AcademiaAgriculture & FarmingHumanitiesSocial SciencesSTEM

Solution to the Three Inscribed Circles Problem

Updated on February 11, 2017
calculus-geometry profile image

TR Smith is a product designer and former teacher who uses math in her work every day.

A classic geometry and trigonometry problem is to find the relative radii of three equal circles inscribed in a larger circle. If the smaller circles are mutually tangent, and they are all tangent to the larger circle, you can use simple geometry and basic trigonometry to find one radius in terms of the other. Sometimes this problem is presented as finding the relative areas, but the method of solution is the same.

The more general problem of finding the missing radius among four mutually tangent circles of arbitrary radii is called the Soddy Problem, or alternatively the Kissing Circles Problem or Osculating Circle's Problem. This particular problem in which three of the circles are equal can be solved using trigonometry and geometry, or algebraically using the Soddy Circle equation, also known as Descartes' Equation. Both methods are given below.

Solution Step 1

The first step in the solution is to mark several line segments connecting a radius of the larger outer circle to some radii of one of the smaller inner circles. In the diagram above, a line segment drawn from where the small and large circle are tangent to center of the small circle is a radius with a length of x. Another line segment drawn from the center of a small circle to where the two small circles are tangent is another radius. The line drawn from the center of a small circle to the center of the large circle is given a length of y.

As you can see from the diagram, the radius of the small circle is x and the radius of the large circle is x + y.

Solution Step 2

The next step is to realize that one of the small triangles formed by the line segments is a 30-60-90-degree triangle. The length of the triangle's hypotenuse is y, the length of the longer leg is x, and the length of the shorter leg is y/2. The ratio of the longer leg to the hypotenuse is sqrt(3)/2. This means

x/y = sqrt(3)/2
x = y*sqrt(3)/2
y = 2x/sqrt(3)

If the radius of the smaller circle is x and the radius of the larger circle is x + y, the ratio of the smaller radius to the larger radius is

x/[x + 2x/sqrt(3)] = 1/[1 + 2/sqrt(3)] = sqrt(3)/[2 + sqrt(3)]

= 2*sqrt(3) - 3 ≈ 0.464102

For example, if the radius of the larger circle is 12 cm, then the radius of each smaller circle is 12*0.464102 = 5.569 cm. If the radius of one of the smaller circles is 20 inches, then the radius of the larger circle is 20/0.464102 = 43.094 inches. The ratio of the radii is also the ratio of the circles' circumferences.

Ratio of the Circles' Areas

Now that we have shown that the radius ratio is 2*sqrt(3) - 3 = 0.464102, we can also find the area ratio. If the larger circle has a radius of R, its area is πR^2. Each of the smaller circles has a radius of [2*sqrt(3)-3]R, so their areas are each π[21-12*sqrt(3)]R^2. Taking the ratio of a small circle's area to the large circle's area you get

π[21-12*sqrt(3)]R^2 / πR^2
= 21 - 12*sqrt(3)
≈ 0.21539

The ratio of all three smaller circles' area to the larger circle's area is 3*[21 - 12*sqrt(3)] ≈ 0.646171.

Soddy's Formula / Descartes' Equation

If four mutually tangent (osculating or "kissing") circles have radii of A, B, C, and D, then these radii are related by the remarkable formula

(1/A + 1/B + 1/C + 1/D)^2 = 2/A^2 + 2/B^2 + 2/C^2 + 2/D^2

which was first discovered by Rene Descartes and later independently rediscovered by Frederick Soddy. If you know the radii of three of the circles, you can use the formula to find the radius of the 4th. The formula actually gives two possible solutions, because for any three mutually tangent circles, there are two ways to draw a fourth that is tangent to the other three.

If we call the radius of each small circle x and the radius of the large circle R, Descartes' Equation gives us

(1/x + 1/x + 1/x + 1/R)^2 = 2/x^2 + 2/x^2 + 2/x^2 + 2/R^2

(3/x + 1/R)^2 = 6/x^2 + 2/R^2

9/x^2 + 6/(xR) + 1/R^2 = 6/x^2 + 2/R^2

3(1/x)^2 + 6(1/x)(1/R) - (1/R)^2 = 0

x^2 - 6Rx - 3R^2 = 0

Solving for x in terms of R using the quadratic formula gives

x = [6R ± sqrt(48R^2)]/2

= [3 ± 2*sqrt(3)]R

≈ -0.464102R and 6.464102R

The first answer, x = -0.464102R matches the result we computed with trigonometry except for the minus sign. The negative value is because the unknown radius corresponds to a circle that surrounds the three given circles. The second answer, x = 6.464102R gives the radius of a cricle set in the middle of the three tangent circles, as in the bottom half of the diagram above.

Why Is It Named After Soddy If Descartes Discovered It First?

Soddy's name is attached to the osculating circle formula because he memorialized it in the form of a poem called "The Kiss Precise" which he composed in 1936:

In the poem, "bends" are the curvatures of the circles, where curvature is defined as the inverse of the radius. The last stanza of the poem also gives the solution to the 5 osculating spheres problem. If 5 mutually tangent spheres have radii of A, B, C, D, and E, then the 5 quantities are related by the equation

(1/A + 1/B + 1/C + 1/D + 1/E)^2 =

3/A^2 + 3/B^2 + 3/C^2 + 3/D^2 + 3/E^2

Comments

    0 of 8192 characters used
    Post Comment

    No comments yet.

    Click to Rate This Article