Collinear points

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Source: Iranian National Olympiad (3rd Round) 2007

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#1 h Aug 27, 2007, 8:28 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ , $l$ and $P$ be arbitrary triangle, line and point. $A',B',C'$ are reflections of $A,B,C$ in point $P$ . $A''$ is a point on $B'C'$ such that $AA''\parallel l$ . $B'',C''$ are defined similarly. Prove that $A'',B'',C''$ are collinear.

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#2 h Aug 27, 2007, 9:42 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Omid Hatami wrote:

Let $ABC$ , $l$ and $P$ be arbitrary triangle, line and point. $A',B',C'$ are reflections of $A,B,C$ in point $P$ . $A''$ is a point on $B'C'$ such that $AA''\parallel l$ . $B'',C''$ are defined similarly. Prove that $A'',B'',C''$ are collinear.

http://www.cut-the-knot.org/Curriculum/Geometry/Zaslavsky.shtml (interchange triangles ABC and A'B'C').

darij

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#3 h Jul 2, 2009, 1:31 PM • 2 Y

Y by Adventure10, Mango247

My solution:

Let $A_3,B_3,C_3$ be points on $BC,CA,AB$ respectively,such that $A_2A_3 \| BC',B_2B_3 \| CA',C_2C_3 \| AB'$
Easy to see that $AB' = A'B = C_2C_3,BC' = B'C = A_2A_3,CA' = C'A = B_2B_3$ .Then apply Thales's theorem and Menelaus's theorem we have: $A_2,B_2,C_2$ are collinear if and only if $A_3,B_3,C_3$ are collinear.
Let $d_1,d_2,d_3$ be lines pass through $A,B,C$ and parallel to $A_2A_3,B_2B_3,C_2C_3$
,respectively.Then $d_1,d_2,d_3$ are concurrent at Q (the proof is simple,just take $Q' = d_1 \cap d_2$ then $QB = CA'$ and $QB \| CA'$ so $QBA'C$ is parallelogram and from it we have $CQ \| BA' \| C_2C_3$ .
$QAA_2A_3,QBB_2B_3,QCC_2C_3$ are parallelograms so $QA_3 \| QB_3 \| QC_3 \| l$ ,then $Q,A_3,B_3,C_3$ are collinear and that complete the proof

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#4 h Nov 7, 2022, 3:25 PM

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Trivial by moving point.Move the point $A''$ on $B'C'$ then define $B''$ and $C''$ such that $AA''//BB''//CC''$
$deg(A'')=deg(B'')=deg(C'')=1$ we want to show that $A''-B''-C''$ collinear so 4 case is enough.Take $A''$
$C''$ , $B''$ , $AA''//BB'$ and $\infty_{BC}$ . $\blacksquare$

This post has been edited 1 time. Last edited by sevket12, Jan 12, 2023, 4:23 PM

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