aa r X i v : . [ m a t h . M G ] J a n TWO DYNAMICAL SYSTEMS IN THE SPACE OF TRIANGLES
YURY KOCHETKOV
Abstract.
Let M be the space of triangles, defined up to shifts, rotations and dilations. We define twomaps f : M → M and g : M → M . The map f corresponds to a triangle of perimeter π the triangle withangles numerically equal to edges of the initial triangle. The map g corresponds to a triangle of perimeter 2 π the triangle with exterior angles numerically equal to edges of the initial triangle. For p ∈ M the sequence { p, f ( p ) , f ( f ( p )) , . . . } converges to the equilateral triangle and the sequence { p, g ( p ) , g ( g ( p )) , . . . } converges to the”degenerate triangle” with angles (0 , , π ). In Supplement an analogous problem about inscribed-circumscribedquadrangles is discussed. Introduction
Dynamical systems in space of triangles are objects of an interest for many years. For example in [2] and [3]the map is studied that corresponds to a triangle its pedal triangle. And in [1] the map is studied, where a newtriangle is constructed from cevians of the given one.We adopt another approach: we interchange roles of edges and angles. Namely, to a triangle with perimeter π we correspond the triangle whose angles are numerically equal to edges of the initial triangle, and to a trianglewith perimeter 2 π we correspond the triangle whose exterior angles are numerically equal to edges of the initialone.Let M be the space triangles defined up to shifts, rotations and dilations. Thus, an element of M is a triple ofpositive numbers with the sum π . Triples ( α, β, γ ), ( β, γ, α ) and ( γ, α, β ) are the same, but mirror symmetrictriangles are different elements in M . We denote by α the smallest angle of a triangle and by γ — the biggest,thus, α β γ .We will consider two maps f : M → M and g : M → M . Let p be a triangle of perimeter π , with angles α, β, γ and let α ′ , β ′ and γ ′ be lengths of edges opposite to angles α, β and γ , respectively. Then f ( p ) = q , where q = ( α ′ , β ′ , γ ′ ). Let now p be the same element in M , but with perimeter 2 π . Let a, b, c be exterior angles,adjacent to α , β and γ and a ′ , b ′ , c ′ be lengths of edges, opposite to α, β and γ , respectively. Then g ( p ) = r ,where r = ( π − a ′ , π − b ′ , π − c ′ ), i.e. ( a ′ , b ′ , c ′ ) are exterior angles of the new triangle. Remark . The triangle inequality is not valid for values of interior angles, but valid for values of exteriorangles. Hence, f is not a bijection, but g is. Theorem 2.1.
Let p ∈ M , then sequence { p, f ( p ) , f ( f ( p )) , . . . } converges to the equilateral triangle. Theorem 4.1.
Let p ∈ M , then the sequence { p, g ( p ) , g ( g ( p )) , . . . } converges to the point (0 , , π ) , which doesnot belong to M , but belong to its boundary. In Supplement we consider the map h that correspond to a inscribed-circumscribed quadrangle of perimeter 2 π the inscribed-circumscribed quadrangle which angles are numerically equal to edges of the initial quadrangle. Theorem 5.1.
Let Q be an inscribed-circumscribed quadrangle then the sequence { Q, h ( Q ) , h ( h ( Q )) , . . . } con-verges to the ”degenerate” quadrangle with angles , , π, π . Properties of the map f Let us remind that in a triangle the bigger edge lies opposite the bigger angle. Thus, α ′ β ′ γ ′ . Lemma 2.1. α ′ = π · sin( α )sin( α ) + sin( β ) + sin( γ ) , β ′ = π · sin( β )sin( α ) + sin( β ) + sin( γ ) , γ ′ = π · sin( γ )sin( α ) + sin( β ) + sin( γ ) . Proof.
It is enough to note that in triangle lengths of edges are proportional to sines of opposite angles. Asperimeter must be π , it remains to find the proportionality coefficient. (cid:3) Lemma 2.2. α ′ > α and equality is satisfied only when α = π ,Proof. π · sin( α )sin( α ) + sin( β ) + sin( γ ) = π · sin( α )sin( α ) + sin( β ) + sin( α + β ) == π · sin( α )2 sin β + α cos β − α + 2 sin β + α cos β + α = π · sin( α )4 sin β + α cos α cos β = π sin α β + α cos β . If 0 < x < π , then sin( x ) > xπ . Hence, π sin α β + α cos β > α β + α cos β , with equality only when α = π . It is enough to prove that3 α β + α cos β > α ⇔ > β + α β α + 2 β α . Now it remains to note that the first summand is not greater, than 2, and the second is not greater, than 1(because α π ). (cid:3) Lemma 2.3. γ ′ γ and equality is satisfied only when γ = π .Proof. As γ ′ = π · sin( γ )sin( α ) + sin( β ) + sin( γ ) = π · sin( γ )2 sin π − γ · cos β − α + sin( γ ) , then γ ′ is maximal when the difference β − α is maximal. If γ > π , then this difference is maximal for α = 0and β = π − γ . But then γ ′ < π · sin( γ )2 sin π − γ · cos π − γ + sin( γ ) π γ. If π γ < π , then the difference is maximal when β = γ and α = π − γ . But then γ ′ = 2 π · sin γ · cos γ · cos γ · cos γ − π + 2 · sin γ · cos γ = π · sin γ sin γ + sin γ = π cos( γ ) + 2 · cos γ + 1 = π γ ) + 2 π γ. (cid:3) Lemma 2.4.
The point ( π , π , π ) is a stationary attracting point of the map f .Proof. Let α = π xε, β = π yε, γ = π zε, x + y + z = 1 , x + y + z = 0 . then in the first approximation α ′ = π x πε √ , β ′ = π y πε √ , γ ′ = π z πε √ . WO DYNAMICAL SYSTEMS IN THE SPACE OF TRIANGLES 3
It remains to note that π √ < (cid:3) The above statements prove the theorem.
Theorem 2.1.
Let p ∈ M , then the sequence { p, f ( p ) , f ( f ( p )) , . . . } converges to the point ( π , π , π ) . Properties of the map g The reasoning here will be in terms of exterior angles. Let p = ( α, β, γ ) ∈ M and a = π − α , b = π − β and c = π − γ — values of exterior angles. In what follows we will use notations a ′ , b ′ , c ′ instead of α ′ , β ′ , γ ′ and wewill assume that a b c . Lemma 3.1. a ′ = 2 π · sin( a )sin( a ) + sin( b ) + sin( c ) , b ′ = 2 π · sin( b )sin( a ) + sin( b ) + sin( c ) , c ′ = 2 π · sin( c )sin( a ) + sin( b ) + sin( c ) . Proof.
It is enough to mention that sin( α ) = sin( a ), sin( β ) = sin( b ) and sin( γ ) = sin( c ). (cid:3) Lemma 3.2.
The point ( π , π , π ) is a stationary repelling point of the map g .Proof. Let a = 2 π xε, b = 2 π yε, c = 2 π zε, x + y + z = 0 , x + y + z = 1 . Then in the first approximation a ′ = 2 π − π √ · xε, b ′ = 2 π − π √ · yε, c ′ = 2 π − π √ · zε. It remains to note that π √ > (cid:3) Lemma 3.3. a ′ > b ′ > c ′ .Proof. a < b < c ⇔ α > β > γ ⇔ a ′ > b ′ > c ′ . (cid:3) Barycentric coordinates
Let us consider an equilateral triangle △ ABC and define a barycentric coordinates a, b, c , a + b + c = 2 π . Asthe value of an exterior angle is < π , then we will work with triangle △ A B C , where A B , C are midpointsof BC, AC and AB , respectively. ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ A BC C A B O C B A Figure 4.1
YURY KOCHETKOV
Here points A , B , C have coordinates (0 , π, π ), ( π. , π ) and ( π, π, A , B , C havecoordinates ( π, π , π ), ( π , π, π ) and ( π , π , π ), respectively. The point O — the center of ABC has coordinates( π , π , π ). g maps • the triangle A OC onto the triangle A OC : A → A , C → C , O → O (and back); • the triangle A OB onto the triangle A OB : A → A , B → B , O → O (and back); • the triangle B OC onto the triangle B OC : B → B , C → C , O → O (and back).In what follows we will always assume that a b and a c . Let g ( g ( a, b, c )) = ( a ′′ , b ′′ , c ′′ ). Our aim is to provethat a ′′ < a .Let I t = { ( a, b, c ) ∈ △ A B C : a = t, t π } and J t = { ( a, b, c ) ∈ △ A B C : a = t, π > t > π } . ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C O ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C OD E K L set I t set J t F M
Figure 4.2
Coordinates of points D , E and F are ( t, π − t, π ), ( t, π, π − t ) and ( t, π − t , π − t ), respectively. Coordinatesof points K , L and M are ( t, t, π − t ), ( t, π − t, t ) and ( t, π − t , π − t ), respectively.We have ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C O ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C O g ( I t ) g ( J t ) G Q PN
Figure 4.3
WO DYNAMICAL SYSTEMS IN THE SPACE OF TRIANGLES 5
Here g ( F ) = G and G = (cid:18) π · cos t cos t + 1 , π cos t + 1 , π cos t + 1 (cid:19) = (2 s, π − s, π − s ) , where s = π · cos t cos t + 1 . Then g ( K ) = P , g ( L ) = Q , g ( M ) = N and P = (cid:18) π − cos( t ) , π − cos( t ) , − π · cos( t )1 − cos( t ) (cid:19) , Q = (cid:18) π − cos( t ) , − π · cos( t )1 − cos( t ) , π − cos( t ) (cid:19) . Remark . When t = π (i.e. when DE coincides with B C ), then altitudes B B and C C are tangent tothe arc B GC at points B and C . ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C O ✡✡✡✡✡✡✡✡✡✡✡✡✡ ❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✡ B C A A B C OU V W X YZ g ( g ( I t )) g ( g ( J t )) Figure 4.4
We will denote by GG ( t ) — the first coordinate of the point W = g ( g ( F )): GG ( t ) = 2 π · cos( s )cos( s ) + 1 = 2 π · cos π · cos t cos t + 1cos π · cos t cos t + 1 + 1 . Formulas for coordinates of points N = g ( M ) and Z = g ( g ( M )) are the same, only here π t π . Thus thefunction GG is defined in the segment [0 , π ]. In the figure below the plot of GG ( t ) is presented. ✲ t ✻ qq π π Figure 4.5
Lemma 4.1. GG ( t ) < t . YURY KOCHETKOV
A sketch of the proof. As GG (0) = 0 and GG ( π ) = π it is enough to prove, that GG is a strictly increasingfunction and its plot is downward convex. We have GG ′ = (cid:18) π · cos( s )cos( s ) + 1 (cid:19) ′ = − π · sin( s ) · s ′ (cos( s ) + 1) = − π · sin( s )(cos( s ) + 1) · − π · sin t (cos t + 1) > GG ′′ = − π · [cos( s ) · ( s ′ ) + sin( s ) · s ′′ ] · (cos( s ) + 1) + 2 · sin ( s ) · ( s ′ ) (cos( s ) + 1) . The numerator of this expression is2 π · (1 + cos( s )) · [ − ( s ′ ) · (2 − cos( s ) − sin( s ) · s ′′ ] . And the expression in square brackets is − π · (2 − cos( s )) · (cid:18) − cos t (cid:19) + sin( s ) · (cid:18) − cos t (cid:19) · (cid:18) t (cid:19) . The proof of the positivity of the above expression is a technical task. (cid:3)
The first coordinate of the point U is π · (1 − cos( t )). In the figure below are presented plots of the firstcoordinates of points U (the upper curve) and W (the lower curve) for 0 < a π . ✲✻ qq t π π Figure 4.6The first coordinate of the point X is π − cos π − cos( t ) , π < t π . WO DYNAMICAL SYSTEMS IN THE SPACE OF TRIANGLES 7
In the figure below are presented plots of the first coordinates of points X (the upper curve) and Z (the lowercurve). ✲ t ✻ q π π qq π q π ✟✟✟✟✟✟✟✟✟✟✟✟ Figure 4.7
Theorem 4.1.
Let p ∈ M , then the sequence { p, g ( p ) , g ( g ( p )) , . . . } converges to the point (0 , , π ) . This pointdoes not belong to the set M , but belong to its boundary.Proof. We see that the map h = g ( g ( p )) decreases the least exterior angle. Thus the sequence { T, h ( T ) , h ( h ( T )) , . . . } ,where T is a triangle, converges to a ”degenerate triangle” with angles (0 , , π ). On the other hand, the sequence { g ( T ) , h ( g ( T )) , h ( h ( g ( T ))) , . . . } also converges to the same ”degenerate triangle” (with another zero angles). (cid:3) Example 4.1.
Let T be a triangle with exterior angles (1 , . , π − . g ( T ) , g ( g ( T )) , g ( g ( g ( T ))) , . . . have the following exterior angles. (3 . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . T be (1 . , . , π − . . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . . , . , . YURY KOCHETKOV Supplement
We will consider plane inscribed-circumscribed quadrangles (ic-quadrangles). Sums of opposite angles of anic-quadrangle are π and sums of lengths of opposite edges are equal. Up to shifts, rotations and dilations suchquadrangle is uniquely defined by its angles and their order in going around the quadrangle. Two angles of aic-quadrangle are acute (and they are adjacent) and two are obtuse (and they are also adjacent).Let α and β be obtuse angles and α > β . ❇❇❇❇❇❇❇❇❇❇ (cid:0)(cid:0)(cid:0)❍❍❍❍❍❍❍❍❍❍❍❍ B AC D
Here α is the value of ∠ A , β is the value of ∠ B , ∠ C = π − α , ∠ D = π − β . Let r be the radius of the inscribedcircle, then | CD | = r · (tan( α/
2) + tan( β/ > | BC | = r · (tan( α/
2) + cot( β/ >> | AB | = r · (cot( α/
2) + tan( β/ > | DA | = r · (cot( α/
2) + cot( β/ . If the perimeter of
ABCD is 2 π , then | CD | = π · sin( α ) · sin( β )cos( α − β ) , | BC | = π · sin( α ) · cos( β )sin( α + β ) . The map h corresponds to an ic-quadrangle of perimeter 2 π the ic-quadrangle of perimeter 2 π with anglesnumerically equal to edges of initial ic-quadrangle. Theorem 5.1.
Let Q be an ic-quadrangle of perimeter π , then the sequence { Q, h ( Q ) , h ( h ( Q )) , . . . } convergesto a ”degenerate quadrangle” with angles (0 , , π, π ) .Sketch of the proof. The sum of obtuse angles in the quadrangle h ( h ( Q )) is strictly greater, than the sum ofobtuse angles in the initial quadrangle Q . (cid:3) Remark . The sum of obtuse angles in h ( Q ) can be less, than the sum of obtuse angles in Q . For example,if α = 1 .
85 and β = 1 ,
75, then
P i · sin( α ) · sin( β )cos( α − β + P i · sin( α ) · cos( β )sin( α + β = 3 . < α + β = 3 . References [1] M. Hajja,
The sequence of generalized median triangles and new shape function , J. Geometry, (2009), 71-79.[2] J.G. Kigston & J.L. Synge, The sequence of pedal triangles , Amer. Math. Monthly, (1988), 609-620.[3] P. Ungar,
Mixing property of pedal mapping , Amer. Math. Monthly, (1990), 898-900.
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