Let $\omega_1$ and $\omega_2$ be two non-intersecting circles. Let one of its internal tangents touches $\omega_1$ and $\omega_2$ at $A_1$ and $A_2$, respectively, and let one of its external tangents touches $\omega_1$ and $\omega_2$ at $B_1$ and $B_2$, respectively. Prove that if $A_1B_2 = A_2B_1$, then $A_1B_2 \perp A_2B_1$.
Problem
Source: 2020 Caucasus Mathematical Olympiad
Tags: geometry
GeoMetrix
16.03.2020 10:42
Easy problem. Proof: Clearly we have that if $I= \overline{A_1A_2} \cap \overline{B_1B_2}$ then $\Delta IB_1A_2 \cong \Delta IA_1B_2$ since $\overline{IA_2}=\overline{IB_2}$ and also $\overline{IA_1} =\overline{IB_1}$ and $\overline{A_1B_2}=\overline{A_2B_1}$. From here we get that $\overline{A_1A_2} \perp \overline{B_1B_2}$ and also $\angle IA_2B_1=\angle IB_2A_1 $ $\implies$ that if $\overline{A_2B_1} \cap \overline{A_1B_2}=G$ then $(IA_2B_2G)$ concyclic. From here the result follows since $\angle A_2GB_2=\angle A_2IB_2=90^\circ$. $\blacksquare$.
Infact the problem even holds in the case of iff. For this just notice that $B_1$ will be the orthocentre of $\triangle{A_2A_1B_2}$ and also we would have that $\angle A_1A_2B_2=45^\circ$ so we coud hend with the formulaes $\overline{A_2B_1}=2R \cos \angle A_1A_2B_2=2R \sin \angle A_1A_2B_2= \overline{A_1B_2}$ and we would be done $\blacksquare$.
Afo
10.09.2020 17:49
Clearly we have that if $I= \overline{A_1A_2} \cap \overline{B_1B_2}$ then $\Delta IB_1A_2 \cong \Delta IA_1B_2$ since $\overline{IA_2}=\overline{IB_2}$ and also $\overline{IA_1} =\overline{IB_1}$ and $\overline{A_1B_2}=\overline{A_2B_1}$ and so $\angle A_2IB_1 = \angle A_2IB_2 \implies A_1A_2 \perp B_1B_2$. Let $A_1B_2 \cap A_2B_1=D$. Then $\triangle A_2A_1D \sim A_2B_1I (AAA) \implies A_2B_1 \perp A_1B_2$.
rafaello
20.02.2021 03:07
Outline. Let $X=A_1A_2 \cap B_1B_2$. Let $B_1'$ be the point on $B_1B_2$, so that $XB_1=XB_1'$. Thus, we have $XA_1=XB_1=XB_1'$ and since $XA_2=XB_2$, we get that $A_1B_1' \parallel A_2B_2$ and since $\angle B_1A_1B_1'=90^\circ$, we get that $A_1B_1\perp A_2B_2$. We also have that $A_2B_2=B_2A_1=A_2B_1'$, therefore $\triangle B_1A_2B_1'$ is isosceles and since $X$ is the midpoint of $B_1B_1'$, we have that $A_1A_2\perp B_1B_2$, hence $A_1$ is the orthocentre of $\triangle B_1A_2B_2$, hence $A_1B_2 \perp A_2B_1$. $ \square$