I think that this problem is also true in its more general configuration of similar rectangles, instead of squares. I will search my notes and if I am not mistaken, I will post here next time the synthetic ( but difficult ) solution I have in mind. Kostas vittas.

We can solve this problem, in its more general configuration of similar rectangles instead of squares, as a direct application of the below two Lemmas: LEMMA 1. – A triangle $ \bigtriangleup ABC$ is given and le $ BCAbAc,$ $ ACBaBc,$ $ ABCaCb$ be, three similar rectangles erected outwardly to it, on its sidesegments $ BC,$ $ AC,$ $ AB$ respectively. Through $ Ba,$ $ Bc,$ we draw two lines perpendicular to $ AbBa,$ $ BcCb$ respectively and we denote as $ Q,$ their intersection point. Prove that the line segment $ BQ,$ passes through the point $ P,$ as the center of the rectangle $ BcBaB'aB'c,$ congruent to $ ACBaBc.$ PROOF OF THE LEMMA 1. Through vertices $ A,$ $ C$ of $ \bigtriangleup ABC,$ we draw two lines perpendicular to $ BcCb,$ $ AbBa,$ respectively. It is easy to show that these lines are coincided with the medians of $ \bigtriangleup ABC,$ through $ A,$ $ C,$ respectively and so, intersect each other at the centroid $ G$ of $ \bigtriangleup ABC.$ $ ($ We denote as $ A',$ $ B',$ $ C',$ the orthogonal projections of $ A,$ $ B,$ $ C$ respectively, on the line segment $ BcCb$ and from cyclic quadrilaterals $ ABCbB',$ $ ACBcC',$ we have that $ \angle AB'C' =\angle ABCb =\angle ACBc =\angle AC'B'$ and so, $ AB' = AC'.$ Hence, the line segment $ AA',$ as the midparallel of the trapezium $ BB'CC',$ is coincided with the median of $ \bigtriangleup ABC,$ through vertex $ A$ $ ).$ We denote as $ M,$ the midpoint of side segment $ AC$ and as $ N,$ the center of the rectangle $ ACBaBc.$ It is easy to show that the points $ M,$ $ N,$ $ P,$ are collinear. Because of $ GA\parallel QBc,$ $ GC\parallel QBa$ and $ AC\parallel = BcBa,$ we conclude that the triangles $ \bigtriangleup GAC,$ $ \bigtriangleup QBcBa$ are congruent and so, we have that $ GQ\parallel = ABc$ $ ,(1)$ It is easy to show that $ MP =\frac{3(ABc)}{2}$ $ ,(2)$ From $ (1),$ $ (2)$ $ \Longrightarrow$ $ MP =\frac{3(GQ)}{2}$ $ ,(3)$ From $ (3),$ because of the collinearity of $ B,$ $ G,$ $ M$ and $ MP\parallel GQ,$ we conclude that the points $ B,$ $ Q,$ $ P,$ are collinear $ ($ because of $ BM =\frac{3(BG)}{2}$ $ )$ and the proof of the Lemma 1, is completed. LEMMA 2. – In the configuration of the Lemma 1, we denote as $ R,$ the intersection point of the line segments $ AcBc,$ $ BaCa.$ Prove that the line segment $ BQ,$ passes through the point $ R.$ $ \bullet$ About the Lemma 2, we can say that it is also true, in its even more general configuration of arbitrary rectangles, erected on the side segments of $ \bigtriangleup ABC,$ instead of similar ones or squares. We remind that it has already been posted as a separate problem, in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=108818, which unfortunately has not been discussed yet. By the way of this Omid Hatami’s problem, I think it is a good time to post there, at least, the proof by a friend of mine Nikos Kyriazis, the one of two synthetic solutions I have in mind. $ \bullet$ Because now, of the collinearities of $ B,$ $ Q,$ $ P$ and $ B,$ $ R,$ $ Q,$ we conclude that the line segments $ BP,$ $ CaBa,$ $ AcBc$ are concurrent at one point and the proof of the proposed problem may be completed. Kostas Vittas. PS. Although it is a difficult way to solve this problem, however, it is based on two Lemmas, useful ( I think ) to all of us in the future. I am looking forward in another sorter and simpler proof and I am interesting if has already been arisen such that a solution, during the contest. Also I am interesting, if the official solution of this problem is available.

Attachments:
t=164548(a).pdf (11kb)
t=164548.pdf (7kb)

Is very easy if we"ll use the complex numbers !

Virgil Nicula wrote: Is very easy if we"ll use the complex numbers ! Although I am not happy because of the "complex numbers" is so much to me, however, I like very much to try for the solutions of the problems, based on classical geometry tools comprehensible to the mid mind and specially, when ( I think ) the only known solutions about some problems, are based on extensive ( to me ) mathematical knowledge. Best regards, Kostas vittas.

Vittasko, the your extension and the its synthetical proof are very nicely and interestingly. You are right ! But usually, the first attempt is with the strong instruments and afterwards find a fine jewel - a synthetical proof. Thank you for this present ! And I like the "butterfly" (synth.) and I don't wish to kill it with the "atomic bomb" (strong instr.) !

I know a very short and nice solution by Ceva's theorem

khashi70 wrote: I know a very short and nice solution by Ceva's theorem I am eager for such a solution ( when I find a complicated proof of a problem, I am almost sure that there is another clever one, by a simpler way, which someone else younger man, has already found or will find later ). Dear Khashi70, would you like to share with us your solution? Thank you, Kostas vittas.

Dear all my friends. I would like to post here the solution I have in mind, of the problem "An interesting collinearity", http://www.mathlinks.ro/Forum/viewtopic.php?t=108818 which It is used as Lemma 2, for the proof of the extension of the problem presented by Omid Hatami. $ ($ I thought that I had find, one year ago, in Paul Viu’s web page, the problem with the erected squares, but now I can’t find it. Probably I am mistaken, but If someone has a reference about it, somewhere else, kindly please inform me $ ).$ PROBLEM (An interesting collinearity). Three rectangles $ ABDE,$ $ ACFH,$ $ BCKL,$ are erected on the side segments $ AB,$ $ AC,$ $ BC$ respectively, of a given triangle $ \bigtriangleup ABC,$ outwardly to it. If $ M$ is the intersection point of the line segments $ EK,$ $ HL$ and $ N,$ the one of the lines through $ K,$ $ L$ and perpendicular to $ KF,$ $ LD$ respectively, prove that the points $ A,$ $ M,$ $ N,$ are collinear. PROOF. 1). – At first, we can say that the problem is true, in the particular case when the line segment $ EH,$ is parallel to $ BC.$ 1a). - $ ($ see the figure t=164548(b) $ ),$ We will prove that the point $ N\equiv EK\cap HL,$ lies on the altitude of $ \bigtriangleup ABC,$ through vertex $ A.$ We denote the points $ B'\equiv AH\cap BL$ and $ C'\equiv AE\cap CK.$ It is easy to show that $ BB' = CC'$ $ ($ From the cyclic quadrilaterals $ ABCC',$ $ ACBB',$ we have that the points $ B',$ $ C',$ are the antidiametric (=diametric opposite) points of $ B,$ $ C$ respectively, with respect to the circumcircle $ (O),$ of $ \bigtriangleup ABC$ $ ).$ From this fact we have $ B'C'\parallel = BC\parallel = KL.$ We see now, that the triangles $ \bigtriangleup AB'C',$ $ NLK,$ are perspective because of the collinearity of the points $ E\equiv AC'\cap NK$ and $ H\equiv AB'\cap NL$ and $ S\equiv B'C'\cap LK,$ where $ S,$ is the point at infinity. $ ($ we say that two parallel lines, are intersected at infinity $ ).$ So, by Desarques’s theorem, we conclude that the line segments $ AN,$ $ B'L,$ $ C'K,$ are concurrent at one point and then we have $ AN\parallel B'L\parallel C'K.$ Hence, $ AN\perp BC$ and the proof of part (1a), is completed. Kostas vittas.

Attachments:
t=164548(b).pdf (8kb)

1b). - $ ($ see the figure t=164548(c) $ ),$ We will prove now, that the point $ N',$ as the intersection point of the lines through $ K,$ $ L$ and perpendicular to $ KF,$ $ LD$ respectively, lie also on the altitude of $ \bigtriangleup ABC,$ through vertex $ A.$ We denote the point $ S\equiv DE\cap FH$ and we will prove that $ EH\parallel DF.$ From $ SE\parallel AB$ and $ SH\parallel AC$ and $ EH\parallel BC,$ we have that the triangles $ \bigtriangleup SEH,$ $ \bigtriangleup ABC,$ are similar and then $ \frac{SE}{AB}=\frac{SH}{AC}$ $ ,(1)$ But $ AB = ED$ $ ,(2)$ and $ AC = HF$ $ ,(3)$ From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{SE}{ED}=\frac{SH}{HF}$ $ \Longrightarrow$ $ EH\parallel DF.$ That is, in the particular case of $ EH\parallel BC,$ we have also $ DF\parallel BC.$ $ \bullet$ We denote as $ Q,$ the orthogonal projection of $ N',$ on $ KL$ and we will prove that the line segment $ NQ,$ passes through vertex $ A$ of $ \bigtriangleup ABC.$ Through $ B,$ $ C,$ we draw two lines and parallel to $ LD,$ $ KF$ respectively and we denote as $ D',$ $ F',$ their intersection points, from the line segment $ DF.$ Also, through $ B,$ $ C,$ we draw two lines and perpendicular to $ BD',$ $ CF'$ respectively and we denote as $ N'',$ their intersection point. It is easy to show that $ N'N''\perp BC$ $ ($ From $ N''B\parallel N'L$ and $ N''C\parallel N'K$ and $ BC\parallel = LK,$ we have the congruence of the triangles $ \bigtriangleup N''BC,$ $ \bigtriangleup N'LK$ and so, from the parallelograms $ BN''N'L$ and $ CN''N'K,$ we conclude tha $ N'N''\equiv QN'N''\perp BC$ $ ).$ $ \bullet$ In order to prove that the line segment $ N'N''$ passes through $ A,$ it is enough to prove that $ A'\equiv A'',$ where $ A'\equiv AB\cap N'N''$ and $ A''\equiv AC\cap N'N''.$ Let $ B',$ $ C'$ be, the intersection points of the line segment $ DF,$ from the line segments $ BL,$ $ CK,$ respectively. From $ D'B\parallel DL$ $ \Longrightarrow$ $ \frac{B'B}{BL}=\frac{B'D'}{D'D}$ $ ,(4)$ From $ F'C\parallel FK$ $ \Longrightarrow$ $ \frac{C'C}{CK}=\frac{C'F'}{F'F}$ $ ,(5)$ From $ B'C'\parallel = BC\parallel = LK$ $ \Longrightarrow$ $ \frac{B'B}{BL}=\frac{C'C}{CK}$ $ ,(6)$ From $ (4),$ $ (5),$ $ (6)$ $ \Longrightarrow$ $ \frac{B'D'}{D'D}=\frac{C'F'}{F'F}$ $ ,(7)$ From the similar right triangles $ \bigtriangleup B'BD,$ $ \bigtriangleup PBA'$ $ ($ $ A'\equiv AB\cap N'N''$ $ )$ $ \Longrightarrow$ $ \frac{B'D'}{D'D}=\frac{PN''}{N''A'}$ $ ,(8)$ $ ($ because of the perpendicularities $ AB\perp BD$ and $ N''B\perp BD'$ and $ PB\perp BB',$ where $ P\equiv BC\cap N'N'',$ we have that $ \angle PBN'' =\angle B'BD'$ and $ \angle N''BA' =\angle D'BD$ $ ).$ By the same way, from the similar right triangles $ \bigtriangleup C'CF,$ $ \bigtriangleup PCA''$ $ ($ $ A''\equiv AC\cap N'N''$ $ )$ $ \Longrightarrow$ $ \frac{C'F'}{F'F}=\frac{PN''}{N''A''}$ $ ,(9)$ From $ (7),$ $ (8),$ $ (9)$ $ \Longrightarrow$ $ \frac{PN''}{N''A'}=\frac{PN''}{N''A''}$ $ \Longrightarrow$ $ A''\equiv A'\equiv A$ Hence, the point $ N,$ lies on the altitude of $ \bigtriangleup ABC,$ through vertex $ A$ and the proof of the part (1b), is completed. Kostas Vittas.

Attachments:
t=164548(c).pdf (8kb)

2). We return now, in to our problem in its general configuration $ ($ see the figure t=164548(d) $ ).$ Through the point $ H,$ we draw a line parallel to $ BC,$ which intersects the line segment $ AE,$ at one point so be it $ E'.$ The line through $ E',$ and parallel to $ AB,$ intersects the line segment $ BD,$ at one point so be it $ D'.$ We denote as $ N,$ the intersection point of $ HL,$ $ E'K$ and as $ N',$ the one of the lines through $ K,$ $ L$ and perpendicular to $ KF,$ $ LD',$ respectively. It has already been proved that the points $ N,$ $ N',$ lie on the altitude of $ \bigtriangleup ABC,$ through vertex $ A.$ We denote as $ M',$ the intersection point of the line segment $ AM,$ $ ($ $ M\equiv EK\cap HL$ $ ),$ from the line through $ L$ and perpendicular to $ LD.$ It is enough to prove that the point $ M',$ lies on the line segment $ KN'\perp KF.$ $ \bullet$ Through $ L,$ we draw the line $ AX,$ perpendicular to $ AL$ and it is easy to see that the homologous rays of the pencils $ L.AN'M'K$ and $ L.XD'DB,$ form equal angles, because of their perpendicularities. $ ($ $ AL\perp LX$ and $ N'L\perp LD'$ and $ M'L\perp LD$ and $ KL\perp LB$ $ ).$ Then, we conclude that these pencils have equal cross ratios, that is we have that $ (L.AN'M'K) = (L.XD'DB)$ $ ,(10)$ The line segment $ BD,$ intersects the pencil $ L.XD'DB$ and so, we have that $ (L.XD'DB) = (P,D',D,B)$ $ ,(11)$ where $ P\equiv BD\cap LX.$ Let $ S$ be, the intersection point of the line segments $ AK,$ $ HL.$ As the pencil $ A.LNMS,$ is intersected from the line segment $ HL,$ we have $ (A.LNMS) = (L,N,M,S) = (K.LNMS)$ $ ,(12)$ But the line segment $ AE,$ intersects the pencil $ K.LNMS$ and so, we have that $ (K.LNMS) = (Q,E',E,A)$ $ ,(13)$ where $ Q\equiv AE\cap KL.$ $ \bullet$ It is easy to show that the points $ P,$ $ Q,$ lie on the circumcircle $ (W),$ of the triangle $ \bigtriangleup ABL,$ because of the cyclic quadrilaterals $ ABLP$ and $ ABLQ.$ So, we conclude that the points $ P,$ $ Q,$ are the antidiametric (=diametric opposite) points of $ A,$ $ B$ respectively and then the quadrilateral $ ABPQ,$ is a rectangle. After this result, it is easy to show that $ PQ\parallel = BA\parallel = DE\parallel = D'E'$ and hence, $ (P,D',D,B) = (Q,E',E,A)$ $ ,(14)$ From $ (10),$ $ (11),$ $ (12),$ $ (13),$ $ (14)$ $ \Longrightarrow$ $ (L.AN'M'K) = (A.LNMS)$ $ ,(15)$ From $ (15)$ and because of the pencils $ L.AN'M'K,$ $ A.LNMS,$ have the line segment $ AL,$ as their common ray, we conclude that the points $ N'\equiv LN'\cap AN$ and $ M'\equiv LM'\cap AM$ and $ K\equiv LK\cap AS,$ are collinear. In other words we conclude that the points $ A,$ $ M,$ $ N,$ are collinear and the proof is completed. $ \bullet$ This proof is dedicated to Nikos Dergiades. Kostas vittas.

Attachments:
T=164548(d).pdf (12kb)

Thank you dear Kostas, for your very nice and hard solution for Lemma 2 . It's really interesting and new to me. And dear Khashi70, would you like to post your solution here?

It's my solution , I prove this problem by Trigonometric form of Ceva's theorem : but first notice that : $ \triangle AB_{a}C_{a}$ is similiar to $ \triangle B_{c}AB$ and if $ \angle C_{a}B_{a}A = x$ then $ \angle BB_{a}A = x$ and similiary if $ \angle BB_{a}C = y$ then $ \angle CB_{c}A_{c}= y$ . we have $ \frac{\sin{PBB_{c}}}{\sin{PB B_{a}}}=\frac{\sin{BB_{c}P}}{\sin{BB_{a}P}}$ and we have $ \angle BB_{c}P = 135-x$ , $ \angle BB_{a}P = 135-y$ thus $ \frac{\sin{PBB_{c}}}{\sin{PB B_{a}}}=\frac{\sin{(135-x)}}{\sin{(135-y)}}$. $ \angle BB_{c}A_{c}= 45-x-y$ , $ \angle A_{c}B_{c}B_{a}= 45+y$ , $ \angle C_{a}B_{a}B_{c}= 45+x$ and $ \angle BB_{a}C_{a}= 45-x-y$ now u can check the Trigonometric form of Ceva's theorem

Omid Hatami wrote: Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent. Solution. Denote $ M=BB_c\cap B_aC_a$, $ N=BB_a\cap B_cA_c$ and $ S=C_aB_a\cap A_cB_c$. Since two triangles $ ABB_c$ and $ AC_aB_a$ are similar, we conclude that $ \angle AB_cB=\angle AB_aC_a$, which implies that four points $ A$, $ B_c$, $ B_a$ and $ M$ are cyclic. Similarly, $ N$ also lies on this circle, let it is $ \Gamma$. Let $ K=MN\cap B_cB_a$. We have $ PB_c$, $ PB_a$ are two tangents of $ \Gamma$, which follows that $ B_cB_a$ is the polar of $ P$ relative to the circle $ \Gamma$. On the other hand, we have $ K$ lies on $ B_cB_a$, it equivalent to $ P$ lies on the polar of $ K$ relative to $ \Gamma$ $ (1)$. Let's consider the cyclic quadrilateral $ MNB_aB_c$, we have $ K$ and $ B$ are the intersections of its opposite sides, and $ S$ is the intersection of its diagonals. Therefore the polar of $ K$ relative to $ \Gamma$ is the line $ BS$ $ (2)$. From $ (1)$ and $ (2)$ we conclude that $ B$, $ S$ and $ P$ are collinear, it equivalent to $ BP$, $ C_aB_a$ and $ A_cB_c$ are concurrent, as desired.

April wrote: Omid Hatami wrote: Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent. Solution. Denote $ M = BB_c\cap B_aC_a$, $ N = BB_a\cap B_cA_c$ and $ S = C_aB_a\cap A_cB_c$. Since two triangles $ ABB_c$ and $ AC_aB_a$ are similar, we conclude that $ \angle AB_cB = \angle AB_aC_a$, which implies that four points $ A$, $ B_c$, $ B_a$ and $ M$ are cyclic. Similarly, $ N$ also lies on this circle, let it is $ \Gamma$. Let $ K = MN\cap B_cB_a$. We have $ PB_c$, $ PB_a$ are two tangents of $ \Gamma$, which follows that $ B_cB_a$ is the polar of $ P$ relative to the circle $ \Gamma$. On the other hand, we have $ K$ lies on $ B_cB_a$, it equivalent to $ P$ lies on the polar of $ K$ relative to $ \Gamma$ $ (1)$. Let's consider the cyclic quadrilateral $ MNB_aB_c$, we have $ K$ and $ B$ are the intersections of its opposite sides, and $ S$ is the intersection of its diagonals. Therefore the polar of $ K$ relative to $ \Gamma$ is the line $ BS$ $ (2)$. From $ (1)$ and $ (2)$ we conclude that $ B$, $ S$ and $ P$ are collinear, it equivalent to $ BP$, $ C_aB_a$ and $ A_cB_c$ are concurrent, as desired. Let us to present a schema, for this beautiful solution. Best regards, Kostas Vittas.

Attachments:
t=164548(e).pdf (9kb)

Dear Mathlinkers, an article concerning the Vecten's figure and its developpement can be seeing on my site http://perso.orange.fr/jl.ayme , la figure de Vecten vol. 5 p. 114 Sincerely Jean-Louis

khashi70 do a good solution

An generalization can be found in https://artofproblemsolving.com/community/c6h1505582p8907331

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