triangle between common tangents of externally tangent circles in terms of r,R

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

w

20 users

TOPICS

No tags match your search

M

No tags match your search

M

triangle between common tangents of externally tangent circles in terms of r,R

G H J

Reveal topic tags

tangent circles

common tangents

area of a triangle

L

G H BBookmark kLocked kLocked NReply

Source: Mexican Mathematical Olympiad 1988 OMM P3

The post below has been deleted. Click to close.

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

30629 posts

#1 h Jul 27, 2018, 9:11 PM • 3 Y

Y by HWenslawski, Adventure10, GeoKing

Two externally tangent circles with different radii are given. Their common tangents form a triangle. Find the area of this triangle in terms of the radii of the two circles.

Z K Y

The post below has been deleted. Click to close.

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

122 posts

#2 h Oct 6, 2021, 7:07 PM • 2 Y

Y by HWenslawski, GeoKing

Let $\omega_R$ be the circle with radius $R$ and center $O_R$ , and $\omega_r$ the circle with radius $r$ and center $O_r$ . WLOG assume $R>r$ .

Let $C$ be the common point of $\omega_R$ and $\omega_r$ . Let $A$ and $E$ be the points of tangency to $\omega_r$ , and $B$ , $D$ the points of tangency to $\omega_R$ , such that $AB$ , $DE$ and the perpendicular to $O_RO_r$ passing through $C$ are the common tangents to $\omega_R$ and $\omega_r$ . Then we will define $X$ as the intersection of $AB$ and $DE$ . $Y$ and $Z$ are the other vertices of the triangle formed by the other tangents, such that $Y$ lies on $AB$ , and $Z$ lies on $DE$ . Now we can begin the solution.

First, some initial observations:
(1) The figure is symmetrical over $O_RO_r$ ,
(2) $X, O_r, C, O_R$ are collinear,
(3) $\bigtriangleup XAO_r, \bigtriangleup XCY, \bigtriangleup XBO_R$ are all similar.

By (3), $\frac{AO_r}{O_rX}=\frac{BO_R}{O_RX}$ . Let $O_rX = x$ , then $\frac{r}{x} = \frac{R}{R+r+x}$ . Solving for $x$ gives $x=\frac{r(R+r)}{R-r}$ . By Pythagoras on $\bigtriangleup XAO_r$ , we obtain $AX=\sqrt{x^2-r^2}=\frac{\sqrt{4r^3R}}{R-r}$ .

Again, because of (3), $\frac{XC}{CY}=\frac{XA}{AO_r}$ . We know $XC = x + r = \frac{2Rr}{R-r}$ . So $\frac{2Rr}{CY\cdot (R-r)}=\frac{\frac{\sqrt{4r^3R}}{R-r}}{r}$ . After symplifying, $CY = \sqrt{Rr}$ .

Since the figure is symmetrical over $O_RO_r$ (1), $CY = CZ$ , so $\frac{1}{2} YZ = CY$ .

We know $YZ$ is perpendicular to $O_RO_r$ , so $YZ$ is perpendicular to $CX$ . This means that the altitude of $\bigtriangleup XYZ$ is $CX$ .

Then $(\bigtriangleup XYZ)= \frac{XC\cdot YZ}{2}= XC\cdot \frac{1}{2} YZ = \frac{2Rr}{R-r} \cdot \sqrt{Rr} = \frac{2Rr\sqrt{Rr}}{R-r}$ , and we are done.

Please notify me if I made a mistake anywhere in the solution

This post has been edited 2 times. Last edited by JuanDelPan, Oct 6, 2021, 9:04 PM
Reason: Mistake

Z K Y

The post below has been deleted. Click to close.

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

49 posts

jampm

#3 h Oct 6, 2021, 8:34 PM

Y by

My solution is kinda different to yours pan, and I also get a different answer.
solution
I'm starting to doubt my solution, did I get smth wrong?

Z K Y

The post below has been deleted. Click to close.

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

122 posts

#4 h Oct 6, 2021, 9:05 PM

Y by

@above, indeed I made a mistake while calculating $AX$ . I fixed it now, and it's the same answer.

Z K Y

The post below has been deleted. Click to close.

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

116 posts

#5 h Jun 21, 2024, 12:10 AM • 2 Y

Y by GeoKing, Jack_w

Solution

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a