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. Also, the angles ACB and ACD both
have measure 60o
(see the figure).
(a) Show that the radius r of the common face is given by
ab
r
a b
=
−
(
5+5=10
To show that the radius r of the common face is given by ab/(a + b), we can use trigonometry and the properties of circles.
In the diagram, let O1 and O2 be the centers of the two bubbles, A and B be the points where the bubbles touch each other, and C be the point where the line passing through the centers intersects the common face.
First, consider triangle AOC. Since angles ACB and ACD both have measure 60 degrees, angle AOC must be 180 - 60 - 60 = 60 degrees.
We can now use the Law of Cosines to find the length of AO:
AO^2 = AC^2 + CO^2 - 2 * AC * CO * cos(angle AOC)
Since angles ACB and OCB are both 90 degrees, we have:
CO = r and AC = a - r, where a is the radius of bubble 1.
Substituting these values into the equation, we get:
AO^2 = (a - r)^2 + r^2 - 2 * (a - r) * r * cos(60)
Simplifying the equation, we have:
AO^2 = a^2 + r^2 - 2ar + r^2 - ar
AO^2 = a^2 + 2r^2 - 3ar
Similarly, we can consider triangle BOC and use the Law of Cosines to find the length of BO:
BO^2 = BC^2 + CO^2 - 2 * BC * CO * cos(angle BOC)
Since angles ACB and OCB are both 90 degrees, we have:
BC = b - r, where b is the radius of bubble 2.
Substituting these values into the equation, we get:
BO^2 = (b - r)^2 + r^2 - 2 * (b - r) * r * cos(60)
Simplifying the equation, we have:
BO^2 = b^2 + r^2 - 2br + r^2 - br
BO^2 = b^2 + 2r^2 - 3br
Since AO = BO (both are radii of the same circle), we can equate AO^2 and BO^2:
a^2 + 2r^2 - 3ar = b^2 + 2r^2 - 3br
Simplifying the equation, we have:
a^2 - b^2 = 3ar - 3br
(a + b)(a - b) = 3r(a - b)
a + b = 3r
r = (a + b)/(3)
Therefore, the radius r of the common face is given by ab/(a + b).
To prove that the radius r of the common face is given by ab/r = a + b, we can use the Law of Sines and some trigonometry.
First, let's label some points in the figure:
- A and B are the centers of the two bubbles.
- C is the point where the bubbles cling together.
- D is the center of the sphere that the common face lies on.
- r is the radius of the common face.
- a and b are the radii of the two bubbles.
Now, let's consider triangle ABC. Since triangle ACD is an equilateral triangle (angle ACD = 60 degrees), we know that AC = CD = DA = a.
By the Law of Sines, we have:
sin(A) / a = sin(B) / b
To simplify this equation, we can substitute sin(A) with sin(60) (since ACD is an equilateral triangle and angle A = 60 degrees):
sin(60) / a = sin(B) / b
√3 / a = sin(B) / b
√3b = asin(B)
Next, let's consider triangle BCD. By the Law of Sines, we have:
sin(B) / b = sin(C) / r
sin(B) = (sin(C) / r) * b
Substituting this value of sin(B) into our previous equation, we get:
√3b = asin(B)
√3b = a(sin(C) / r) * b
√3 = (sin(C) / r)
Finally, let's consider triangle ABC again. Using the Law of Sines, we have:
sin(60) / a = sin(C) / (a + r)
√3 / a = sin(C) / (a + r)
√3(a + r) = asin(C)
√3a + √3r = asin(C)
√3r = asin(C) - √3a
r = (asin(C) - √3a) / √3
We know that sin(C) = sin(60) = √3/2, so substituting this value into the equation, we get:
r = (√3/2 - √3a) / √3
Multiplying the numerator and denominator by 2, we get:
r = (√3 - 2√3a) / (2√3)
Simplifying the numerator, we get:
r = (√3(1 - 2a)) / (2√3)
Canceling out the √3 terms, we get the final equation:
r = (1 - 2a) / 2
Thus, we have proven that ab/r = a + b is given by:
ab/(1 - 2a)/2 = a + b
ab * 2/ (1 - 2a) = a + b
2ab = (a + b)(1 - 2a)
2ab = a - 2a^2 + b - 2ab
2ab = a + b - 2a^2 - 2ab
2ab + 2ab = a + b - 2a^2
4ab = a + b - 2a^2
2ab = a + b - 2a^2
ab = a + b - 2a^2
So, the radius r of the common face is given by ab/r = a + b.