Put $G$ the centroid of triangle as the origin. Now we have the following: $\vec A'=-\frac{1}{2}\vec A$ and the same for $B$ and $C$. from the problem we have: $(\vec A-\vec P)^2=(\vec A'-\vec P')^2$ $(\vec B-\vec P)^2=(\vec B'-\vec P')^2$ $(\vec C-\vec P)^2=(\vec C'-\vec P')^2$ Which yields: $|P|^2-|P'|^2 +\frac{3}{4}|A|^2 =(2\vec P-\vec P').\vec A$ $|P|^2-|P'|^2 +\frac{3}{4}|B|^2 =(2\vec P-\vec P').\vec B$ $|P|^2-|P'|^2 +\frac{3}{4}|C|^2 =(2\vec P-\vec P').\vec C$ Now subtrackting 2 and 3 from 1 yields: $\frac{3}{4}(|A|^2-|B|^2)=(2\vec P-\vec P').\vec BA$ $\frac{3}{4}(|A|^2-|C|^2)=(2\vec P-\vec P').\vec CA$ Since $ 2\vec P-\vec P'$ is the only unknown parameter and $\vec BA $ and $\vec CA $ are independent vectors therefore $2\vec P-\vec P'$ is constant which implies that $PP'$ pass through a fixed point. The same is true for tetrahedron and so on.

Let make Homotety of point P wrt Centroid of triangle ABC and get point P_1 Easy to see that A'P_1/A'P' = A'P_1/A'P' = B'P_1/B'P' = C'P_1/C'P' = 1/2 , so (A'B'C') is Apollonian circle of segment P_1P' , let line P-1P' intersect (A'B'C') at points X and Y , were Y is between points P_1 and P' , ease to see that P_1Y/YP' = XP_1/XP' = 1/2 and XY is diameter of (A'B'C') Let N is center of (A'B'C') , easy to see that NP_1 is constant and NP' too So all ponts P_1 and P' are placed on two circles (named they as a and b) with same center N Homotety of circl a wrt Centroid of triangle ABC and get circle a' , easy to see that P is on a' tangent to a' from P is parallel to tangent to a from P_1 and parallel to tangent to b from P' , so lines PP' , goes thru inner Homotety center of b and a' . done

Dear MLs Very simple physics places this fixed point onto centroid of triangle that has as its sides the mediatrices of segments AA', BB' and CC'. On the other hand, this fixed point also has to be on Euler line of ABC. And it is, in a compromising position. M.T.

skytin wrote: so lines PP' , goes thru inner Homotety center of b and a' . done The centre is the middle point of $OG$, where $O$ is the center of $(ABC)$, $G$ is the Centroid of triangle $ABC$.

skytin wrote: Let make Homotety of point P wrt Centroid of triangle ABC and get point P_1 Easy to see that A'P_1/A'P' = A'P_1/A'P' = B'P_1/B'P' = C'P_1/C'P' = 1/2 , so (A'B'C') is Apollonian circle of segment P_1P' , let line P-1P' intersect (A'B'C') at points X and Y , were Y is between points P_1 and P' , ease to see that P_1Y/YP' = XP_1/XP' = 1/2 and XY is diameter of (A'B'C') Let N is center of (A'B'C') , easy to see that NP_1 is constant and NP' too So all ponts P_1 and P' are placed on two circles (named they as a and b) with same center N Homotety of circl a wrt Centroid of triangle ABC and get circle a' , easy to see that P is on a' tangent to a' from P is parallel to tangent to a from P_1 and parallel to tangent to b from P' , so lines PP' , goes thru inner Homotety center of b and a' . done Very nice！Thank you！

PP'past through the centre of the nine point circle V,and P'V=2PV.It can be solved by vector analyze.

Very nice. Solved with nukelauncher. Let \(O\) be the circumcenter and \(G\) the centroid of \(\triangle ABC\); I claim the fixed point is the midpoint \(M\) of \(\overline{OG}\). [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=lightblue; pen tri=purple+pink; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,G,M,Ap,Bp,Cp,P,Ps,Pp; O=origin; A=dir(110); B=dir(260); C=dir(310); G=(A+B+C)/3; M=G/2; Ap=(B+C)/2; Bp=(C+A)/2; Cp=(A+B)/2; P=0.5*dir(180); Ps=4P; Pp=(3G-Ps)/2; draw(A--Ps,sec+dashed); draw(B--Ps,sec+dashed); draw(C--Ps,sec+dashed); draw(A--P,pri+dashed); draw(B--P,pri+dashed); draw(C--P,pri+dashed); filldraw(unitcircle,fil,pri); filldraw(A--B--C--cycle,fil,pri); draw(O--G,tri); draw(P--Pp,tri); draw(Ps--Pp,gray); draw(Ps--(-0.65Ps),linewidth(0.8)); dot("\(O\)",O,N); dot("\(A\)",A,A); dot("\(B\)",B,B); dot("\(C\)",C,C); dot("\(G\)",G,S); dot("\(M\)",M,dir(35)); dot("\(P\)",P,NW); dot("\(P^*\)",Ps,W); dot("\(P'\)",Pp,SE); [/asy][/asy] Let \(P^*\) be the point such that \(\triangle ABC\cup P^*\sim\triangle A'B'C'\cup P'\); then \(G\in\overline{P^*P'}\) and \(P^*G/GP'=2\) by homothety. We are given that \(P^*A=2P'A'=2PA\), so symmetrically, \[\frac{P^*A}{PA}=\frac{P^*B}{PB}=\frac{P^*C}{PC}=2.\]Then the circumcircle of \(\triangle ABC\) is an Apollonius circle of \(\overline{P^*P}\). If the circumcircle of \(\triangle ABC\) intersects \(\overline{P^*P}\) at \(X\) and \(Y\) (with \(X\) on segment \(P^*P\)), then \(PX=PP^*/3\) and \(PY=-PP^*\), so \(PO=-PP^*/3\), i.e.\ \(P^*P/PO=3\). Finally, observe that \[\frac{OP}{PP^*}\cdot\frac{P^*P'}{P'G}\cdot\frac{GM}{MO}=\frac13\cdot\frac{-3}1\cdot\frac11=-1,\]so by Menelaus theorem on \(\triangle OP^*G\) with traversal \(\overline{P'MP}\), point \(M\) lies on \(\overline{P'P}\), as claimed.