Leon Bankoff's Surprises

    The late Leon Bankoff described (Alexanderson [1]) some elementary properties of the arbelos (also known as the Shoemaker's Knife) as his "surprises". We shall reorganize them into theorems, with proofs. The proofs are at an advanced European high school level, and involving some mundane algebraic simplifications as exercises. Later on I shall redo some of the proofs using circle inversions, which will involve much less algebra and have more geometric insight and beauty. That should attract you to come to Biola.

    Let C1, C2, C3 be 3 semicircular arcs with diameters AC, AB, BC, on the same side of a straight line ABC. Let their radii be 1, a, b. WLOG (Without loss of generality), assume a > b. Then a + b = 1, so that we shall assume:
        (a - b)² = 1 - 4 a b,
        1 - 2 b = a - b,
        a² + b² = 1 - 2 a b,
        1 - a b - a² = b, etc.
     Let BD be a line perpendicular to AC, cutting C1 at D. The arbelos is the figure bounded by the arcs C1, C2, C3. We shall denote the center and radius of any circle Cn by O_n and r_n. Note also that O₁O₂ = 1 - a = b, O₁B = 2 a - 1 = | a - b |.

    Theorem 1 Let M'MNN' be the chord in the arbelos touching C2, C3 at M, N. Then BMDN is a rectangle, whose circumcircle C9 will have the same area as the arbelos. (Bankoff's Surprises 1, 2, 3)
    Proof: Let BD cut MN at K, then from property of tangents to circles, KM = KB = KN. Let C9 be the circle on MN as diameter, then it goes through B, hence ÐAMB = ÐMBN = ÐBNC = 90° Therefore the lines AM and CN when extended will meet at right angles at some point D', so that MBND' is a rectangle. Since K is the midpoint of MN, it lies on BD', so that D' lies on line BD. Since ÐAD'C = 90°, D' must coincide with D. (Note the sly proof.)
     Area of the arbelos = (½)p 1² - (½)p a² - (½)p b² = (½)p (1 - (1 - 2 a b)) = p a b.
     On the other hand, BD² = AB . BC = 2 a . 2 b, so that the radius of C9 is Ö (a b), and then C9 has area p a b, equal to that of the arbelos. QED

    Lemma 1. Let AD be an altitude of any triangle ABC, then AB² - AC² = DB² - DC² .
    Proof: AB² - AC² = (AD² + DB²) - (QD² + DC²). QED

    Theorem 2. (Archimedes' twin circles) Let C5 be the circle touching C1, C2, and BD. Let C6 be the circle touching C1, C3, and BD. Then C5 and C6 have the same radius, namely, a b. (Bankoff's Surprise 4). The minimum sized circle C13 containing and touching C5 and C6 have the same size as C9. (Bankoff's Surprise 5)
    Proof: Let O₅X be the perpendicular from O₅ to line ABC. Let r be the radius of C5.
Then O₅O₂² - O₅O₁² = XO₂² - XO₁² gives
(a + r)² - (1 - r)² = (a - r)² - (a - b - r)²
\ (a + r)² - (a - r)² = (a + b - r)² - (a - b - r)²
Factorizeing each side using the formula p² - q² = (p - q) (p + q),
\ 4 a r = 4 (a - r) b. \ a r + b r = a b, i.e., r = a b.
     Similarly, we can prove C6 has radius = a b = that of C5.
     Furthermore, O₅X² = O₅O₂² - O₂X² = (a + a b)² - (a - a b)² = 4 a²b,
so that O₅ is at 2 aÖb distance from line AC.
Similarly, O₆ is at 2 bÖa distance from line AC, so
O₅O₆² = (2 a b)² + (2 aÖb - 2 bÖa )² = (Exercise!) . . .
= 4 a b (1 - Ö(a b))² \ O₅O₆ = 2Ö(a b) - 2 a b,
\ Diameter of C13 = this + 2 a b = 2Ö(a b) = Ö(AB . BC) = BD = diameter of C9. QED

    Lemma 2. (See-Saw Lemma) Let AOC be the diameter of a semicircle, center O. Let DEF be any line tangent to the semicircle at a point E, and cutting the tangents at A and C at D and F respectively. Let EB be the perpendicular from E to AC. Then BE, AF, CD are concurrent, EB bisects ÐDBF, and EB and OA are respectively the H.M. and G.M. (harmonic mean and geometric mean) of AD and CF. It also proves that G.M. of any two positive numbers is no less than the H.M.
    Proof: Let h = AD = DE, k = CF = EF, r = AO. Then OD ^ OF because they bisects ÐAOE and ÐCOE. so DADO is similar to DCOF, so that h / r = AD / OA = OC / CF = r / k. Hence h k = r² and r is the G.M. of h, k.
     Let AF cut CD at G, then DADG is similar to DFCG, so that AG / GF = AD / FC = h / k = DE / EF = AB / BC. Hence G lies on BE. BE = BG + GE = CF . h / (h + k) + AD . k / (h + k) = 2 h k / (h + k) = the H.M of h, k. Obviously BE <= r. Since AB / BC = DE / EF = h / k, DBAD is similar to DBCF. \ ÐABD = Ð CBF, and BE bisects ÐDBF. QED

    Corollary. The largest circle C8 contained in the segment between the arc C1 and the chord M'N' of C1 touching C2 and C3, has radius a b. (See Diagram 2)
    Proof: Since O₂ and O₃ are at distances a, b from M'N', the distance of O₁ from M'N' is (a O₁O₃ + b O₂O₁) / (O₂O₃) = a² + b² = 1 - 2 a b. \ the diameter of C8 is 2 a b. QED

    Lemma 3. (Radical Center) Let 3 circles touch one another at 3 non-collinear points, B,P,Q. Then the common tangents at P, Q, B are concurrent. That point is called the radical center of the 3 circles.
    Proof: Let the tangents at P, Q meet at T. Let the tangents at P, B meet at T'. Since the tangents at P, Q, B are not parallel, if T is different from T', then BT' must cut TQ at a point T'' different from T or T'. WLOG, assume T is between P and T', then since ÐPTQ and ÐPT'B < 180°, T'' is between T and Q. But then BT'' > BT' = PT' > PT = TQ > T''Q, which is absurd. \ T, T' T'' are the same point. QED (I am still looking for a better proof.)

    Theorem 3. Let a circle C4 contained in the arbelos touch C1, C2, C3 at R, P, Q respectively. Then the circumcircle C7 of P,Q, R has the same radius as C4 and C5, and touches line AC. (Bankoff's Surprise 9). The distance between C4 and the line AC is the same as the radius of C4. (Bankoff's Surprise 7). Derive formulas for its radius r₄, and the coordinates of its center O₄ and P, Q, R.
    Proof: We choose B as origin, BC and BD as positive x- and y- axes. Then A is (-2 a, 0), C is (2 b, 0). By Lemma 3,
tangents at P, Q, B concur at some point T,
tangents at P, R, A concur at some point U, and
tangents at R, Q, C concur at some point V. Let m = AU, n = BT, p = CV. Then by the See-Saw Lemma,
m n = a², n p = b², and m p = 1². Multiplying, we get
n² = (m n) (n p) = a²b², so that n = a b. Hence
m = a² / n = a / b, p = b² / n = b / a. Also the circle C7 with T as center and a b as radius will pass through B, P, Q. It touches AC because BT ^ AC.
    Let (x_R, y_R) be the coords of R, then y_R = H.M.(m,p)
= 2 (a/b) (b/a) / (a/b + b/a) = 2 a b / (1 - 2 a b). x_R =
(2 b m + (- 2 a) p)/(m + p) = 2 a b (a - b) / (1 - 2 a b).
    Similarly, the coords of P are
( - 2 a . n / (m + n), 2 m n / (m + n)) = . . . = ( - 2 a b²/( 1 + b²), 2 a b / (1 + b²))
    and coords of Q are (2 a²b/ (1 + a²), 2 a b/(1 + a²)).
Semi-perimeter of DTUV = s = m + n + p = (a² + b² + a²b²) / a b
= (1 - 2 a b + a²b²) / a b = (1 - a b)²/a b.
Hence the radius r₄ of C4 = area of DTUV / s =, by Heron's formula, Ö( m n p / s) = a b / (1 - a b).
Let (x₄, y₄) be the coords of O₄, then
y₄ = y_R (1 - r₄) = . . . = 2 a b / (1 - a b) = 2 r₄.
Hence the minimum distance between C4 and line AC is also r₄. QED

    Theorem 4. The circle C10 with radius 2 a b touching ABC at B will touch C1 and C4 at R. RB bisects ÐARC. (Bankoff's Surprise 8).
    Proof: Once can verify that the point O₁₀(0, 2 a b) is at distance 2 a b from the point R, and lies on the line O₁R.
Now AR² = (2 a + 2 a b (a - b) /(1 - 2 a b))² + (2 a b / (1 - 2 a b))² = (Exercise!) . . . = 4 a²/(1 - 2 a b).
Similarly, CR² = (2 b - 2 a b (a - b) / (1 - 2 a b))² + (2 a b / (1 - 2 a b))² = . . . = 4 b²/(1 - 2 a b).
\ AR / CR = a / b = AB / CB. Hence BR bisects ÐARC. QED

    Theorem 5. Let M₁, M₂, M₃ be the midpoints of the arcs C1, C2, C3 respectively, then M₂ is the center of a circle C11 that goes through A, B, Q, R, and M₃ is the center of a circle C12 that goes through C, B, P, R. (Bankoff's Surprise 6).
    Proof: Slope of BQ is y_Q/x_Q = 1/a. Slope of M₂O₃ is -a /O₂O₃ = -a. Hence M₂O₃ is the perpendicular bisector of BQ, so that M₂ is equidistant from A, B, Q.
    Similarly, since BR bisects ÐARC, ÐARB = 45° = (½)ÐAM₂B. \R lies on the circle through A,B, with M₂ as center. QED