Exercise 0.9 from Dan Pedoe's Geometry A Comprehensive Course

The following problem was posed as Dan Pedoe's Exercise 0.9:

If (z-a)/(z-b) = r(cos(θ) + i sin(θ), show that the curves r = constant, and θ = constant, for fixed complex numbers a and b and varying Z, are circles of orthogonal coaxal systems, the one which has the points a and b as limiting points, and the other of circles which pass through these two points.

Since this problem occurred in an introductory chapter of the text, my expectation was that it was going to be a straightforward application of his quick review on complex numbers. It proved to be something more interesting.

In fact, establishing the part when r is constant is pretty straightforward once you recognize that the set of points whose distance to two fixed points is a constant ratio r is in fact a circle.

But the case for fixed θ proved the more interesting/complicated.

I recommend that anyone who has read this much should try to work it out before reading the solution linked below. I would love to see another approach to the solution if you find one.

For anyone who is interested, here is a link to a paper which provides what I hope will be a pretty self-contained solution intended for a reader who has studied senior high school level math. It includes a little background material for anyone who might be a little rusty.

Note: One point not explicitly mentioned in the paper is that when θ is held constant and r varies, θ presents itself as the angle between the perpendicular bisector of the segment ab and the line joining the center of the so-called θ-circle and the point a (or b). Thus the changing θ-circles can be visualized for θ in [0,π/2].

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.

The Circles of a Triangle

Let a circle with center C and radius ρ be denoted (C,ρ). For any triangle ΔABC, we can associate five circles: the incircle (I,r), the circumcircle (O,R),and the three excircles (I_a,r_a), (I_b,r_b),(I_c,r_c). In this blog we will define these circles and show how they are interrelated.

We follow the discussion in Coxeter 1.5 (Introduction to Geometry).

Let the sides of ΔABC have lengths AB = c, AC = b and BC = a. Denote the area of the triangle by Δ (or sometimes ΔABC). Let the semiperimeter of the triangle be defined as s = (a + b + c)/2 and let h be the altitude from B to the side AC. We will show that:

• Δ = [s(s-a)(s-b)(s-c)]^1/2
• h = [c² - [(b² + c² - a²)/2b]²]^1/2
• Δ = rs = abc/4R = r_a(s-a) = r_b(s-b) = r_c(s-c) where r and R are the radii of the incircle and the circumcircle respectively
• r = [(s - a)(s - b)(s - c)/s]^1/2 ; and r_a = [s(s - b)(s - c)/(s - a)]^1/2; etc which give the radii in terms of a,b,c
• 4R = r_a + r_b + r_c - r

Review of some basic geometric results

Two definitions:

• In a circle (C, ρ) a central angle is an angle formed by two radii and the measure of the angle is defined to be the measure of the corresponding intercepted arc. The measure of arc is just the length of the arc measured in units of ρ. Thus the complete circumference has measure 2π (length = 2πρ). Thus a central angle whose arc is a quarter of the circle has a measure of π/2 (regardless of the value of ρ). If central angle ∠ACB has measure x then the intercepted arc AB opposite the angle has measure x as well. In practice we usually drop the phrase "the measure of" in describing the measure of an angle or arc when the meaning is clear. In this vernacular, the sum of the measures of the angles of a triangle is π.
• We also say that a line is tangent to a circle, (C,ρ), with point of tangency, P, if the line is perpendicular to the radius CP at P. We will establish a few basic results needed in our discussion.

Now the proofs of some basic results

• An angle inscribed in a circle has a measure half that of its intercepted arc. Proof: Let ∠ABC be inscribed in circle (O,r) at point B and have measure x. (So B is on the circle.) Case I. Suppose one ray, say BC, contains the center, O. Then (the measure of) arc AC equals (the measure of) ∠AOC. Draw radius OA. Then OA = OB so ΔAOB is isosceles and ∠OAB = ∠OBA = x. ∠AOC = π - ∠AOB and ∠AOB = π - 2x. Thus ∠AOC = π - (π - 2x) = 2x. Case II. Suppose neither AB nor CB contain O and the diameter from B is outside ∠ABC. Draw the diameter containing B (and O) meeting the circle (O,r) at D. Let the measure of ∠CBD = y. Then using Case I, the measure of arc AD is 2(x+y) and the measure of arc CD is 2y, so the measure of arc AC is 2x. A similar argument holds for the case that the diameter from B is between AB and BC.
• Note: The result above holds if AB is tangent to the circle at B. Proof: Draw the diameter BD. The measure of arc BD is π. Let the measure of ∠CBD = y and ∠ABC = x. We know AB is perpendicular to BD so (x + y) = π/2, so 2(x + y) = π. Arc BD = arc BC + arc CD. We know that arc CD has measure 2y (by argument above), so arc BC has measure π - 2y = 2x.
• Given circle (O,r) and point A outside the circle, AF tangent at D (D between A and F), and ABC a straight line intersecting the circle first at B and then at C, the measure of ∠DAC is equal to half the difference in the measures of the intercepted arcs DC and DB. Proof: Draw chord DC. The exterior angle ∠CDF is the sum of the remote interior angles ∠DCA and ∠DAC. If x is the measure of ∠DAC, y is the measure of ∠DCA, and z is the measure of ∠CDF, then z = x + y, x = z - y = half arc DC - half arc DB. That is ∠DAC = (arc DC - arc DB)/2.

A look at the circles associated with ΔABC

• Incircle (I,r). Given ΔABC, suppose circle (I,r) is inscribed in the triangle such that sides AB, AC, and BC are all tangent to (I,r). Let the radii to the points of tangency be ID to BC, IE to AC and IF to AB, all of length r. Triangles ΔAIE and ΔAIF are right triangles (radii perpendicular to tangents) with two sides equal (hypotenuse AI in common and IE = IF). Thus they are congruent. This means AI bisects the angle at A. Similarly, BI and CI bisects the angles at B and C. This means the angle bisectors of a triangle meet at a point I which is the center of the incircle (I,r).
• Notice (using ΔABC = ΔAIB + ΔBIC + ΔAIC) that Δ = ar/2 + br/2 + cr/2 so Δ = rs
• Circumcircle (O,R). Given ΔABC, suppose circle (O,R) is circumscribed about the triangle touching only at the points A, B, and C. Let B' be the midpoint of side AC.The triangles ΔOCB' and ΔOAB' are congruent by SSS (OA = OC = R). Thus ∠OB'A and ∠OB'C are right angles. The same holds for the other two sides (by symmetry) so O is the intersection of the perpendicular bisectors of the sides of ΔABC and R is the distance from O to each vertex.
• The three excircles (I_a,r_a), (I_b,r_b),(I_c,r_c). We will discuss one excircle, (I_b,r_b), the others being defined similarly. Suppose in ΔABC we extend the sides AC, AB, BC into lines and consider the bisectors of the exterior angles formed at each vertex. Let EACF be the line containg AC. Let HI, and JK be the bisectors of angles ∠EAB and ∠BCF, respectively. Let HI and JK meet at I_b. Draw perpendiculars I_bU, I_bV, and I_bW to the lines containing AC, AB, and BC, respectively. Since the exterior angles are bisected, ΔUCI_b and ΔWCI_b are conguent right triangles with CI_b in common. So I_bW = I_bU. Similarly, I_bU = I_bV. So the three perpendicular segments are equal in length, call it r_b, so circle (I_b,r_b) is tangent to the three (extended) sides of ΔABC. We call this an excircle (or the B-excircle) of ΔABC.
Now draw BI_b intersecting AC at B'. Let AU = b_x (= AV), UC = b_y (= CW). Looking at the diagram, using the results just established for r_b, and considering the triangles ΔBI_bV and ΔBI_bW, we have:
• (1/2)r_bb_x + (1/2)r_bb_x + (1/2)r_bb_y + (1/2)r_bb_y + Δ = (1/2)r_b(c+b_x) + (1/2)r_b(a+b_y )
• r_bb_x + r_bb_y +Δ = (1/2)r_b(a+b+c)
• Δ = r_bs - r_bb = r_b(s-b)
• Similarly, Δ = r_a(s-a) and Δ = r_c(s-c)
• Combining this with the result from the incircle, we have Δ = rs = r_a(s-a) = r_b(s-b) = r_c(s-c).
• Now consider the altitude BD of length h from B to AC in ΔABC. We know that Δ = bh/2. We will get an expression for h in terms of the lengths of the sides of the triangle. Let AD = x, so DC = b-x. The Pythagorean theorem gives us
  • c² = x² + h² and a² = (b-x)² + h².
  • Eliminating the h² and x² terms we get x = (b² + c² - a²)/2b
  • Notice that x/c = cos A = (b² + c² - a²)/2bc which is the Law of Cosines.
  • Then h² = c² - x² = c² - [(b² + c² - a²)/2b]² which gives the altitude in terms of the lengths of the sides.
  • Δ = hb/2 = (1/4)[4b²c² - (b² + c² - a²)²]^1/2
  • = (1/4)[-a⁴ - b⁴ - c⁴ + 2a²b² + 2a²c² + 2b²c²]^1/2
  • = (1/4)[(a + b + c)(-a + b + c)(a - b + c)(a + b - c)]^1/2
  • = [s(s - a)(s - b)(s - c)]^1/2
  • That is Δ = √[s(s - a)(s - b)(s - c)] which gives the area of a triangle in terms of the lengths of the sides.
• Combining the results of the last two sections, r² = Δ²/s² = (s - a)(s - b)(s - c)/s which gives the radius of the incircle in terms of a, b, and c and (r_a)² = Δ²/(s - a)² = s(s - b)(s - c)/(s - a), etc which gives the radii of the excircles in terms of a, b, and c.
• Finally, in circumcircle (O,R) of ΔABC we have (from the discussions above) the central angle ∠AOC equal to twice the inscribed angle ∠ABC (or simply ∠B) and ΔAOB' congruent to ΔCOB' where B' is the midpoint of AC. Since OA = OC = R and AB' = B'C = b/2, and ∠AOB' = ∠COB' = ∠B. Thus sin B = (b/2)/R = b/2R. Thus 2R = b/sin B. Since the argument is symmetric for any side we have 2R = b/sin B = a/sin A = c/sin C, which is the Law of Sines. Now draw altitude BD, diameter BK, and chord KC. Since ∠BCK intercepts a semicircle its measure is π/2, that is, it is a right angle. Since ∠BKC and ∠BAD intercept arc AC, they are equal angles. Thus the two right triangles ΔBAD and ΔBKC are similar. So BD/BA = BC/BK, that is,
  • BD = ac/2R
  • Thus Δ = (1/2)bac/2R = abc/4R
  • But abc = s(s-b)(s-c) + s(s-c)(s-a) + s(s-a)(s-b) - (s-a)(s-b)(s-c)
  • = Δ²/(s-a) + Δ²/(s-b) + Δ²/(s-c) - Δ²/s
  • = Δ(r_a + r_b + r_c - r)
  • Therefore 4R = r_a + r_b + r_c - r

Drawing the circles of a triangle

In this section we will consider the triangle ΔABC where A is the point (0,0), C is the point (b,0), b > 0, and B is the point (c₂,c₂), C₁ > 0 and C₂ > 0. Let m₁ = c₂/c₁ and m₂ = c₂/(c₁-b). We will construct the 5 circles for this triangle. Here are links to two such diagrams: A(0,0)C(8,0)B(8,8) and A(0,0)C(10,0)B(5,5) which were constructed using the Desmos graphing tool at desmos.com.

• The sides of ΔABC
  • Line AB is y = m₁ x
  • Line AC is y = 0
  • Line BC is y = m₂(x-b) if b ≠ c₁ and x = b if b = c₁
• The exterior bisectors of ΔABC
  • Exterior bisector at ∠A is y = tan(-α)x where α = (1/2)(π - tan^-1m₁)
  • Exterior bisector at ∠C is y = tan(β)(x-b) where β = (1/2)[π + tan^-1m₂] if c₁ < b, β = (1/2)[tan^-1m₁] if c₁ > b, and β = π/4 if c₁ = b.
  • Exterior bisector at ∠B is y = kx + (c₂ - kc₁) + (m₁c₁ - c₂)/(cosθ - m₁sinθ) where k = (m₁cosθ + sinθ)/(cosθ - m₁sinθ), θ = - γ and γ = (1/2)(tan^-1m₁ - tan^-1m₂) if c₁ < b, γ = (1/2)(π + tan^-1m₁ - tan^-1m₂) if c₁ > b, and γ = (1/2)(π + tan^-1m₁ ) if c₁ = b.
  Note: To get the angle γ we just analyze the three cases at B. Then, to get the equation of the bisector at B we rotate the line through AB around the point B by the angle θ = - γ. Thus we apply the general result that if the line y = mx (recall A is taken to be the origin so the y-intercept of line AB is zero) containing the point P = (p,q) is rotated by the angle θ about P, then the equation of the rotated line is y = kx + (q - kp) + (mp - q)/(cosθ - msinθ) where k = (mcosθ + sinθ)/(cosθ - msinθ). The details of this are as follows:
  • y= mx, P = (p,q) is a point on the line and we want the equation of the line obtained by rotating the line by θ around P
  • Translate P to the origin: X = x - p, Y = y - q. Thus x = X + p, y = Y + q. The new line is Y + q = m(X + p), so Y = mX + (mp - q)
  • Rotate by θ about the origin: X = cosθX' + sinθY' and Y = -sinθX' + cosθY' so we get Y' = kX' + (mp-q)/(cosθ - msinθ) where k = (mcosθ + sinθ)/(cosθ - msinθ).
  • Now translate back to P: X'' = X' + p, Y'' = Y' + q. Thus X' = X'' - p and Y' = Y'' - q. We substitute again to get Y'' = kX'' + (q - kp) + (mp-q)/(cosθ - msinθ)
  • Note: the denominator of k is not zero because that would imply the bisector was vertical which cannot happen the way we have defined ΔABC
• The Centers of the excircles [Let c₃ = (c₂ - kc₁) + (m₁c₁ - c₂)/(cosθ - m₁sinθ)]
  • I_b = (x₁,y₁) is the intersection of bisectors at A and C: -tanα x = tanβ x - btanβ so x₁ = btanβ/(tanα + tanβ), y₁ = -tanα x₁
  • I_c = (x₂,y₂) is the intersection of bisectors at A and B: -tanα x = kx + c₃; so x₂ = -c₃/(k + tanα), y₂ = -tanα x₂
  • I_a = (x₃,y₃) is the intersection of bisectors at B and C: tanβ(x - b) = kx + c₃ so x₃ = (c₃ + btanβ)/(tanβ - k), y₃ = tanβ(x₃ - b)
• The radii of the excircles
  In our setup line AC is the x-axis (y=0) and since all excircles are tangent to AC the radii can be obtained as |y_i| from the centers (x_i,y_i). Thus (I_a,r_a) = ((x₃,y₃), |y₃|), (I_c,r_c) = ((x₂,y₂), |y₂|), (I_b,r_b) = ((x₁,y₁), |y₁|)
Note: In general the distance from the line y = mx + b to the point (p,q) is given by d = |mp - q + b|/[1 + m²]^1/2 unless the line is vertical, x = b, in which case d = |p-b|. The formula is obtained by finding the intersection point Q of y = mx + b and the line perpendicular to this through P, y - q = -(x - p)/m, and then finding the distance between P and Q. This could be used to find the radii to the other (extended) sides of the triangle.
• The Incircle (I, r)
  We need to get the intersection of two (interior) angle bisectors. Let a₁ = (1/2)tan^-1m₁, so y = tan(a₁)x is the bisector at A. Let a₂ = (1/2)tan^-1(m₂) if C₁ < b, a₂ = (1/2)(π - tan^-1(m₂)) if C₁ > b, and a₂ = -π/4 if c₁ = b. Then y = tan(a₂)(x-b) is the bisector at C. Then tan(a₁)x = tan(a₂)(x-b) gives x₀ = btana₂/(tana₂ - tana₁), y₀ = x₀tana₁, and r₀ = |y₀|. (I,r) = ( (x₀, y₀), r₀).
• The Circumcircle (O, R)
  We need to get the intersection of two perpendicular bisectors. The perpendicular bisector of AB is y - c₂/2 = -(x - c₁/2)/m₁ which intersects perpendicular bisector of AC at x = b/2. Thus we take x₄ = b/2, y₄ = c₂/2 -(b/2 - c₁/2)/m₁, and r₄ = [x₄² + y₄²]^1/2. (O, R) = ( (x₄, y₄), r₄).

Mutually Tangent Circles

This article will examine circles that are mutually tangent to each other with distinct points of tangency. We build up from two, to three, then to four mutually tangent circles. This will prepare the way for the next article on Descartes' Circle Theorem.

Let R denote the set of real numbers, R⁺ the non-negative real numbers, RxR = R² = {X=(x,y): x∈ R,y ∈R} the real plane. That is R² is the set of all ordered pairs of real numbers.

A relation in R² is any subset of R². Thus a line, {X:y=ax+b}, and a circle, {X: XA=r}, are relations in R², the latter having radius r and center A.

Two circles are tangent to each other (mutually tangent) if the relations defining them have exactly one point in common.

Two Tangent Circles

Given circle E_A = {X:XA = r_A}, find circle E_C = {X:XC = r_C} tangent to E_A. There are two cases. Case A: E_A surrounds E_C ; Case B: E_A and E_C are externally tangent to each other.

• Case A: Let C ∈ {X: XA < r_A}, i.e. C is in the interior of E_A. Then E_C = {X: XC = r_A-AC} is tangent to E_A. Proof: First we show they have a point in common. Let the radial segment from A through C hit E_A at P. Then P ∈ E_A and r_A =AP=AC+CP. So CP=r_A-AC and P ∈ E_C. Now we show P is unique. Suppose there is another point Q ≠ P such that Q ∈ E_A and E_C. Then AQP form a triangle. AQ = r_A and CQ = r_A - AC. Thus CQ = AQ-AC. But in triangle ΔACQ we must have CQ + AC > AQ. Thus there is no such Q. That is P is the unique intersection of the two circles. Note: In particular, the line joining the centers A and C includes the point of tangency, P.
• Case B: Let C ∈ {X: XA > r_A}, i.e. C is in the exterior of E_A. Then E_C = {X: XC = AC-r_A} is tangent to E_A. Proof: Similar to case A. Draw the segment AC and let it intersect E_A at P. Then PC=AC-r_A, so P is also in E_A. Again suppose there is another point Q ≠ P such that Q ∈ E_A and E_C. If Q is collinear with A,P, and C then QC = r_A + AC so Q ∉ E_C. In ΔAQC, AQ + QC > AC so QC > AC - AQ = AC - r_A, so Q ∉ E_C. Thus P is the unique point of intersection of the two circles so they are mutually tangent.

Three Mutually Tangent Circles with Three Distinct Points of Tangency

For each of the cases of two tangent circles we now look at how to include a third circle which is mutually tangent to the two at three distinct points of tangency. In Case A, in order for the points of tangency to be distinct, the new circle E_B must be surrounded by E_A. In Case B, E_B might surround E_A and E_C or it might be externally tangent, to both of them.

• Case A: We can find the third tangent circle in two ways. Case A1: Choose any point B ∈ {X:XA < r_A} ∩ {X:XC > r_C} as the center of E_B and then find r_B. Case A2: Choose any 0 < r_B < r_A-r_C, the range of possible radius measurements for circles with centers inside E_A and outside E_C, and then find the center B.
• Case B: We have two cases and each can be done in two ways (as with Case A). Case B1: E_B surrounds E_A and E_C. Case B2: E_A, E_B, and E_C are externally tangent to one another. We divide these into four cases. Case B11: Choose B in the exterior of both E_A and E_C then find r_B. Case B12: Choose a radius 0 < r_B and find a center B. Case B21: Choose B in the exterior of both E_A and E_C then find r_B. Case B22: Choose a radius 0 < r_B and find a center B.

First we examine cases A1, B11, and B21 where we are given A, r_A, C, r_C and B and we need to find r_B. Note, we will also get expressions for r_A and r_C which will be useful throughout. In each case we consider ΔABC with sides of length a=BC, b=AC, and c=AB and semi-perimeter s = (a + b + c)/2.

• Case B11: We can write r_B = c + r_A = a + r_C and b = r_A + r_C. Thus a+ B + c = 2s = 2r_B. Thus
  • r_B = s
  • r_A = s-c
  • r_C = s-a.
• Case B21: Using the fact that the segment joining the centers of two tangent circles includes the point of tangency (from the analysis of two tangent circles), we have a = r_B + r_C, b = r_A + r_C, and c = r_A + r_B. Thus (a + b + c) = 2s = 2(r_A + r_B + r_C). Therefore,
  • r_B = s - (r_A + r_C) = s-b
  • r_A = s-a
  • r_C = s-c.
• Here is a picture of the two constructions using A(0,0), C(3,0) and B(1,1). The black trio is case B11 and the orange trio is case B21.
• Case A1: Choose B in the interior of E_A and the exterior of E_C. From Case B11 above we know that r_B = s-c.

Next we will look at the cases A2, B21, and B22 where we are given A, r_A, C, r_C and r_B and we need to find B. In each case we will set up a system of two quadratic equations parameterized with a linear function of r_B which we can solve for B.

• Case A2: B must satisfy BA = r_A - r_B and BC = r_B + r_C. Notice that E_B is at its maximum radius when B is collinear with A and C. In this case 2r_B + 2r_C = 2r_A. Thus the maximum value for r_B is r_A - r_C.
• Case B12: E_B surrounds. B must satisfy BA = r_A - r_B and BC = r_B - r_C.
• Case B22: E_A, E_B, E_C externally tangent. B must satisfy BA = r_A + r_B and BC = r_B + r_C.

Since A and C are arbitrary, we assume A is the origin of a rectilinear coordinate system, so A = (0,0). Also without loss of generality we can take C = (b,0) so that AC = b in ΔABC. Then our system, in general, becomes

• x² + y² = ƒ₁²(r_B)
• (x-b)² + y² = ƒ₂²(r_B)

Solve in terms of parameter r_B by subtracting the second equation from the first to get 2bx - b² = ƒ₁² - ƒ₂² or,

x = (b² + ƒ₁² - ƒ₂²)/2b and y = ±√[ƒ₁² - x²]. We get the solutions for our three cases by specifying b, ƒ₁ and ƒ₁for each.

• Case A2: b = (r_A - r_C); ƒ₁(r_B) = r_A - r_B; ƒ₂(r_B) = r_C + r_B, 0 < r_B < r_A - r_C.
Case B12: b = (r_A - r_C); ƒ₁(r_B) = r_A - r_B; ƒ₂(r_B) = r_B - r_C.
• Case B22: b = (r_A - r_C); ƒ₁(r_B) = r_A + r_B; ƒ₂(r_B) = r_C + r_B.

In each case we get two solutions for B, one above the line through A and C and one below. We must have r_A ≠ r_B to get a solution for B.

Four Mutually Tangent Circles with Six Distinct Points of Tangency

From the last section, we have two distinct cases of three mutually tangent circles with distinct points of tangency to extend to a fourth circle. Case A: One of the three surrounds the other two. Case B: the three circles are externally tangent to each other. In these cases, to add a 4th tangent circle, we will get systems of three quadratic equations in three unknowns, the coordinates of the center D and the radius d of the fourth tangent circle E_D. As in the case of three circles, the difference in the systems are the linear functions of d in those systems. We will call A = (0,0), C(b₁,0), and B(c₁,c₂). The sides of ΔABC are then a = √[(c₁-b₁)² + c₂²], b = b₁, and c = √[c₁² + c₂²]. Again let s = (a + b + c)/2.

• Case A:

  In this case E_A surrounds the other three circles and there are two placements for E_D; one on the same side of the line through B and C as A and one on the far side of this line from A.

  • AD = ƒ₁(d) = r_A - d
  • BD = ƒ₂(d) = r_B + d
  • CD = ƒ₃(d) = r_C + d
• Case B:

  In the case that E_D surrounds E_A, E_B, and E_C, then AD = ƒ₁(d) = d - r_A; BD = ƒ₂(d) = d - r_B; CD = ƒ₃(d) = d - r_C, where d is the radius of E_D. In particular, AD = ƒ₁(d) = | d - r_A|; BD = ƒ₂(d) = | d - r_B|; CD = ƒ₃(d) = | d - r_C| > 0, where d is greater than r_A, r_B, and r_C and d is the radius of E_D.

  In the case that E_D is externally tangent to E_A, E_B, and E_C, then AD = ƒ₁(d) = r_A + d; BD = ƒ₂(d) = r_B + d; CD = ƒ₃(d) = r_C + d, where d is the radius of E_D. We can rewrite these equations as AD = ƒ₁(d) = r_A - (-d); BD = ƒ₂(d) = r_B - (-d); CD = ƒ₃(d) = r_C - (-d) and again as AD = ƒ₁(d) = |(-d) - r_A|; BD = ƒ₂(d) = |(-d) - r_B|; CD = ƒ₃(d) = |(-d) - r_C|, where (-d) < 0.

  Thus in general we have:

  • AD = ƒ₁(d) = | d - r_A|
  • BD = ƒ₂(d) = | d - r_B|
  • CD = ƒ₃(d) = | d - r_C|

  where this d ∈ R-{0}. We will see below that in solving for d we get a quadratic in d which yields solutions D₁, D₂ such that if D_i < 0 then | D_i| is the radius of an externally tangent E_D and if D_i > 0 then D_i is the radius for E_D that surrounds the other three circles.

Now consider the general solution (for both cases A and B) involving the general ƒ₁(d), ƒ₂(d), ƒ₃(d), and then look at the differences when the actual values of these functions are inserted. For ease of notation we will use f = ƒ₁, g = ƒ₂, h = ƒ₃(d). The general problem is to solve the system:

• E_A(d): x² + y² = f²(d)
• E_B(d): (x-c₁)² + (y-c₂)² = g²(d)
• E_C(d): (x-b₁)² + y² = h²(d)

We solve the system in three steps.

• Find the line of intersection for E_A(d) and E_C(d), L_AC(d), as a function of d.
• Find the line of intersection for E_A(d) and E_B(d), L_AB(d), as a function of d.
• Find the points where the lines of intersection intersect our circles of interest, say E_A(d).
• The lines of intersection represent the points where the distances to E_A(d) and E_C(d) or E_A(d) and E_B(d) are the same for a given d. We want to find d where all three distances are the same for a given d. That is, we want to find the value(s) of d where the lines L_AC(d) and L_AB(d) intersect. These will give us center points for E_D with radius d.

Find L_AC(d):

• x² + y² - f²(d) = (x-b₁)² + y² - h²(d)
• x = (b₁² + f² - h²) / 2b₁

Find L_AB(d):

• x² + y² - f²(d) = (x-c₁)² + (y-c₂)² - h²(d)
• 2c₁x + 2c₂y = c₁² + c₂² + f² - g² = c² + f² - g².
• Letting u = (c² + f² - g²) / 2c₂, we have y = (-c₁/c₂)x + u.

Find the intersections with E_A(d).

• L_AC(d) meets E_A(d) at x₁(d) = (b₁² + f² - h²) / 2b₁ and y₁(d) = ±√[f² - x₁²(d)].
• L_AB(d) meets E_A(d) when x² + ((-c₁/c₂)x + u)² = f².
• This leads to the quadratic equation (c/c₂)²x² - (2c₁u/c₂)x + u² -f² = 0
• Use the quadratic formula to get two solutions for x:
• x₂(d) = (c₁c₂u + c₂√[(cf)² - (d₁u)²]) / c²
• x₃(d) = (c₁c₂u - c₂√[(cf)² - (d₁u)²]) / c²
• and the corresponding y terms y₂(d) = (-c₁/c₂)x₂ + u and y₃(d) = (-c₁/c₂)x₃ + u

To find point(s) where all three equations of the system are satisfied we have to find a value(s) for d such that x₁(d) = x₂(d) or x₁(d) = x₃(d). That is where

• (b₁² + f² - h²) / 2b₁ = (c₁c₂u ± c₂√[(cf)² - (d₁u)²]) / c²
• We isolate the radical, square both sides, substitute for u, and then collect terms with common differences of squares of functions f, g, and h to get:
• (f² - h²)²[c/b₁]² + (f² - g²)² - (f² - h²)(f² - g²)[2c₁/b₁] + (f² - h²)[2c²(1-c₁/b₁)] + (f² - g²)[2(c² - c₁b₁)] - 4c₂²f² + c²[b₁² + c² - 2c₁b₁] = 0

Notice that f, g, and h only show up as squares so the absolute value does not come into play when we make the substitutions. By using the values from the beginning of this section we can compute:

• Case A: f² - g² = w(2d-c) and f² - h² = v(2d-b) where w = -(a + b) and v = -(a + c). Also, f² = (s-d)².
• Case B: f² - g² = w(2d-c) and f² - h² = v(2d-b) where w = (a - b) and v = (a - c). Also, f² = (d-(s-a))².

We substitute these values into the equation in the last paragraph and collect terms with the same degree of d to get a quadratic in d: αd² + βd + κ = 0, where the coefficients for each case are as follows. Recall, b = b₁.

• Case A: w = -(a + b) and v = -(a + c)
  • α = 4[(cv/b)² - 2c₁vw/b + w² - c₂²]
  • β = 4[-(cv)²/b + c₁vw(b+c)/b - cw² + vc²(1-c₁/b) + w(c² - bc₁) + 2sc₂²]
  • κ = (cv)² - 2c₁cvw + (cw)² - 2bvc²(1-c₁/b) - 2cw(c² - bc₁) - 4s²c₂² + c²[b² + c² - 2c₁b]
• Case B: w = (a - b) and v = (a - c)
  • α = 4[(cv/b)² - 2c₁vw/b + w² - c₂²]
  • β = 4[-(cv)²/b + c₁vw(b+c)/b - cw² + vc²(1-c₁/b) + w(c² - bc₁) + 2(s-a)c₂²]
  • κ = (cv)² - 2c₁cvw + (cw)² - 2bvc²(1-c₁/b) - 2cw(c² - bc₁) - 4(s-a)²c₂² + c²[b² + c² - 2c₁b]

Notice that besides the values for v and w, the only difference in these cases is the term "s" in Case A is replaced by "s-a" in case B in the β and κ terms. We solve for d = (-β ±√[β² - 4ακ])/2α. Let D₁ be the root with positive radical and let D₂ be the root with the negative radical. We interpret the roots as follows.

• Case A: In this case the roots are positive and each gives the radius of one of the two possible 4th tangent circles.
• Case B: If a root is negative, its absolute value is the radius of an externally tangent fourth circle. If a root is positive it is the radius of fourth tangent circle that surrounds the original three.
• In either case, for i = 1, 2 and j = 2, 3: If x₁(D_i) = x_j(D_i) then (x - x_j(D_i))² + (y - y_j(D_i))² = (D_i)² is a fourth tangent circle to the original three.

!--

Examples of four mutually tangent circles

The final part of this article is to show some examples of four tangent circles for different values of b₁, c₁, and c₂. Given the ΔABC we can have the three mutually tangent circles E_A, E_B, E_C with either radii s, s-c, s-b (E_A surrounds, case A) or s-a, s-b, s-c (externally tangent, Case B) respectively. We will label our examples below with the notation: [Case,0,0,c₁,c₂,b₁,0] with Case = A or B followed by coordinates of centers A, B, C. For example B,0,0,1,1,1,0 means we are looking at three mutually tangent circles with centers at A(0,0), B(1,1) and C(1,0). In these examples we have included the lines of intersection L_AB(d) and L_AC(d) for each of the values d = D₁ and d = D₂. Also, we use the subscript notation D_1iji on the D-radii to mean the radius obtained when x₁(D_i) = x_j(D_i) for i = 1,2; j = 2,3. The values are given to two decimal places only. Note: Case A 4th tangent circles are shaded purple and Case B circles are shaded red.

• A,0,0,1,1,1,0

  D₁₂₂₂ = 0.14 D(1.37,0.77)
  D₁₁₃₁ = 0.71 D(0,1)

• B,0,0,1,1,1,0

  D₁₂₃₂ = 1.71 D(0,1)
  D₁₁₂₁ = -0.08 D(0.74,0.26)

• Notice the dramatic difference with the large surrounding circle at C(2.8,0) and the externally tangent circle at C(2.9,0)
  B,0,0,1,1,2.8,0

  D₁₂₂₂ = 479.71 D(111.57,-465.45)
  D₁₁₂₁ = -0.11 D(1.05,0.55)

• B,0,0,1,1,2.9,0

  D₁₂₃₂ = -24.00 D(-4.98,24.59)
  D₁₁₂₁ = -0.11 D(1.06,0.56)

• A,0,0,1,1,,4.3,0

  D₁₂₂₂ = 0.21 D(4.34,0.49)
  D₁₁₂₁ = 0.38 D(4.15,-0.65)

• B,0,0,1,1,,4.3,0

  D₁₂₃₂ = -2.03 D(0.17,3.16)
  D₁₁₂₁ = -0.11 D(1.08,0.62)

Lebesgue Integral - Homework Problem 3 April 1975 APM404

This was a favorite homework problem I kept around for a long time in a folder so I thought I'd include it here for posterity.

Let (X,M,μ) = ([a,b], Lebesgue sets, Lebesgue measure) and α(x) = x ∀ x ∈[a,b]. Show that (i) ƒ is μ measurable if and only if ƒ ∈ H₂^α and (ii) if ∫ ƒ dμ < ∞ ( or ∫ ƒ dα < ∞) then ∫ ƒ dμ = ∫ ƒ dα

1. Let ƒ ∈ H₂^α. Then there exist continuous (and therefore μ measurable) functions ƒ_n which converge to ƒ ae (almost everywhere). Since μ was obtained from an outer measure, (X,M,μ) is a complete space and ƒ is μ measurable.[(i)

2. Assume E ⊂ [a,b], E ∈ M. Then there exists a measurable set C = ∪_n=1,∞ C_n, where the C_n are closed, C ⊂ E and μ(E ∩ C^∼) = 0.

• Let B_m = ∪_n=1,m C_n. Then B_m is closed (a finite union of closed sets) and ∪_m=1,∞ B_m = C
• Thus the characteristic function X_Bm ∈ H₂^α (since B_m complement - B_m^∼ - is open; a result established on APM403 exam) and X_Bm = X_{∪n=1,m Cn} ≥ X_{∪n=1,m-1 Cn} = X_Bm-1
• Also, |X_Bm| ≤ X_[a,b] ∈ L₁ and ∫ X_Bm dα < ∫ X_[a,b] dα = b-a and X_Bm ↑ X_E ae, since μ(E ∩ C^∼) = μ(E ∩ (∪_m=1,∞ B_m )^∼) = 0 so by the Lebesgue Dominated Convergence Theorem, X_E ∈ L₁
• Finally, μ(E) ≡ ∫ X_Eμ(dx) = ∫ X_E dx ≡ α(E) by our choice of μ and α. Thus, if E ∈ M, E ⊂ [a,b] is compact, then μ(E) = α(E) ≤ b-a.

3. Let ƒ be simple, say ƒ(x) = ∑_i=1,n α_iX_Ei(x) where E_i ∩ E_j = ∅ for i ≠ j and [a,b] ⊃ E_i ∈ M, ∀_i.

• Since H₂^α is a vector space, ƒ ∈ H₂^α and
  ∫ ƒ dμ = ∑_i=1,n α_iμ(E_i) = ∑_i=1,n α_iα(E_i) = ∫ ƒ dα
• The integrals make sense since ∀_i, α_i < ∞ and μ(E_i) = α(E_i) ≤ b-a.

4. Let ƒ be bounded and measurable (non-negative), say ƒ ≤ N, ∀ x ∈ [a,b]. There exist simple functions φ_n ↑ ƒ ae (φ_n ∈ H₂^α, ∀ n by (3) above) and

• ∫ φ_n dα = ∫ φ_n dμ ≤ ∫ ƒ dμ ≤ N(b-a)
• so φ_n ∈ L₁, φ_n ↑ ƒ ae, |φ_n| ≤ N, ∫_a,bN dx < ∞ implies f ∈ L₁ and ∫ φ_n converge to ∫ ƒ
• Also, by the Monotone Convergence Theorem, ∫ φ_n dμ → ∫ ƒ dμ
• Therefore, ∫ ƒ dα = lim ∫ φ_n dα = lim ∫ φ_n dμ = ∫ ƒ dμ

5. Let ƒ be measurable, non-negative and consider g_N = ƒ ∧ N as N → ∞

• g_N ∈ H₂^α (by (4)) and g_N ↑ ƒ and ∫ g_N dα = ∫ g_N dμ ≤ ∫ ƒ dμ
• If ∫ ƒ dμ < ∞, then ƒ ∈ H₂^α and ∫ ƒ dμ = lim ∫ g_N dμ = lim ∫ g_N dα = ∫ ƒ dα

6. Let ƒ be measurable (μ). ∫ ƒ dμ exists if and only if ∫ ƒ⁺ dμ < ∞ and ∫ ƒ^- dμ < ∞

• From (5), ∫ ƒ⁺ dμ < ∞ and ∫ ƒ^- dμ < ∞ so ƒ⁺ and ƒ^- are in H₂^α and
• ∫ ƒ dμ = ∫ ƒ⁺ dμ - ∫ ƒ^- dμ = ∫ ƒ⁺ dα - ∫ ƒ^- dα = ∫ ƒ dα

Thus, ƒ ∈ H₂^α implies ƒ is measurable (μ) and if ƒ is μ-integrable then ƒ ∈ L₁^α and ∫ ƒ dμ = ∫ ƒ dx.

Corollary: Let ƒ ∈ H₂^α and c ∈ R^*. If E = {x ∈ [a,b]: ƒ(x) > c} then X_E ∈ H₂^α.

1. If c = ∞ then E = ∅ and X_∅ = 0 ∈ H₂^α
2. If c = -∞ then E = [a,b] so X_E ∈ H₂^α [X_E = | 1 - X_E^∼| ∈ H₂^α ]
3. Assume c ∈ R. ƒ ∈ H₂^α ⇒ ƒ is μ-measurable (Lebesgue) ⇔ E is μ-measurable (Lebesgue) and by step (2) of the problem, X_E ∈ H₂^α