -2sv dt+2 v^2 (dt)^2 ds/dt = (-2sv+dt)/(2s+ds) as dt approaches 0, ds approaches 0 too, so ds/dt = -v. In other words, the rate that the beetles are approaching each other is the rate that they crawl, so the answer is 2m.

Click HERE for a diagram of how they meet at the centre.

Let's consider a $n$ -sided regular polygon, each of side length $d$ units Angle subtended at the centre of the polygon by any two adjacent vertices is $\frac{2 \pi}{n}$ . By symmetry centre does not change. Let the beetles be $A, B, C \ldots N$ and each beetle is crawling towards the next beetle with a constant velocity $v$ units

relative velocity of $A$ w.r.t $B$

= $v - v \cos \frac{2 \pi}{n}$

= $v[1 - \cos \frac{2 \pi}{n}]$

= $2v \sin^{2} \frac{\pi}{n}$

Time required to meet at the centre is $t$ units = $\frac{\text{Relative distance}}{\text{Relative velocity}}$

But the $\text{Relative distance}$ will be = $d$

Hence, time required to meet at the centre $t = \frac{d}{2vsin^{2} \frac{\pi}{n}}$

In this problem, $n=4$ $\Rightarrow$ $t = \frac{d}{v}$ $\ldots \ldots 1$

Suppose the distance travelled in reaching the centre is $x^{\prime}$ units, $t= x^{\prime}v \Rightarrow x^{\prime} = d = 2$ (by equating the two equations of time.)

By symmetry, the beetles will always make a square and the center of those squares will always be the same. Therefore, the angle between their direction (toward another beetle) and the direction to the center is always $45°$ .

That means the radial speed (speed toward the center) is constant and is $v_r=0.01·\cos(45°)=\frac{0.01}{\sqrt{2}}$ m/s.

The distance from the starting point and the center is $d=\sqrt{2}\,$ m. Approaching it at $\frac{0.01}{\sqrt{2}}\,$ m/s, that gives a total time $t=\frac{d}{v_r}=200\,$ s.

Total distance crawled is $v·t=0.01·200=2\,$ m.

Let $s$ be the distance traveled by the beetle John ( $J$ ) and $r(s)$ its distance to the middle of the room $M$ , as a function of $s$ . John follows beetle Paul ( $P$ ). By symmetry, its distance to the $M$ is the same, $|MP| = |MJ| = r(s)$ . Furthermore, the angle $\angle JMP$ is always straight.

Now let John take an infinitesimal step $ds$ toward Paul. This step is along the line $JP$ , and thus makes a $45^\circ$ angle with $JM$ . We can easily establish that the corresponding change in $r$ is

$dr = -ds \cos 45^\circ = -\frac{ds}{\sqrt2}$

Upon integration, we find

$r(s) = \int dr = r_0 + \int_0^s -\frac{ds}{\sqrt2} = \sqrt2 \,\mathrm{m} - \frac{s}{\sqrt2}$

Therefore, John reaches the middle ( $r(s)=0$ ) for $s = \boxed{2\,\mathrm{m}}$

To get the complete picture, let me derive what the path looks like. Let $\phi$ be the angle that the line $MJ$ makes with the line $MJ$ as it was at $s=0$ . From the same geometrical picture we can establish the relation

$r\,d\phi = ds \sin 45^\circ = \frac{ds}{\sqrt2}$

from which we can deduce $d\phi = -\frac{dr}{r}$ , and from that $\phi(r) = \ln \frac{r_0}{r}$ , or $r(\phi) = r_0 e^{-\phi}$ (note that $\phi$ runs from zero to infinity). It's thus a logarithmic spiral.

By symmetry, we can conclude that the center of the quadrilateral at any point of time before the beetles converge will stay fixed, and the beetles will converge at the center of the quadrilateral. Let $O$ be the center of the quadrilateral. Initially, let $JR$ be the distance between John and Ringo (we can consider any two consecutive beetles). Let $M$ be the midpoint of $JR$ . Note that $JR$ subtends an angle $\frac{pi}{2}$ at $O$ . So $M$ subtends an angle $\frac{pi}{4}$ at $O$ . Now consider the radial component of the velocity of a beetle. Let $P$ be a point on $JR$ and $Q$ on $OR$ , such that points $P$ and $Q$ describe the trajectory of John. The radial component of the velocity will be then $V \sin \frac{pi}{4} = \frac{V}{\sqrt{2}}$ . From the reference frame of $O$ , the beetles will approach at a straight line with velocity $\frac{V}{\sqrt{2}}$ (tangential velocity omitted). The time needed to converge will then be $\frac{K}{V \sin \frac{\pi}{4}} = \frac{S}{V(1-\cos \frac{pi}{2})} = \frac{S}{T}$ (where $K$ is the radius of the polygon and $S= AB$ ). Then, the total distance covered by each beetle will simply be $VT= S= 2 \ meter$ . Note that the final answer is independent of the velocity of each beetle. I don't know why it was provided in the question.

Intuitively, it should be clear that we don't need the speed to solve this, only the fact that they are all travelling at the same speed.

BONUS: how many degrees of rotation about the centre does each beetle completes before they meet? As with the original problem (see @Josh Silverman 's solution), no calculation is required for the answer, only logical reasoning.

its simple! just use Pythagorean theorem. we know that the room is a SQUARE with the sides all being 2m long. we know we only need 2 beetles that are opposite and forget about the other 2. we split the room into a diagonal half to form a right angled triangle. Pythagorean theorem helps use to find the hypotenuse's length in meters. we need to know that both sides are 2m. then we square both sides. in this case 2 X 2 for one side and 2 X 2 for the other leaving both answers to be 4. we then add the two answers together and square root the sum leaving you with the final answer of 2.

TAKE RELATIVE VELOCITY OF ONE WITH RESPECT TO OTHER. THEN USE RELATIVE DISTANCE / RELATIVE SPEED TO CALCULATE TIME. THEN USE SPEED X TIME TO CALCULATE DISTANCE TRAVELED

At every moment, the four beetles will be in a square. The square will continually strink as bugs move inward. We look at one side of the square and we see something like this:

O>>>----------X

When beetle O meets beetle X, all beetles must've met in the middle by symmetry. That means O traveled along the length of one side of the square, traveling $2 m$ .

The path of the beetles can be found with polar coordinates, but that is unnecessary to obtain this distance.