Will They Ever Meet?

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It is easy to see that they all will meet at $O$ , which is the center of circumcirle of $\Box ABCD$

See at the component of $A$ along AO, which is $10 \times \sin (45^\circ )$ (km\hr)

and even when the direction changes as per $B$ 's movement, the component along $AO$ towards $O$ is not going to change! and because all of them are at same speed, they will meet at $O$

The distance $AO$ is $\dfrac{100}{\sqrt{2}}$ (in meters), and the speed there is $\dfrac{10}{\sqrt{2}}$ km/hr.

Thus it will take $\dfrac{0.1}{10}$ = $0.01$ hours to meet, that is $\boxed{36}$ seconds.

In my solution, I will generalize the problem. Suppose the four runners are running on a Cartesian plane, where the runners at A,B,C,D start at the points $(-s,s), (-s,-s), (s,-s), and (s,s)$ respectively. Suppose also that their speed is $v$ units per second.

Given the symmetry of the problem, we can assume that the runners' positions always form a perfect square which is centred at the origin. That is, if Runner D is at position $(x,y)$ , then Runner A is at $(-y,x)$ , Runner B is at $(-x,-y)$ , and Runner C is at $(y,-x)$ . (I can prove this fact if needed; it's just that the proof is long and not very interesting).

So, let's define $t$ to be the time elapsed (in seconds). Define $(x,y)$ to be the position of Runner D at time $t$ . Then the position of A is $(-y,x)$ . Therefore, the slope of the line from A to D is $\frac{y-x}{y+x}$ . We assume that D runs in the direction of A, so therefore $\frac{dy}{dx} = \frac{y-x}{y+x}$ .

Now let's use the fact that $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ . We get that $(y+x) \frac{dy}{dt} = (y-x) \frac{dx}{dt}$ . Call this Equation $(1)$ .

Furthermore, since the runners' speed is $v$ , we get the following equation: $\left( \frac{dy}{dt} \right)^2 + \left( \frac{dx}{dt} \right)^2 = v^2$ Using $(1)$ , we get the following: $(y+x)^2 \left( \frac{dy}{dt} \right)^2 + (y+x)^2 \left( \frac{dx}{dt} \right)^2 = (y+x)^2 v^2$ $(y-x)^2 \left( \frac{dx}{dt} \right)^2 + (y+x)^2 \left( \frac{dx}{dt} \right)^2 = (y+x)^2 v^2$ $(2x^2 + 2y^2) \left( \frac{dx}{dt} \right)^2 = (y+x)^2 v^2$ $\frac{dx}{dt} =- \frac{(y+x) v}{\sqrt{2x^2 + 2y^2}} *** (2)$

In the last step above, I know it must be negative because D is getting closer to the origin, and therefore his x-coordinate must be decreasing (at first, anyway). From $(1)$ and $(2)$ , we see that $\frac{dy}{dt} = - \frac{(y-x) v}{\sqrt{2x^2 + 2y^2}} ***(3)$

Now, the distance from D to the origin is $\sqrt{x^2+y^2}$ . Let's show that this distance decreases at a constant rate: $\frac{d \left( \sqrt{x^2+y^2} \right)}{dt} = \frac{1}{2 \sqrt{x^2+y^2}} \left( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \right)$ $= \frac{1}{ \sqrt{x^2+y^2}} \left( -\frac{x(y+x)v}{\sqrt{2x^2 + 2y^2}} - \frac{y(y-x)v}{\sqrt{2x^2 + 2y^2}} \right)$ $= \frac{-v (x^2 + y^2)}{\sqrt 2 (x^2 + y^2)} = \frac{-v}{\sqrt{2}}$

Therefore, D approaches the origin at a constant speed of $\frac{-v}{\sqrt{2}}$ . He starts out at the point $(s,s)$ which is a distance of $s \sqrt 2$ away from the origin. Therefore, the amount of time it will take D to reach the origin is $\frac{s \sqrt{2}} {v / \sqrt{2}} = \frac{2s}{v}$ . By symmetry, the other runners will meet him there.

In the given problem, we have $s = 50$ (in metres, so that the square has a side length of 100) and $v = \frac{25}{9}$ (in metres per second). Therefore the time, in seconds, that it will take for the runners to meet each other is: $\frac{2*50}{25/9} = \boxed{36}$

A is moving towards B with a constant speed of 10 m/s. At each and every time B has a velocity perpendicular to A. Therefore distance between A and B decresses at a rate if 10km/hr. Hence answer is 36seconds.

You may want to take a look at @Michael Mendrin 's comment here .