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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?

$\textbf{Added later}:$

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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    $\begingroup$ I think obtuse angled triangle would be a counterexample $\endgroup$
    – vidyarthi
    Commented Aug 25, 2020 at 23:36
  • $\begingroup$ Have you tried $O_1=O_2$ be one of the triangle centers? Also, do you require $a$ between $B$ and $C$ or just on the line they determine? SInce the conditions on $a,b$ and $c$ are independent you should usually be able to perturb two perimeter points and get another pair. $\endgroup$
    – Aaron Meyerowitz
    Commented Aug 26, 2020 at 22:46
  • $\begingroup$ But if $O_1=O_2$, we have one triangle inside the original triangle (on the fixed edge say $AB$) and the perimeter is less than the original one. The point, say $a$, must be between $B$ and $C$. $\endgroup$
    – Shahrooz
    Commented Aug 27, 2020 at 14:49

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