Pursuit and Chase : Irodov problem 1.12

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 aa.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 vv.

Well for a regular nn sided polygon ,

T=av0(1cos2πn)\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 vrelv_{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 (n2)πn\frac{(n-2)\pi}{n} therefore the exterior angle will be 2πn\frac{2\pi}{n}.Resolving the velocity component along this direction,we get

vrel=v_{rel} = vvcosv - v cos θ\theta where θ\theta == 2πn\frac{2\pi}{n}

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

Note that for ,

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

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

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

n=12n = 12 we have t=t = 2a(23)v\frac{2a}{(2-\sqrt 3)v}

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

and so on.

Thanks everyone for reading!

Note by Ayon Ghosh
3 years, 9 months ago

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1 vote

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

Md Zuhair - 3 years, 5 months ago

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Great problem and extension!

Rick Zhou - 3 years, 5 months ago

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Thanks a lot !

Ayon Ghosh - 3 years, 5 months ago

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

Rick Zhou - 3 years, 5 months ago

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Great generalization!

Kushal Thaman - 3 years, 1 month ago

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Even though this is well known. XD

Kushal Thaman - 2 years, 8 months ago

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