AoPS - Problem

Source: Sharygin Correspondence Round 2024 P20

Tags: geometry, isogonals, circumcircle

starchan

06.03.2024 18:53

Lines $a_1, b_1, c_1$ pass through the vertices $A, B, C$ respectively of a triange $ABC$; $a_2, b_2, c_2$ are the reflections of $a_1, b_1, c_1$ about the corresponding bisectors of $ABC$; $A_1 = b_1 \cap c_1, B_1 = a_1 \cap c_1, C_1 = a_1 \cap b_1$, and $A_2, B_2, C_2$ are defined similarly. Prove that the triangles $A_1B_1C_1$ and $A_2B_2C_2$ have the same ratios of the area and circumradius (i.e. $\frac{S_1}{R_1} = \frac{S_2}{R_2}$, where $S_i = S(\triangle A_iB_iC_i)$, $R_i = R(\triangle A_iB_iC_i)$)

Here is my solution: Claim 1: $\measuredangle{ A_1AC}=\measuredangle{A_2AB}$ Proof: We easily get $ \frac{sin \measuredangle{A_1AC}}{sin \measuredangle{A_1AB}} = \frac{A_1C}{A_1B}.\frac{A_2B}{A_2C}$ and similarly $ \frac{sin \measuredangle{A_2AB}}{sin \measuredangle{A_2AC}} = \frac{A_1C}{A_1B}.\frac{A_2B}{A_2C}$ and so $\measuredangle{A_2AB} = \measuredangle{A_1AC}$ Let $B_1C_1 = a_1 , B_2C_2 = a_2$. Let the altitude of $A_1$ onto $a_1$ be $h_1$ and the altitude of $A_2$ be $h_2$. We have: $\frac{R_1}{R_2} = \frac{a_1}{a_2}.\frac{sin \measuredangle{BA_2C}}{sin \measuredangle{BA_1C}}$ and $\frac{S_1}{S_2} = \frac{a_1h_1}{a_2h_2}$ Now by Claim 1 we get that $\frac{h_1}{AA_1} = \frac{h_2}{AA_2}$ so $\frac{S_1}{S_2} = \frac{a_1h_1}{a_2h_2} = \frac{AA_1}{AA_2}.\frac{a_1}{a_2}$ To finish notice that $\frac{AA_1}{AA_2} = \frac{CA_1}{BA_2} . \frac{sin \measuredangle{ACA_1}}{sin \measuredangle{ABA_2}} = \frac{CA_1}{BA_2} . \frac{sin \measuredangle{BCA_2}}{sin \measuredangle{CBA_1}} = \frac{sin \measuredangle{BA_2C}}{sin \measuredangle{BA_1C}}$ Which gives $\frac{R_1}{R_2} = \frac{S_1}{S_2}$