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The simplified version of the problem introduced...
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Shahrooz
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Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

$\textbf{Added later}:$

Is it true that each triangle has perimeter points just for it's one side? It means that each ofIf we relax the three conditions true independently and $O_1$problem and $O_2$ can be determined just for each side of triangle.the below condition be true, does this problem have a solution?

$$p(BO_1a)+p(aO_2C)=p(ABC).$$

Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

$\textbf{Added later}:$

Is it true that each triangle has perimeter points just for it's one side? It means that each of the three conditions true independently and $O_1$ and $O_2$ can be determined just for each side of triangle.

Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

$\textbf{Added later}:$

If we relax the problem and just the below condition be true, does this problem have a solution?

$$p(BO_1a)+p(aO_2C)=p(ABC).$$

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YCor
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Simplification
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Shahrooz
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Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

$\textbf{Added later}:$

Is it true that each triangle has perimeter points just for it's one side? It means that each of the three conditions true independently and $O_1$ and $O_2$ can be determined just for each side of triangle.

Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$

My question is:

Is it true that each triangle has perimeter points?

$\textbf{Added later}:$

Is it true that each triangle has perimeter points just for it's one side? It means that each of the three conditions true independently and $O_1$ and $O_2$ can be determined just for each side of triangle.

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Shahrooz
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Shahrooz
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