All PP' lines pass through a fixed point

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All PP' lines pass through a fixed point

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Source: Iran TST 2011 - Day 4 - Problem 2

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843 posts

shoki

#1 h May 14, 2011, 3:51 PM • 5 Y

Y by narutomath96, mathmaths, Adventure10, Mango247, and 1 other user

Let $ABC$ be a triangle and $A',B',C'$ be the midpoints of $BC,CA,AB$ respectively. Let $P$ and $P'$ be points in plane such that $PA=P'A',PB=P'B',PC=P'C'$ . Prove that all $PP'$ pass through a fixed point.

This post has been edited 1 time. Last edited by Amir Hossein, May 14, 2011, 5:28 PM

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147 posts

sumita

#2 h May 14, 2011, 8:29 PM • 2 Y

Y by Adventure10, Mango247

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.

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418 posts

skytin

#3 h May 15, 2011, 8:42 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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

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527 posts

#4 h Sep 15, 2011, 6:50 PM • 2 Y

Y by Adventure10, Mango247

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.

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571 posts

yunxiu

#5 h Sep 29, 2011, 2:25 PM • 2 Y

Y by Adventure10, Mango247

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 $ABC$ .

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36 posts

#6 h Oct 22, 2011, 1:26 PM • 1 Y

Y by Adventure10

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！

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P-H-David-Clarence

92 posts

P-H-David-Clarence

#7 h Jul 28, 2016, 12:32 PM • 1 Y

Y by Adventure10

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

This post has been edited 1 time. Last edited by P-H-David-Clarence, Jul 28, 2016, 12:33 PM

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1740 posts

#8 h Feb 20, 2021, 1:11 AM • 4 Y

Y by mijail, guptaamitu1, khina, ihatemath123

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] 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]$

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.

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155 posts

Ritwin

#9 h Jan 29, 2025, 2:57 AM

Y by

We will solve a more general version of the problem:

Problem (generalization): Let $ABC$ be a fixed triangle with vertices on the unit circle. Fix a positive real number $k$ , and let $A' = k(B+C)$ , $B' = k(C+A)$ , and $C' = k(A+B)$ , viewing points as vectors.
As points $P$ and $P'$ vary such that $PA = P'A'$ , $PB = P'B'$ , and $PC = P'C'$ , line $PP'$ passes through a fixed point on the Euler line of $ABC$ .

Proof. Let $R$ be the point such that $P' = k(A+B+C) - kR$ . Then we have $PA = |{P'-A'}| = |{k(A+B+C)-kR-k(B+C)}| = k |{A-R}| = k RA.$ Similar equations for $B$ and $C$ give $\frac{PA}{RA} = \frac{PB}{RB} = \frac{PC}{RC} = k.$ Therefore, $A$ , $B$ , and $C$ lie on the same Apollonian circle with respect to $P$ and $R$ .

In particular, this means $R$ is the inverse of $P$ around $(ABC)$ . Turning to complex numbers, this means $r = 1/\overline{p}$ . Now $k = \frac{PA}{RA} = \left\lvert{\frac{p-a}{1/\overline{p}-a}}\right\rvert = |\overline{p}| \cdot \left\lvert{\frac{p-a}{\overline{a}-\overline{p}}}\right\rvert = |p|,$ so in fact $r = p/k^2$ . Substituting back, we find $p' = k(a+b+c) - p/k$ , so line $PP'$ passes through $\frac{kp'+p}{k+1} = \frac{k^2}{k+1} (a+b+c).$ This point is fixed, and does indeed lie on the Euler line of $ABC$ . $\square$

The special case $k = \tfrac12$ is the original problem. The fixed point is then the midpoint of the circumcenter $O$ and the centroid $G$ . $\blacksquare$

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