# Hi Brilliantians!

Today I have developed a general formula for the classic Irodov problem-

3 particles are kept at the vertices of an equilateral triangle of side $$a$$.The velocity vector of the first particle continually points towards the second,the second towards the third and the third towards the first.After what time will they meet.Each moves with a constant speed $$v$$.

Well for a regular $$n$$ sided polygon ,

# $$\boxed{T = \frac{a}{v_{0}(1-cos\frac{2\pi}{n})}}$$

Working :

Well the usual stuff,figuring out relative velocities and then dividing the initial distance by the $$v_{rel}$$ to get the time.Becuase the trajectory forms logarithmic spirals and maintains its symmetricity all throughout the motion.

Note that each angle of the polygon is given by $$\frac{(n-2)\pi}{n}$$ therefore the exterior angle will be $$\frac{2\pi}{n}$$.Resolving the velocity component along this direction,we get

$$v_{rel} =$$ $$v - v cos$$ $$\theta$$ where $$\theta$$ $$=$$ $$\frac{2\pi}{n}$$

hence $$t =$$ $$\frac{a}{v_{rel}}$$ giving us our desired expression.

Note that for ,

$$n = 3$$ we have $$t = 2a/3v$$.

$$n = 4$$ we have $$t = a/v$$.

$$n = 6$$ we have $$t = 2a/v$$

$$n = 12$$ we have $$t =$$ $$\frac{2a}{(2-\sqrt 3)v}$$

$$n = infinity$$ we have $$t =$$ $$infinity$$ because now the path is a circle.

and so on.

Note by Ayon Ghosh
10 months, 3 weeks ago

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Great generalization!

- 3 months ago

Great problem and extension!

- 7 months, 2 weeks ago

Thanks a lot !

- 7 months, 2 weeks ago

Wow that was fast. Helped me on Aops Physics, was doing this problem :)

- 7 months, 2 weeks ago

This problem was already posted by Deepanshu Gupta in 2015 or 2014. :)

- 7 months ago