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 ,

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.

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## Comments

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

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Great problem and extension!

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Thanks a lot !

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

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Great generalization!

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Even though this is well known. XD

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