BdMO National Higher Secondary 2019/7
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Given three cocentric circles $\omega_1$,$\omega_2$,$\omega_3$ with radius $r_1,r_2,r_3$ such that $r_1+r_3\geq {2r_2}$.Constrat a line that intersects $\omega_1$,$\omega_2$,$\omega_3$ at $A,B,C$ respectively such that $AB=BC$.
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Re: BdMO National Higher Secondary 2019/7
Diagram
This diagram is for \[r_1+r_3=2r_2\]- math_hunter
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Re: BdMO National Higher Secondary 2019/7
Have you got the solution of this problem???
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Re: BdMO National Higher Secondary 2019/7
May be not for all cases $AB=BC$ can't be drawn even if $r_1+r_3\geq 2r_2$. I think $2r_{2}^{2} \geq r_{1}^{2}+r_{3}^{2}$ also must hold
- Anindya Biswas
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Re: BdMO National Higher Secondary 2019/7
This problem is just asking for a triangle whose $2$ sides and median on the third side is given.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Anindya Biswas
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Re: BdMO National Higher Secondary 2019/7
Draw a parallelogram with side lengths $r_1$ and $r_3$ and with a diagonal of length $2r_2$. The other diagonal of this parallelogram is the required segment.samiul_samin wrote: ↑Mon Mar 04, 2019 9:40 amGiven three cocentric circles $\omega_1$,$\omega_2$,$\omega_3$ with radius $r_1,r_2,r_3$ such that $r_1+r_3\geq {2r_2}$.Constrat a line that intersects $\omega_1$,$\omega_2$,$\omega_3$ at $A,B,C$ respectively such that $AB=BC$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
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Re: BdMO National Higher Secondary 2019/7
Can you show the proof?
- Anindya Biswas
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Re: BdMO National Higher Secondary 2019/7
Diagonals of parallelogram bisects each other.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann