Finding the common face of clinging soap bubbles using trigonometric functions of angles

Question

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 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

Answer 1

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 $r=12$.

Answer 2

Hint.

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

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

Answer 3

Considerations of surface tension physics.. differential pressure on either side of wall determine radii of spherical bubble segments

$$ \frac{1}{r}=\frac{1}{a}-\frac{1}{b},~ a<b ~ $$ which is the relation between curvatures. It is well explained in the link:

Two bubbles interphase circle radius

$$ \angle BCD= 120^{\circ}$$

and CA is angle bisector at C.

Inserting $(a=3, b=4)$ we get $r=12.$ The angles if required are computed from the Cosine Rule.

If there are three coalescing bubbles it will be further interesting to derive and find along similar lines, that each circle tangent at the common vertex makes the same $ \angle 120^{\circ}$ to one another.

Answer 4

As this is a homework problem, I will just give an outline of the answer. Denote the length of $\overline{AD}$ by $\ell$ and the length of $\overline{BA}$ by $s$.

Step 1: Find nice formulae for $\ell^2, s^2$ in (and at worst quadratic in) $a,b,r$ by using the law of cosines.

Step 2: Use the law of sines, and that $\sin( \angle BAC) = \sin(\angle DAC)$, to show $a/s = r/\ell$.

Step 3: The equation $r^2 = a^2 \ell^2/s^2$ is quadratic in $r$. Show that it has as solutions $r = a, ab/(a-b)$.

Step 4: Figure out why the solution $r=a$ is unnecessary.