Generalized Archimedean Arbelos Twins

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1 Forum Geometricorum Volume 4 (204) FORUM GEOM ISSN Generalized rchimedean rbelos Twins Nikolaos Dergiades bstract We generalize the well known rchimedean arbelos twins by extending the notion of arbelos, and we construct an infinite number of rchimedean circles rchimedean arbelos On a segment we take an arbitrary point P and with diameters P, P, we construct the semicircles O (R ), O 2 (R 2 ), O(R), where R = R + R 2 If we cut from the large semicircle the small ones then the resulting figure is called from antiquity (rchimedes) arbelos (the shoemaker s knife) The perpendicular at P to meets the large semicircle at Q and divides the arbelos in two mixtilinear triangles with equal incircles Theorem (rchimedean arbelos twins) The two circles K (r ) and K 2 (r 2 ) that are inside the arbelos and are tangent to the arbelos and the line PQ have equal radii r = r 2 = R R 2 R +R 2 (see Figure ) Equivalently, r = R + R 2 Q K 2 K O P O O 2 Figure circle in the arbelos with radius r is called an rchimedean circle ([6, p6]) We can find infinitely many such twin incircles where the above case is a limit special case We generalize the notion of the arbelos as a triangle whose sides are in general circular arcs We will investigate some special cases of these arbeloi Publication Date: December 6, 204 Communicating Editor: Paul Yiu

2 40 N Dergiades 2 Generalized arbelos 2 Soddy type arbelos with 3 vertices tangency points of the arcs Let U X, U Y, U Z (counter clockwise direction) be the vertices of the arbelos whose sides are arcs of circles tangent to each other at the vertices, with centers X, Y, Z and radii r X, r Y, r Z If the rotation from U Y to U Z on the arc with radius r X is clockwise relative to U X, then r X is positive, otherwise it is negative Similarly we characterize the radii r Y, r Z relative to the movement on the appropriate arcs from U Z to U X, and from U X to U Y Hence, we can have 8 different cases of arbeloi with the same vertices and the same radii in absolute value If we change the sign of the radii of the arbelos, then we get the complementary arbelos Denote by Δ the area of the triangle XY Z and r the radius of a circle tangent to the arcs-sides of the arbelos Theorem 2 The radius r of the circle that is tangent to the sides of the above arbelos is given by r = Δ, r X r Y r Z r X r Y r Z or r = + + 2Δ r X r Y r Z r X r Y r Z One of these corresponds to the complementary arbelos X K X Z U X Y U Y K Z U Y U X Y U Z U Z Figure 2a Figure 2b Proof If K(r) is the circle tangent to the sides of the arbelos, then in Figure 2a, the radius r X is negative, the sides of triangle XY Z are a = r Y +r Z, b = r Z +r X, c = r X + r Y, and the tripolar coordinates of K relative to XY Z and the area Δ of triangle XY Z are KX = λ = r X + r, KY = μ = r Y + r, KZ = ν = r Z + r, Δ= r X r Y r Z (r X + r Y + r Z )

3 Generalized rchimedean arbelos twins 4 If we substitute these values into the equality that the tripolar coordinates satisfy ([3]): (μ 2 + ν 2 a 2 ) 2 λ 2 +(ν 2 + λ 2 b 2 ) 2 μ 2 +(λ 2 + μ 2 c 2 ) 2 ν 2 we get Therefore, (μ 2 + ν 2 a 2 )(ν 2 + λ 2 b 2 )(λ 2 + μ 2 c 2 )=λ 2 μ 2 ν 2, ( r ) 2 = 4Δ2 r X r Y r Z rx 2 r2 Y r2 Z r = + + ± 2Δ r X r Y r Z r X r Y r Z Remarks () The rcimedean arbelos is of Soddy type with collinear vertices where r X = R, r Y = R 2, r Z = R = (R + R 2 ), Δ=0 In this case, there is a double solution Hence, the inradius of the rchimedean arbelos is given by r = + R R 2 R + R 2 (2) If the radii r X, r Y, r Z are positive, then the radii r refer to the inner and outer Soddy circles ([2]) In the sequel, we shall adopt the following notation: For a line l, Π l denotes the orthogonal projection map onto l Lemma 3 Let K(r) be a circle tangent externally or internally to the circles O (R ) and O 2 (R 2 ), where R, R 2 may assume any real values For a point F on the radical axis of the circles, r = O O 2 Π O O 2 ( FK) () R R 2 K F F O M K O 2 Figure 3

4 42 N Dergiades Proof Let M be the midpoint of O O 2 (Figure 3) We have R2 2 R 2 = FO2 2 FO 2 =( FO 2 FO ) ( FO 2 + FO )=2O O 2 FM, R + r 2 R 2 + r 2 = KO 2 KO2 2 =( KO KO 2 ) ( KO + KO 2 )=2O O 2 MK y addition, we get 2r(R R 2 )=2 O O 2 ( FM + MK)=2 O O 2 Π O O 2 ( FK), and the result follows Remark For the rchimedean twins () gives r = (2R r )R 2 = = + R + R 2 + R r R R 2 Theorem 4 Let U X U Y U Z be an arbelos with collinear centers X, Y, Z on a line L, and K(r) be the incircle of the arbelos r = Π L ( U Y U Z ) r X r Y XY r X r Z XZ Proof Since U Z, U Y are points on the radical axes of the circle pairs X(r X ), Y (r Y ), and X(r X ), Z(r Z ) respectively, by Lemma 3 we have XY ΠL ( U Z K) XZ Π L ( U Y K) r = = r X r Y r X r Z Since X, Y, Z are all on the line L, Π L ( U Z K)= r(r X r Y ) XY j and ΠL ( U Y K)= r(r X r Z ) XZ j, where j is a unit vector on L Hence, Π L ( U Y U Z )= Π L ( U Y K) Π L ( U Z K) ( rx r Z = r XZ r ) X r Y j XY From this the result follows 22 rbelos of type On a line L we take the consecutive points U Z, U Y, U Z, U Y, and construct on the same side of the line the semicircles (U Z U Z), (U Y U Y ), and (U Y U Z ) Let U X be the intersection of the first two semicircles (see Figure 4) The arbelos U X U Y U Z is of type It has arc U Y U Z positive, and U Z U X, U X U Y both negative The diameter U Y U Z is the base of the arbelos 22 The incircle Let K(r) be the incircle of this arbelos and,, C the points of tangency If S is the external center of similitude of the semicircles (U Z U Z) and (U Y U Y ), then the line C passes through S, and the inversion with pole S and power d 2 = SU Y SU Z = SUX 2 = S SC swaps the semicircles (U Z U Z) and (U Y U Y ), and leaves the circle K(r) and the semicircle (U Y U Z ) invariant Hence, S 2 = S SC, and the S is tangent at to both K(r) and (U Y U Z ) If the line S meets the perpendiculars to L at U Y, U Z at D and E respectively, then D is the radical center of K(r), (U Y U Z ), (U Y U Y ), and E is the radical center of K(r), (U Y U Z ), (U Z U Z) Hence, DC = D and E = E

5 Generalized rchimedean arbelos twins 43 U X C K E D S U Z U Y U Z U Y Figure 4 Construction of the incircle We construct the external center of similitude S of the semicircles (U Z U Z) and (U Y U Y ), and take a point on the semicircle (U Y U Z ) such that S = SU X Let the line S meet the perpendiculars to L through U Y, U Z at D, E respectively We take on (U Z U Z) the point such that E = E, and on (U Y U Y ) the point C such that DC = D The circumcircle of C is the incircle of the arbelos The radius of the incircle If we take U Z U Y =2y, U Y U Z =2R, U Z U Y =2z, then r X = R, r Y = R y, r Z = R z, XY = y, and XZ = z From Theorem 4, we have 2R r = = Ryz R(y + z)+yz, or 2R+y y 2R+z z r = R + y + z (2) 23 rbelos of type On a line L we take the consecutive points U Y, U Z, U Y, U Z, and construct on the same side of the line the semicircles (U Z U Z), (U Y U Y ), and (U Y U Z ) Let U X be the intersection of the first two semicircles (see Figure 5) The arbelos U X U Y U Z is of type It has arc U Y U Z positive and the arcs U Z U X, U X U Y of different signs The base of the arbelos is U Y U Z Construction of the incircle The construction is the same as in type, but now S is the internal point of similitude of the semicircles (U Z U Z) and (U Y U Y ) The radius of the incircle We have the same formula as (2), but now since U Y is not on the right hand side of U Z, the distance z must be negative, and so we have r = R + y z (3)

6 44 N Dergiades E U X K C D U Y U Z S U Y U Z Figure 5 Remark For the incircle K(r) of the rchimedean arbelos, since U X =, U Y = P, U Z = with base 2R 2, y = R, z = R R 2, (3) gives r = + R R 2 R + R 2 3 Generalized rchimedean arbelos twins We shall construct in the rchimedean arbelos a generalized pair of inscribed equal circles Theorem 5 In the rchimedean arbelos (Figure 6) where P =2R, P = 2R 2, =2R +2R 2, we extend (to left and right) with equal segments =2x = The semicircle ( P ) divides the arbelos in two arbeloi with incircles K (r ), K 3 (r 3 ), and the semicircle (P ) divides the arbelos in two arbeloi with incircles K 2 (r 2 ), K 4 (r 4 ) We have a couple of twin circles: r = r 2 and r 3 = r 4, with r = x + R + R 2 and = + r 3 R R 2 R + R 2 + x Q K 2 K 4 K 3 K 2x 2R P 2R 2 2x Figure 6 Proof The circle K (r ) is the incircle of an arbelos of type with base P Hence, r = R + x + R 2

7 Generalized rchimedean arbelos twins 45 The circle K 2 (r 2 ) is the incircle of an arbelos of type with base P Hence, r 2 = R 2 + x + R The circle K 3 (r 3 ) is the incircle of an arbelos of type with base P Hence, r 3 = R 2 + R R +R 2 +x The circle K 4 (r 4 ) is the incircle of an arbelos of type with base P Hence, r 3 = R + R 2 R +R 2 +x Therefore, r = r 2 and r 3 = r 4 Remark If x, then the semicircles ( P ) and (P ) tend to the perpendicular semiline PQ, and the four incircles tend to the rchimedean twin circles 4 isectors of the rchimedean rbelos We have seen that in the rchimedean arbelos, the semiline PQ divides the arbelos in two arbeloi with equal incircles This semiline as a degenerate semicircle is like a bisector from P of the arbelos that produces the twins (rchimedean circles) with radius r P such that r P = R + R 2 We find the other two bisectors of the arbelos from the vertices and (Figure 7) Q Q 2 Q O P 2x O 2 Figure 7 Let the semicircle (O 2 ) bisector of the arbelos meet the semicircle (P) at Q, and (r ), 2 (r 2 ) be the incircles of the arbeloi Q P, Q The arbelos Q P with base P is of type, so we have r = R + x R +R 2 The arbelos Q ( with base is the ) complement of an arbelos of type ; so we have r 2 = R +R 2 R R 2 x In order to have r = r 2 = r, we set 2x = R 2 Hence the point O 2 is the midpoint of P, and the bisector semicircle (O 2 )meets PQ at the point Q such that PQ = P PO 2 = 2R R 2, and r = R + 2 R 2 R +R 2 Similarly, if O is the midpoint of P, then the semicircle (O ) is the bisector of the arbelos from that passes also from Q, and r = 2 R + R 2 R +R 2 Hence, the three bisectors of the rchimedean rbelos are concurrent at Q

8 46 N Dergiades 5 Infinitely many rchimedean circles There are many exciting constructions of rchimedean circles or families of these; see [4, 5] Here we construct a more natural family of rchimedean circles that contains the original rchimedean twins In the rcimedean arbelos with diameters P, P,, and semicircles O (R ), O 2 (R 2 ), O(R), we take at the left of the point X, and at right of the point Y such that X = Y =2x We rotate clockwise P around by an angle π 2 at, and counterclockwise P around by an angle π 2 at The line X meets the line PQat D The line Y meets the line PQ at C We take at the left of the point C such that C = PC =2d, and at right of the point D such that D = PD =2d 2 The semicircles (CP), (Y ) meet at M, and the semicircles (PD), (X) meet at N We show that the incircles K (r ), K 2 (r 2 ) of the arbeloi MP, NP are both rchimedean circles Q M Q N Q K 2 K C X O P O 2 D Y D C Figure 8 Proof Since in triangle XPD, P is a bisector and is parallel to PD (Figure 8), we know that PD + XP = Hence, + = (4) d 2 x + R R Similarly, from triangle PYC,wehave + = (5) d x + R 2 R 2

9 Generalized rchimedean arbelos twins 47 The arbelos MP is of type ; so we have r = R + d + x+r 2 = R + R 2 from (5) The arbelos NP is also of type ; so we have r 2 = R 2 + d 2 + x+r = R 2 + R from (4) Hence r = r 2 is the radius of the rchimedean circle For x =0the circles K (r ), K 2 (r 2 ) coincide with the original rchimedean twin circles since the semicircles (Y ), (X) coincide with the semicircle (), and the semicircles (CP), (PD) coincide with the line PQ If M, M 2 are the midpoints of CP, Y, then applying Stewart s theorem to triangle MM M 2,we have O M 2 M M 2 = M M 2 O M 2 + M 2 M 2 O M O M O M 2 M M 2, or (d +R 2 +x)o M 2 =(d +R ) 2 (R 2 +x)+(r +R 2 +x) 2 d d (R 2 +x)(d +R 2 +x) Substituting d = R 2(x+R 2 ) x,weget O M 2 = R 2 +4R R 2 = O P 2 + PQ 2 = O Q 2 Hence, the locus of M is the circular arc O (Q) from the point Q to Q on the perpendicular Q to Similarly, we can prove that the locus of N is the circular arc O 2 (Q) from the point Q to Q on the perpendicular Q to 6 special generalization of arbelos with arcs of angle 2φ If we substitute the semicircles in rchimedean arbelos with arcs of angle 2φ, ie, P =2R sin φ, P =2R 2 sin φ, =2(R + R 2 )sinφ (Figure 9) [], the points, O, O are collinear; so are the points O, O 2, The tangent to the arc (P ) at P meets the arc () at Q, and the tangent to the arc (P) at P meets the arc () at Q Let K (r ), K 2 (r 2 ) be the incircles of the arbeloi Q P and Q PIf2φ = π, then we have the classical arbelos, and Q, Q coincide with the point Q We prove that r = r 2 Proof The point is on the radical axis of O (R ), O (R + R 2 )) From (), we have O O Π O O( K ) r = = R 2 Π O O(K ) r O r O R + R + R 2 lso, r =Π O O(K P ) Hence, 2R sin 2 φ = Π OO(K )+Π OO(K P )= r 2R + R 2 R 2 + r = r 2(R + R 2 ) R 2, or r = R R 2 sin 2 φ R +R 2 Similarly, OO 2 Π OO2 ( K 2 ) r 2 = = R Π OO2 (K 2 ), r O r O2 R R 2 R 2

10 48 N Dergiades Q Q K K 2 2φ O F 2 C P D F E 2φ 2φ O 2 and r 2 =Π OO2 (PK 2 ) Hence, O Figure 9 2R 2 sin 2 φ = Π OO2 (PK 2 )+Π OO2 (K 2 )= r 2 + r 2 R +2R 2 R = r 2 2(R + R 2 ) R, or r 2 = R R 2 sin 2 φ R +R 2 If 2φ = π, then we have the radius of the rcimedean circle Construction of the twins The perpendicular from P to meets O O 2 at the point C, and the parallel from C to PO meets PO 2 at D Since PC is a bisector in triangle PO O 2,wehave CD = R + R 2 Hence we need the construction of r = CD sin 2 φ The perpendicular from D to meets at E and the perpendicular from E to PO meets this line at the point F The symmetric of F in PC is the point F 2 on the line PO 2 The perpendicular at F 2 to PF 2 meets the circle O (F ) at the point K and the line EF meets the circle O 2 (F 2 ) at the point K 2 These points are the centers of the twin incircles and the construction of these circles is obvious References [] D elev, generalization of the classical arbelos and the rchimedean circles, available at archivegeogebraorg [2] N Dergiades, The Soddy circles, Forum Geom, 7 (2007) 9 97 [3] Hatzipolakis, F M van Lamoen, Wolk, and P Yiu, Concurrency of four Euler lines, Forum Geom, (200) [4] F M van Lamoen, rchimedean adventures, Forum Geom, 6 (2006) [5] F M van Lamoen, Online catalogue of rchimedean circles, lamoen/wiskunde/rbelos/cataloguehtm [6] P Yiu, Euclidean Geometry, Florida tlantic University Lecture Notes, 998; available at Nikolaos Dergiades: I Zanna 27, Thessaloniki 54643, Greece address: ndergiades@yahoogr

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