The incircle of a triangle $A_0B_0C_0$ touches the sides $B_0C_0,C_0A_0,A_0B_0$ at the points $A,B,C$ respectively, and the incircle of the triangle $ABC$ with incenter $ I$ touches the sides $BC,CA, AB$ at the points $A_1, B_1,C_1$, respectively. Let $\sigma(ABC)$ and $\sigma(A_1B_1C)$ be the areas of the triangles $ABC$ and $A_1B_1C$ respectively. Show that if $\sigma(ABC) = 2 \sigma(A_1B_1C)$ , then the lines $AA_0, BB_0,IC_1$ pass through a common point .
Problem
Source: 2010 Balkan Shortlist G3 BMO
Tags: concurrency, concurrent, area of a triangle, areas, geometry, incenter, incircle
Suntafayato
11.05.2020 01:26
parmenides51 wrote: The incircle of a triangle $A_0B_0C_0$ touches the sides $B_0C_0,C_0A_0,A_0B_0$ at the points $A,B,C$ respectively, and the incircle of the triangle $ABC$ with incenter $ I$ touches the sides $BC,CA, AB $at the points $A_1, B_1,C_1$, respectively. Let $\sigma(ABC)$ and $\sigma(A_1B_1C)$ be the areas of the triangles $ABC$ and $A_1B_1C$ respectively. Show that $\sigma(ABC) = 2 \sigma(A_1B_1C)$ , then the lines $AA_0, BB_0,IC_1$ pass through a common point . Are you sure that $[ABC] = 2 [A_1B_1C]$? I don't think this can be possible.
parmenides51
11.05.2020 03:00
I forgot the if word, this is the correct wording parmenides51 wrote: The incircle of a triangle $A_0B_0C_0$ touches the sides $B_0C_0,C_0A_0,A_0B_0$ at the points $A,B,C$ respectively, and the incircle of the triangle $ABC$ with incenter $ I$ touches the sides $BC,CA, AB$ at the points $A_1, B_1,C_1$, respectively. Let $\sigma(ABC)$ and $\sigma(A_1B_1C)$ be the areas of the triangles $ABC$ and $A_1B_1C$ respectively. Show that if $\sigma(ABC) = 2 \sigma(A_1B_1C)$ , then the lines $AA_0, BB_0,IC_1$ pass through a common point .
MathsLion
11.05.2020 16:21
Calculations maybe look boring and ugly, but the problem is pretty straightforward in my opinion.
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