$L$ is a Line that intersect with the side $AB,BC,AC$ of triangle $ABC$ at $F,D,E$ The line perpendicular to $BC$ from $D$ intersect $AB,AC$ at $A_{1},A_{2}$ respectively Name $B_{1},B_{2},C_{1},C_{2}$ similarly Prove that the circumcenters of $AA_{1}A_{2},BB_{1}B_{2},CC_{1}C_{2}$ are collinear
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Tags: geometry, circumcircle
K.titu
10.01.2017 19:49
Any ideas? Please help me I'm getting confused with this problem
anhtaitran
11.01.2017 12:10
Very nice problem Here is my solution: Let (O1);(O2);(O3) are the center of (AA1A2);(BB1B2);(CC1C2). By a easy angle chasing we have:O1;O2;O3 line on the tagent of (O) from A;B;C. Let 3 tagent of (O) from A;B;C cut each other at X;Y;Z. Now,it is easy to see that:O3C2 is paralel to BX and O3C1 is paralel to AY.(1) We have:XO3/YO3=XO3/O3C*O3C/YO3 =BC2/C2C*C1C/C1A(because of (1).) =BF/FA(by Menelaus). It is the same we have:YO1/ZO1=CD/DB and ZO2/XO2=AE/EC. So XO3/YO3*YO1/ZO1*ZO2/XO2=BF/FA*AE/EC*CD/DB=1(By Menelaus ) So O1;O2;O3 are collinear.(2) P/S:Moreover we have:OA;OB;OC is tagent to (AA1A2);(BB1B2);(CC1C2) so O have the same power to (O1);(O2);(O3).(3) Then (O1);(O2);(O3) have the same radical axis.(because of (2);(3).)
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buratinogigle
11.01.2017 12:13
This is problem 13 in Sharygin contest 2016 correspondence round, see problem and solution here http://geometry.ru/olimp/2016/zaochsol-e.pdf
yayups
22.11.2019 10:12
This problem is an easy consequence of linearity. We first claim that $O_A$, the circumcenter of $AA_1A_2$ lies on the tangent to $(ABC)$ at $A$. This is because $AO_A$ must be the isogonal to the line through $A$ perpendicular to $A_1A_2$, or the line through $A$ parallel to $BC$. Now, we see that as $D$ varies on $BC$ linearly, the point $A_1$ varies linearly on $AB$. Thus, the perpendicular bisector of $AA_1$ varies linearly, so $O_A$ varies linearly on $QR$ (here $PQR$ is the triangle with intouch triangle $ABC$). When $D=B$, we see that $O_A=R$ as the perpendicular bisector of $AB$ intersected with $QR$ is $R$. Similarly, when $D=C$, we have $O_A=Q$. Thus, $QO_A/RO_A=CD/DB$, so we are done by Menalaus.