At heights $AA_0, BB_0, CC_0$ of an acute-angled non-equilateral triangle $ABC$, points $A_1, B_1, C_1$ were marked, respectively, so that $AA_1 = BB_1 = CC_1 = R$, where $R$ is the radius of the circumscribed circle of triangle $ABC$. Prove that the center of the circumscribed circle of the triangle $A_1B_1C_1$ coincides with the center of the inscribed circle of triangle $ABC$. E. Bakaev
Problem
Source: Tournament of Towns 2020 oral p2 (15 March 2020)
Tags: circumcircle, geometry, incenter, circumradius, altitudes
ak_47
18.05.2020 16:44
Let $I$ be the incenter of $\triangle ABC$ $AI=4R sin \frac{B}{2} sin \frac{C}{2}, BI=4R sin \frac{A}{2} sin \frac{C}{2}$ Hence, $\angle A_0AI=\angle A_1AI=\frac {|B-C|}{2} \implies cos \angle A_1AI = cos \frac{B}{2} cos \frac{C}{2} - sin \frac{B}{2} sin \frac{C}{2}$ Therefore, applying cosine rule in $\triangle A_1AI$, We get $(IA_1)^2=R^2 + AI^2 - 2R.AI cos (\frac{|B-C|}{2})$ Similarly, $(IB_1)^2 = R^2 + BI^2 -2R.BI cos (\frac{|A-C|}{2})$ We wish to prove that $IA_1 = IB_1$ Hence, it is enough to prove that $AI^2- 2R.AI cos (\frac{|B-C|}{2})=BI^2 -2R.BI cos (\frac{|A-C|}{2})$ That is, $8sin^2 \frac{C}{2} sin^2 \frac{B}{2} - 8sin \frac{C}{2} sin \frac{B}{2} cos \frac{B}{2} cos \frac{C}{2}=8sin^2 \frac{C}{2} sin^2 \frac{A}{2}-8sin \frac{C}{2} sin \frac{A}{2} cos \frac{A}{2} cos \frac{C}{2}$ That is, $sin^2 \frac{C}{2} sin^2 \frac{B}{2} - sin^2 \frac{C}{2} sin^2 \frac{A}{2} = sin \frac{C}{2} cos \frac{C}{2}(sin \frac{B}{2} cos \frac{B}{2} - sin \frac{A}{2} cos \frac{A}{2})$ That is, $sin \frac{B}{2}(cos \frac {B}{2} cos \frac {C}{2} - sin \frac{B}{2} sin \frac {C}{2}) = sin \frac{A}{2}(cos \frac {A}{2} cos \frac {C}{2} - sin \frac{A}{2} sin \frac {C}{2})$ That is, $sin \frac{B}{2} cos \frac{B+C}{2} = sin \frac{A}{2} cos \frac{A+C}{2}$ That is, $sin \frac{B}{2} sin \frac{A}{2} = sin \frac{A}{2} sin \frac {B}{2}$ which is true Hence, $IA_1=IB_1$ Similarly, $IB_1=IC_1 \implies IA_1=IB_1=IC_1$ Hence, the incenter of $\triangle ABC$ which is $I$ is also the circumcenter of $\triangle A_1B_1C_1$
ak_47
18.05.2020 17:13
$AA_1=AO$ and $AI$ is angle bisector of $\angle A_1AO$
Hence $AI$ is perpendicular bisector of $A_1O$
Hence $IO=IA_1$
Similarly,$IB_1=IC_1=IO \implies IA_1=IB_1=IC_1$
Hence, $I$ is circumcenter of $\triangle A_1B_1C_1$
srirampanchapakesan
18.05.2020 19:02
This is a special case of the following result: Let P and Q are isogonal conjugates w.r.t triangle ABC.Take points X,Y and Z on AQ, BQ and CQ such AX=AP, BY = BP and CZ = CP. Then then X,Y and Z lie on a circle with centre at the incenter I of triangle ABC. This circle also passes through circumcenter P of triangle ABC and its radius is IP. Similarly a circle for Q
dungnguyentien
19.09.2020 09:09
Let $(O)$ be the circumcircle of $\Delta ABC$. Since $\angle BAO=\angle CAA_1$, $AI$ is the angle bisector of $\angle OAA_1$. By symmetry through $AI$, $IA_1=IO$. Similarly, $IA_1=IB_1=IC_1=IO$.