Let $ ABC $ be a triangle and $ A_1,B_1,C_1 $ the projections of $ A,B,C $ on their opposite sides. Let $ A_2,A_3 $ be the projection of $ A_1 $ on $ AB, $ respectively on $ AC. B_2,B_3,C_2,C_3 $ are defined analogously. Moreover, $ A_4 $ is the intersection of $ B_2B_3 $ with $ C_2C_3; B_4, $ the intersection of $C_2C_3 $ with $ A_2A_3; C_4, $ the intersection of $ A_2A_3 $ with $ B_2B_3. $ Show that $ AA_4,BB_4 $ and $ CC_4 $ are concurrent.
Problem
Source: Stars of Mathematics 2007, Day 1, Problem 3
Tags: Pure geometry, geometry
Pluto1708
28.09.2018 12:43
Easy to note that $A_4B_4C_4 \sim ABC$.So there exists a center of homothety $A_4B_4C_4 \mapsto ABC $which is precisely $ AA_4 \cap BB_4 \cap CC_4 $ Done!
govind7701
28.09.2018 12:54
Let $X,Y$ be reflections of $A_1$ over $AB$ and $AC$. Note $AB$ is angle bisector of $\angle A_1C_1B_1$, $X$ lies on $B_1C_1$ and similarly $Y$ too. Thus, $A_2A_3, B_2B_3, C_2C_3$ are midlines of $\Delta A_1B_1C_1$ $A_4B_4C_4$ is the medial triangle of the orthic triangle. From here, line 141, we are done.
parmenides51
28.09.2018 12:55
posted 3 days ago here
pro_4_ever
28.09.2018 12:57
Pluto1708 wrote: $A_4B_4C_4 \sim ABC$.So there exists a homothety $A_4B_4C_4 \mapsto ABC $ Done! Your result is very nice. Please include a proof Looking forward to a reply...
Pluto1708
28.09.2018 15:01
pro_4_ever wrote: Pluto1708 wrote: $A_4B_4C_4 \sim ABC$.So there exists a homothety $A_4B_4C_4 \mapsto ABC $ Done! Your result is very nice. Please include a proof Looking forward to a reply... Since $A_2A_3 \parallel BC \implies B_4C_4 \parallel BC $ and similarly others so $A_4B_4C_4 \sim ABC $ Have I done something wrong ?
muditbhatt58
28.09.2018 15:45
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pro_4_ever
28.09.2018 19:00
Pluto1708 wrote: Since $A_2A_3 \parallel BC \implies B_4C_4 \parallel BC $ and similarly others so $A_4B_4C_4 \sim ABC $ Have I done something wrong ? No, but we mathlinkers would like to know about another interesting result quoted by you... Why is $A_2A_3$ parallel to $BC$? If this were true it may lead to some nice theorems like every triangle is isosceles and therefore equilateral. This further reduces the requirement of projective transformation in some cases. But I will be happy if I get the proof
muditbhatt58
28.09.2018 19:33
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Pluto1708
28.09.2018 19:58
pro_4_ever wrote: Pluto1708 wrote: Since $A_2A_3 \parallel BC \implies B_4C_4 \parallel BC $ and similarly others so $A_4B_4C_4 \sim ABC $ Have I done something wrong ? No, but we mathlinkers would like to know about another interesting result quoted by you... Why is $A_2A_3$ parallel to $BC$? If this were true it may lead to some nice theorems like every triangle is isosceles and therefore equilateral. This further reduces the requirement of projective transformation in some cases. But I will be happy if I get the proof Oops I am really sorry All because of bad drawing