Let $M$ be the point of intersection of the diagonals of a cyclic quadrilateral $ABCD$. Let $I_1$ and $I_2$ are the incenters of triangles $AMD$ and $BMC$, respectively, and let $L$ be the point of intersection of the lines $DI_1$ and $CI_2$. The foot of the perpendicular from the midpoint $T$ of $I_1I_2$ to $CL$ is $N$, and $F$ is the midpoint of $TN$. Let $G$ and $J$ be the points of intersection of the line $LF$ with $I_1N$ and $I_1I_2$, respectively. Let $O_1$ be the circumcenter of triangle $LI_1J$, and let $\Gamma_1$ and $\Gamma_2$ be the circles with diameters $O_1L$ and $O_1J$, respectively. Let $V$ and $S$ be the second points of intersection of $I_1O_1$ with $\Gamma_1$ and $\Gamma_2$, respectively. If $K$ is point where the circles $\Gamma_1$ and $\Gamma_2$ meet again, prove that $K$ is the circumcenter of the triangle $SVG$.
Problem
Source: 2012 Balkan Shortlist G4 BMO
Tags: geometry, incenter, circumcircle, Cyclic, Circumcenter, circles
26.11.2020 21:00
Proof of $KS=KV$: Since $$\angle LI_2I_1=180^\circ-\angle CI_2I_1=90^\circ-\angle CBD=90^\circ-\angle CAO=180^\circ-\angle I_2I_1D=\angle I_2I_1L,$$we infer that triangle $LI_1I_2$ is isosceles. Obviously, point $K$ is the midpoint of $LJ$, hence $$\angle SVK=\angle O_1VK=\angle O_1LK=\angle O_1JL=\angle O_1SK,$$which implies $KS=KV$. Proof of $KS=KG$: We prove now the following Claim. Claim: $LJ$ is perpendicular to $I_1N$. Proof: Let $R$ be the midpoint of $NI_2$. Then, $FR \parallel TI_2$, hence $FR \perp LT$, and since $TN \perp LI_2$ we obtain that $LJ \perp TR$. To conclude, we have $TR \parallel I_1N$, therefore $I_1N \perp LJ$, as desired $\blacksquare$. To the problem, from the Claim $\angle I_1SJ=\angle I_1GZ=90^\circ$ so $SI_1JG$ is cyclic, giving $$\angle SGK=\angle O_1I_1J=90^\circ-\frac{\angle I_1O_1J}{2}=90^\circ-\frac{\angle SKG}{2},$$hence $\angle SGK=90^\circ-\frac{\angle SKG}{2}$, implying that triangle $KSG$ is cyclic with $KS=KG$, as desired.
06.08.2024 08:22
Orestis_Lignos wrote: Proof of $KS=KV$: Since $$\angle LI_2I_1=180^\circ-\angle CI_2I_1=90^\circ-\angle CBD=90^\circ-\angle CAO=180^\circ-\angle I_2I_1D=\angle I_2I_1L,$$we infer that triangle $LI_1I_2$ is isosceles. Obviously, point $K$ is the midpoint of $LJ$, hence $$\angle SVK=\angle O_1VK=\angle O_1LK=\angle O_1JL=\angle O_1SK,$$which implies $KS=KV$. Proof of $KS=KG$: We prove now the following Claim. Claim: $LJ$ is perpendicular to $I_1N$. Proof: Let $R$ be the midpoint of $NI_2$. Then, $FR \parallel TI_2$, hence $FR \perp LT$, and since $TN \perp LI_2$ we obtain that $LJ \perp TR$. To conclude, we have $TR \parallel I_1N$, therefore $I_1N \perp LJ$, as desired $\blacksquare$. To the problem, from the Claim $\angle I_1SJ=\angle I_1GZ=90^\circ$ so $SI_1JG$ is cyclic, giving $$\angle SGK=\angle O_1I_1J=90^\circ-\frac{\angle I_1O_1J}{2}=90^\circ-\frac{\angle SKG}{2},$$hence $\angle SGK=90^\circ-\frac{\angle SKG}{2}$, implying that triangle $KSG$ is cyclic with $KS=KG$, as desired. can you explain me how to get $LJ$ perpendicular to $I_1N$? i'm quite confused