Let $AL_a$, $BL_b$, $CL_c$ be the bisecors of triangle $ABC$. The tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at point $K_a$, points $K_b$, $K_c$ are defined similarly. Prove that the lines $K_aL_a$, $K_bL_b$ and $K_cL_c$ concur.
Problem
Source: Sharygin CR 2019 P19 (Grade 10 - 11)
Tags: geometry
06.03.2019 16:01
I still can't believe they decided to place this in grade 10 - 11. Note that $AL_a, AL_b, AL_c$ concur at the incenter of triangle $ABC$. Note that $AK_a, AK_b, AK_c$ concur at the symmedian point of triangle $ABC$. Thus, by Cevian Nest Theorem, $K_aL_a, K_bL_b, K_cL_c$ concur.
06.03.2019 19:01
This was actually a troll. I am not giving the solution, too lazy. But mine is identical to Vrangr. (I haven't seen TDP's). But I took the cevian nest with respect to $K_aK_bK_c$, and used that Gergonne point and incenter of intouch triangle(Bleh!) exists.
07.03.2019 04:21
Alternatively, note that $K_aL_a$ is the polar of $M_a$ w.r.t. $(ABC)$, where $M_a$ is the foot of the $A$-external angle bisector. But $M_a, M_b, M_c$ are collinear by Desargue on $\triangle L_aL_bL_c$ and $\triangle ABC$, so it follows that their polars concur.
10.03.2019 09:09
Lol, was I the only one who bashed it Bary!
Attachments:
Sharygin- 19.pdf (30kb)
10.03.2019 10:14
Vrangr wrote: Let $AL_a$, $BL_b$, $CL_c$ be the bisecors of triangle $ABC$. The tangents to the circumcircle of $ABC$ at $B$ and $C$ meet at point $K_a$, points $K_b$, $K_c$ are defined similarly. Prove that the lines $K_aL_a$, $K_bL_b$ and $K_cL_c$ concur. very nice problem
10.06.2019 06:45
Let $(I)$, $O$ be incircle, circumcenter of $\triangle ABC$, $S$ be midpoint of $BC$; $M$ $\equiv$ $AI$ $\cap$ $(ABC)$; $D$ is orthogonal projection of $I$ on $BC$; $J$ be internal homothetic center of $(I)$ and $(ABC)$; $N$ be reflection of $M$ through $O$ Since: $O$ is midpoint of $MN$ and $\overline{K_aS} . \overline{K_aO} = K_aB^2 = K_aC^2 = \overline{K_aM} . \overline{K_aN}$ then: $\dfrac{\overline{K_aO}}{\overline{K_aM}} = \dfrac{\overline{K_aN}}{\overline{K_aS}} = \dfrac{\overline{K_aO} - \overline{K_aN}}{\overline{K_aM} - \overline{K_aS}} = \dfrac{\overline{NO}}{\overline{SM}}$ We have: $\dfrac{\overline{JI}}{\overline{JO}} . \dfrac{\overline{K_aO}}{\overline{K_aM}} . \dfrac{\overline{L_aM}}{\overline{L_aI}} = \dfrac{\overline{JI}}{\overline{JO}} . \dfrac{\overline{NO}}{\overline{SM}} . \dfrac{\overline{SM}}{\overline{DI}} = 1$ So: $K_a$, $L_a$, $J$ are collinear Similarly: $K_b$, $L_b$, $J$ are collinear, $K_c$, $L_c$, $J$ are collinear Hence: $K_aL_a$, $K_bL_b$, $K_cL_c$ concurrent at a point $J$ lie on $IO$
01.10.2020 22:51
Same as any solution but fine. See that $\triangle L_a, L_b, L_c$ is the incentral triangle of $\triangle ABC$ and $\triangle K_aK_bK_c$ is the tangential triangle of $\triangle ABC$. By Cevian Nest Theorem, $\triangle K_aK_bK_c$ and $\triangle L_aL_bL_c$ are perspective
01.07.2022 17:06
21.07.2022 14:54
Hmm I skipped cevian nest in EGMO because of its complicated diagram, so for now here's a solution that doesn't use cevian nest Trivial/Troll 10 seconds problem. Notice that by sine rule in $\triangle BK_aL_a$ and $\triangle CK_aL_a$ we get $$\frac{\sin \angle BK_aL_a}{\sin \angle CK_aL_a}=\frac{BL_a}{CL_a}=\frac{AB}{AC}$$Thus $$\prod_{\text{cyclic}}{\frac{\sin \angle BK_aL_a}{\sin \angle CK_aL_a}}=\prod_{\text{cyclic}}{\frac{AB}{AC}}=1$$and so we are done by the Trigonometric form of ceva's theorem $\blacksquare$