Two equal common tangents are perpendicular

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Two equal common tangents are perpendicular

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2020 Caucasus Mathematical Olympiad

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

111 posts

#1 h Mar 16, 2020, 7:03 AM

Y by

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 $A_1B_2 \perp A_2B_1$ .

This post has been edited 1 time. Last edited by bigant146, Mar 16, 2020, 8:49 AM
Reason: delete ~

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

924 posts

#2 h Mar 16, 2020, 7:42 AM

Y by

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 $\angle A_2GB_2=\angle A_2IB_2=90^\circ$ . $\blacksquare$ .

PS

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1002 posts

Afo

#3 h Sep 10, 2020, 2:49 PM

Y by

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 $\triangle A_2A_1D \sim A_2B_1I (AAA) \implies A_2B_1 \perp A_1B_2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1079 posts

#4 h Feb 20, 2021, 12:07 AM

Y by

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 $A_1B_2 \perp A_2B_1$ . $\square$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a