I am trying to help my daughter with a problem from Stewart's Precalculus book.This problem comes right after law of sines.

When two bubbles cling together in midair, their common surface is part of a sphere whose center D lies on the line passing through the centers of the bubbles (please refer to the figure below) also angles ACB and ACD each have measure 60 degrees

  • Show that the radius r of the common surface is given by r = ab / (b - a)
  • Find the radius of the common face if the radii of the bubbles are 3cm and 4cm

I could do the second one but after using law of cosines to find length of the segment AB in triangle CBA. That came out as \sqrt{13} Then I used law of sines in triangle ABC to find angle CAB = 73.897 degrees

Angle CAD = 180 - angle CAB = 106.1 degrees angle CDA = 180 - 106.1 - 60 = 13.897 degrees

Then I used law of sines in triangle CAD to find the value of r

But I couldn't make any headway for the first one. Also it seems to me that I don't need law of cosines to solve this problem.

Any help will be appreciated. Thanks Diagram

  • $\begingroup$ When I saw the image, my first guess was that $r$ was the radius of the Apollonius circle defined by the points $A,B$ and the ratio $k = \frac{a}{b}$. However, this would give $$r = \frac{ab}{b^2-a^2}\sqrt{a^2+b^2-ab}$$ I find Stewart's choice of $r$ somewhat arbitrary. $\endgroup$ – mechanodroid Jan 1 at 1:06

By the law of cosines, $$BD=\sqrt{a^2+r^2-2ar\cos(120)}$$ $$BA=\sqrt{a^2+b^2-2ab\cos(60)}$$ $$AD=\sqrt{b^2+r^2-2br\cos(60)}$$ Since $BD=BA+AD$ we now have $$\sqrt{a^2+r^2-2ar\cos(120)}=\sqrt{a^2+b^2-2ab\cos(60)}+\sqrt{b^2+r^2-2br\cos(60)}$$ Note that $\cos(60)=1/2$ and $\cos(120)=-1/2$. Hence we obtain $$\sqrt{a^2+r^2+ar}=\sqrt{a^2+b^2-ab}+\sqrt{b^2+r^2-br}$$ WolframAlpha now gives the solution $r=ab/(a-b)$, although you can prove it by hand if necessary by squaring both sides, isolating the remaining root and then squaring both sides again.

For the second one, just plug in $a=4$ and $b=3$ to obtain $r=12$.

  • $\begingroup$ thank you for the prompt answer! The hint in the problem (which I forgot to mention) is to use law of sines along-with angle θ and 180 - θ have the same sine). Just wondering if there is any easier way to get to r = ab/(a - b) $\endgroup$ – user3138594 Jan 1 at 1:39


The line $CA$ is angle $\angle{DCB}$ bisector so

$$ \frac{BA}{AD} = \frac{a}{r} $$


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