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

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

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Source: Mexican Mathematical Olympiad 1988 OMM P3

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#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.

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#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

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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?

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

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

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#5 h Jun 21, 2024, 12:10 AM • 2 Y

Y by GeoKing, Jack_w

Solution

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