Descartes Circle Theorem - Beecroft Configurations

This entry is an expanded exposition of the rest of the material in Chapter 1.5 of Coxeter's "Introduction to Geometry" where he uses Beecroft's observations to prove the Descartes Circle Theorem.

Given four circles E_i, i = A, B, C, D mutually tangent to each other at six distinct points of tangency, we define the bend , e_i, of each circle as plus or minus the reciprocal of the corresponding radius r_i, i = A, B, C, D where bend is negative if the circle surrounds the other three and positive otherwise. That is:

e_i = ±1/r_i i = A, B, C, D where e_i < 0 if E_i surrounds the other three circles and positive otherwise. The bend of a straight line is taken as 0.

Descartes Circle Theorem: 2(Σe_i²) = (Σe_i)²

This tells us that if we have the radii of three tangent circles we can obtain the fourth using a simple quadratic in the unknown bend.

It is Coxeter's genius to mark the trail of a proof using the most concise and well thought out diagrams and text to make his exposition fully informative to one who is willing to use the guideposts to complete the trail. In this case the proof is not direct. It uses a result of an English "amateur", Philip Beecroft, who rediscovered Descartes work some 200 years later and carried it further. In particular, Beecroft observed that the quartet of circles E_i give rise to another quartet of circles H_i which are also tangent to each other at the same six distinct points and intricately related to the originals. Define H_i for i= A, B, C, D, as the circumcircle of the triangle made up of the three intersections points of its three complementary E-circles, that is, the triplet other than the E_i^th, and let n_i be the bend of H_i. Then H_i is either the incircle or an excircle of the triangle formed by the centers of the complementary circles. In particular:

• Case A: Suppose E_B surrounds E_C and E_D; then H_A is the B-excircle of ΔBCD. (Of course case A might have E_C or E_D) surrounding in which case H_A would be the C-excircle or D-excircle of ΔBCD.)
• Case B: Suppose E_B, E_C and E_D are externally tangent; then H_A is the incircle of ΔBCD

Similarly, we define H_B, H_C and H_D. Then these H-circles are four mutually tangent circles with the same six points of tangency as the corresponding E-circles. Let n_i, i = A, B, C, D be the bends of the H-circles. Here is a jump to a diagram of one such configuration.

Consider case A: We learned in the previous two blog entries that e_B = -1/s, e_C = 1/(s-d), e_D = 1/(s-c) and n_A = ±(1/r_b) where s is the semiperimeter of ΔBCD with sides of length b, c, d and r_b is the radius of the B-excircle. (Whether n_A is positive or negative depends on the particular configuration of circles we are working with. See note CASEA.)

Now consider case B: Since E_B, E_C and E_D are externally tangent we know that e_B = 1/(s-b), e_C = 1/(s-c), e_D = 1/(s-d). Also we know that n_A = ±(1/r) where r = radius of the incircle of ΔBCD. See note CASEB.

Recall from our previous blog entry that r² = (s-b)(s-c)(s-d)/s and r_b ² = s(s-c)(s-d)/(s-b).

We use the identity: uv + uw + vw = (1/u + 1/v + 1/w)uvw to get e_Be_C + e_Be_D + e_Ce_D =

• Case A: (-s + s - d + s - c)(-1/(s)(s-d)(s-c)) = (s - c - d)(-1/(s-b)r_b²) = 1 / r_b²
• Case B: (s - b + s - c + s - d)(1/(s-b)(s-c)(s-d)) = s (1/r²s) = 1/r²

Since in case A, n_A² = 1 / r_b² and in case B n_A² = 1 / r², we have in both cases that:

(1)

e_Be_C + e_Be_D + e_Ce_D = n_A²

By symmetry between the E-circles and the H-circles, the relationship holds in the other direction:

(2)

n_Bn_C + n_Bn_D + n_Cn_D = e_A²

Both hold (again by symmetry in the indices) for all combinations of the subscripts. Thus

(e_A + e_B + e_C + e_D)² = e_A² + e_B² + e_C² + e_D² + 2(e_Ae_B + ... + e_Ce_D) = e_A² + e_B² + e_C² + e_D² + n_A² + n_B² + n_C² + n_D²

(3)

(Σe_i)² = Σe_i² + Σn_i² and by symmetry (4) (Σn_i)² = Σe_i² + Σn_i²

Thus (5)

(Σe_i)² = (Σn_i)² and since the sum of the bends of four mutually tangent circles is positive, this gives us (6) Σe_i = Σn_i > 0

Note that

(e_A + e_B + e_C - e_D)(e_A + e_B + e_C + e_D) = (e_A + e_B + e_C + e_D)² - e_D² = e_A² + e_B² + e_C² + 2(e_Ae_B + e_Ae_C + e_Be_C) - e_D² = e_A² + e_B² + e_C² - e_D² + 2n_D² = (n_Bn_C + n_Be_D + n_Ce_D) + (n_An_C + n_Ae_D + n_Ce_D) + (n_An_B + n_Ae_D + n_Be_D) - (n_An_B + n_Ae_C + n_Be_C) + 2n_D² = 2(n_An_D + n_Be_D + n_Ce_D) - 2n_D² = 2n_D(n_A + n_B + n_C + n_D) But Σe_i = Σn_i So (7) e_A + e_B + e_C - e_D = 2n_D This holds for all combinations of the subscripts and switching the e's and n's. Thus, (e_A + e_B + e_C - e_D)² = 4n_D² (e_B + e_C + e_D - e_A)² = 4n_A² (e_C + e_D + e_A - e_B)² = 4n_B² (e_D + e_A + e_B - e_C)² = 4n_C² Adding all four of these equations and simplifying we get 4Σe_i² = 4Σn_i² So (8) Σe_i² = Σn_i² Thus (9) (Σe_i)² = Σe_i² + Σn_i² = 2Σe_i² And this establishes the Descartes Circle Theorem.

CASEA: H_A is the B-excircle of ΔBCD.

• Case A.1: Suppose E_A is on the far side of line CD from B. Then H_B, H_C, and H_D are surrounded by H_A so n_A = -1/r_b
• Case A.2: Suppose E_A is on the same side of line CD as B. Then H_B, H_C, and H_D are externally tangent to H_A so n_A = 1/r_b

CASEB: H_A is the incircle of ΔBCD.

• Case B.1: Suppose E_A surrounds E_B, E_C and E_D. Then H_B, H_C, and H_D are all externally tangent to H_A so n_A = 1/r_b
• Case B.2: Suppose E_A is externally tangent to E_B, E_C, and E_D. Then H_A surrounds H_B, H_C, and H_D so n_A = -1/r_b

An exercise from Coxeter section 1.5

Coxeter refers to the octet of E-circles and associated H-circles as a Beecroft configuration. We now consider Exercise 9. For any four numbers satisfying k + l + m + n = 0, there is a "Beecroft configuration" having bends e₁ = k(k + l), e₂ = l(k + l), e₃ = n² - kl, e₄ = m² - kl, n₁ = l² - mn, n₂ = k² - mn, n₃ = m(m + n), n₄ = n(m + n). [Hint: Express e₃, e₄, n₁, n₂ as rational functions of e₁, e₂, n₃, n₄]

Solution to Exercise 9

We have seen above that equations such as

e_Be_C + e_Be_D + e_Ce_D = n_A²

hold for all such combinations of e's and n's. Thus, we can write (identifying subscripts A, B, C, and D with 1, 2, 3, 4 and assuming the variable indices i, j, k, l are distinct elements in {1, 2, 3, 4})

n_i(n_j + n_k) = - n_jn_k + e_l²

n_i = (e_l² - n_jn_k)/ (n_j + n_k)

By using (7) twice we see that e_i + e_j = n_k+ n_l, so we have

n_i = (e_l² - n_jn_k)/ (e_i + e_l)

Using this and the corresponding equation for e_i, we get the rational functions suggested in the hint.

n₁ = (e₂² - n₃n₄)/ (e₁ + e₂)

n₂ = (e₁² - n₃n₄)/ (e₁ + e₂)

e₃ = (n₄² - e₁e₂)/ (e₁ + e₂)

e₄ = (n₃² - e₁e₂)/ (e₁ + e₂)

Since we are trying to show the configuration is possible as stated, we can assume, without loss of generality, that e₁ = k(k + l), e₂ = l(k + l), and n₃ = m(m + n). We know k + l + m + n = 0 and e₁ + e₂ = n₃ + n₄. Therefore, k(l + k) + l(l + k) - m(m + n) = n₄. That is (l + k)² - m(m + n) = n₄ and since (l + k) = -(m + n), we have (m + n)² - m(m + n) = n₄ so n(m + n) = n₄. Now using the rational functions established previously we have n₁ = (l²(k + l)² - mn(m + n)²)/(l(k + l) + k(k + l)) = l² - mn. Similarly, n₂ = k² - mn, e₃ = n² - kl, e₄ = m² - kl.

Note: Suppose (l + k) = 0. Then e₁ = e₂ = 0 by assignment and (m + n) = 0 because k + l + m + n = 0. Therefore, n₃ = n₄ = 0. In this case, E₁, E₂, H₃ and H₄ are circles with bend 0, i.e., straight lines with E₁ parallel to E₂ (tangent at infinity) as are H₃ and H₄. In this case e₁ = k(k + l) = 0 = e₂ = l(k + l) and n₃ = m(m + n) = 0 = n₄ = n(m + n). Now k = -l and m = -n mean that e₃ = e₄ = n₁ = e₂ = k² + n² so circles of radius 1/(k² + n²) as half the distance between E₁, E₂ and H₃, H₄. Using the point at infinity as one of the points of tangency we get a (degenerate) configuration where E₃, E₄ are tangent to each other as well as E₁, E₂. H₃ contains the intersections of E₁E₄, E₂E₄ and E₁E₂ (infinity), etc. Similarly H₁ and H₂ can be drawn. Note: if k² + n² = 0 then k = l = m = n = 0.

Finally, (as in Coxeter's solution) the parameterization of the problem can be obtained directly from a configuration as follows: Given e₁, e₂, n₃, n₄, set e₁ = k(l + k), e₂ = l(k + l), so e₁ + e₂ = (k + l)², (k + l) = ±√[e₁ + e₂]. Then k = ±e₁/√[e₁ + e₂] and l = ±e₂/√[e₁ + e₂]. Similarly m = ±n₃/√[e₁ + e₂] and n = ±n₄/√[e₁ + e₂] where we used e₁ + e₂ = n₃ + n₄. By choosing, say, +, +, -, - we get k + l + m + n = [(e₁ + e₂) - (n₃ + n₄)]/√[e₁ + e₂] = 0.

The following diagram shows a Beecroft configuration with k = 5, l = 4, m = 3, and n = -12 It has values e₁ = 45, e₂ = 36, e₃ = 124, e₄ = -11, n₁ = 52, n₂ = 61, n₃ = -27, n₄ = 108. E₁ has center at the (0,0) and E₂ has center on the x-axis. E₄ surrounds the other E-circles.