Well Centered Spherical Quadrangles

Transcription

1 Beiträge zur Algebr und Geometrie Contributions to Algebr nd Geometry Volume 44 (003), No, Well Centered Sphericl Qudrngles An M d Azevedo Bred 1 Altino F Sntos Deprtment of Mthemtics, University of Aveiro, Aveiro, Portugl e-mil: mbred@mtupt Deprtment of Mthemtics, UTAD, Vil Rel, Portugl e-mil: folgdo@utdpt Abstrct We introduce the notion of well centered sphericl qudrngle or WCSQ for short, describing geometricl method to construct ny WCSQ We shll show tht ny sphericl qudrngle with congruent opposite internl ngles is congruent to WCSQ We my clssify them tking in ccount the reltive position of the sphericl moons contining its sides Proposition describes the reltions between well centered sphericl moons nd WCSQ which llow the refereed clssifiction Let L be sphericl moon We shll sy tht L is well centered if its vertices belong to the gret circle S {(x, y, z) IR 3 : x = 0} nd the semi-gret circle bisecting L contins the point (1, 0, 0) If L 1 nd L re two sphericl moons with orthogonl vertices then L 1 nd L re sid to be orthogonl Let us consider the clss Ω of ll sphericl qudrngles with ll congruent internl ngles or with congruent opposite internl ngles Proposition 1 Q Ω if nd only if Q hs congruent opposite sides Proof It is obvious tht ny sphericl qudrngle, Q, with congruent opposite sides is n element of Ω Suppose now, tht Q is n rbitrry element of Ω Then Q hs congruent opposite internl ngles sy, in cyclic order, (α 1, α, α 1, α ), with α i (0, π), i = 1,, α 1 + α > π Lengthening two opposite sides of Q we get sphericl moon, L, s illustrted in Figure 1 1 Supported in prt by UI&D Mtemátic e Aplicções /93 $ 50 c 003 Heldermnn Verlg

2 540 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles Figure 1 The moon L includes Q nd two sphericl tringles, T 1 nd T As T 1 nd T hve congruent internl ngles, they re congruent, nd so the sides of Q common to respectively T 1 nd T re congruent The result now follows pplying the sme resoning to the other pir of opposite sides of Q Proposition Let L 1 nd L be two well centered sphericl moons with distinct vertices of ngle mesure θ 1 nd θ, respectively nd let Q be the sphericl qudrngle Q = L 1 L Then Q hs internl ngles nd sides in cyclic order of the form, (α 1, α, α 1, α ) nd (, b,, b), respectively Moreover, L 1 nd L re orthogonl if nd only if α 1 = α, nd θ 1 = θ if nd only if = b Proof Let L 1 nd L be two well centered sphericl moons with distinct vertices of ngle mesure θ 1 nd θ, respectively L 1 nd L divide the semi-sphere into 8 sphericl tringles, lbelled s indicted in Figure, T i, i = 1,, 8 nd sphericl qudrngle Q = L 1 L Let E nd N be vertices of L 1 nd L, respectively, α 1, α, α 3, α 4 nd, b, c, d be, respectively, the ngles nd sides of Q in cyclic order (see Figure ) Figure The tringles T 5 nd T 6 re congruent (it is enough to verify tht they hve one congruent side nd two congruent ngles) nd so α 1 = α 3 Also T 7 nd T 8 re congruent nd so α = α 4 Since T 5 nd T 6 re congruent nd T 7 nd T 8 re congruent we my conclude tht T 1 nd T re congruent s well s T 3 nd T 4 nd so = c nd b = d Now, θ 1 = θ if nd only if T 1 nd T 4 re congruent, tht is, if nd only if = b Besides, L 1 nd L re orthogonl iff E N = 0, where denotes the usul inner product in IR 3, iff T 6 nd T 7 re congruent iff α 1 = α

3 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles 541 Corollry 1 Using the sme terminology s before one hs i) If θ 1 = θ nd E N = 0 then Q = L 1 L hs congruent internl ngles nd ll congruent sides; ii) If θ 1 = θ nd E N 0 then Q = L 1 L hs ll congruent sides nd distinct congruent opposite pirs of ngles; iii) If θ 1 θ nd E N = 0 then Q = L 1 L hs congruent internl ngles nd distinct congruent opposite pirs of sides; iv) If θ 1 θ nd E N 0 then Q = L 1 L hs distinct congruent opposite pirs of ngles nd distinct congruent opposite pirs of sides By well centered sphericl qudrngle (WCSQ) we men sphericl qudrngle which is the intersection of two well centered sphericl moons with distinct vertices Proposition 3 Let Q be sphericl qudrngle with congruent internl ngles, sy α ( π, π), nd with congruent sides, Then is uniquely determined by α Proof The digonl of Q divides Q in two congruent isosceles tringles of ngles (α, α, α ) Thus, if is the side of Q one hs cos = cos α (1 + cos α) sin α sin α = 1 + cos α 1 cos α We cn observe tht this reltion defines n incresing continuous bijection between α ( π, π) nd (0, π ) Proposition 4 Let Q be sphericl qudrngle with congruent internl ngles, sy α ( π, π), nd with congruent sides Then Q is congruent to WCSQ Proof Let Q be sphericl qudrngle with congruent internl ngles, α ( π, π), nd with ll congruent sides Consider two sphericl moons well centered nd orthogonl, L 1 nd L with the sme ngle mesure θ (0, π) such tht cos θ = cos α + 1 nd Q = L 1 L, see Figure 3 Let us show tht Q is congruent to Q By Corollry 1, Q hs congruent internl ngles nd congruent sides Figure 3 Denoting by α ( π, π) the internl ngle of Q one hs, cos α (π θ) (π θ) = cos + sin cos π = θ sin = cos θ 1 = cos α

4 54 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles Thus α = α nd consequently Q nd Q re congruent, since they hve internl congruent ngles nd by the previous proposition they lso hve congruent sides It should be pointed out tht the reltion cos θ = cos α + 1 defines n incresing continuous bijection between α ( π, π) nd θ (0, π) Proposition 5 Let Q be sphericl qudrngle with congruent internl ngles, sy α ( π, π), nd with distinct congruent opposite pirs of sides, sy nd b Then nyone of the prmeters α, or b is completely determined by the other two Proof Let Q be sphericl qudrngle in the bove conditions For α ( π, π) nd (0, π), b is determined by the system of equtions: where θ is the ngle indicted in Figure 4 Therefore, { cos b = cos (π ) + sin (π ) cos θ cos θ = cos (π α) + sin (π α) cos b cos b = cot cos α Figure 4 In similr wy, cn be expressed s function of b nd α We shll show in next lemm tht α cn lso be expressed s function of nd b Lemm 1 Let Q be sphericl qudrngle with distinct congruent opposite pirs of sides, sy nd b nd with congruent internl ngles, sy α Then cos α = tn tn b Proof Let Q be sphericl qudrngle in the bove conditions Lengthening the vertices of two djcent edges, one gets two isosceles tringles with sides, π b π, nd b, respectively, see Figure 5

5 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles 543 Figure 5 Let θ 1 nd θ be the internl ngle mesure of these tringles, see Figure 5 Then cos α = cos π θ 1 On the other hnd, cos π θ + sin π θ 1 sin π θ cos π = sin θ 1 sin θ nd Thus cos θ 1 = cos θ = cos cos π b sin π b cos b cos π sin π = cos sin b cos b = cos b sin cos cos α = 1 cos θ1 1 cos θ = tn tn b Proposition 6 Let Q be sphericl qudrngle with congruent internl ngles, sy α ( π, π) nd with distinct congruent opposite sides, sy (0, π) nd b = b(, α) Then Q is congruent to WCSQ Proof Suppose tht Q is sphericl qudrngle in the bove conditions We shll show tht for two orthogonl well centered moons of ngle mesure, respectively, θ 1 nd θ, the unique solution of the system of equtions, { cos α = sin θ 1 sin θ (1) cos = cos θ 1+cos α sin α defines well centered qudrngle (the moon s intersection) congruent to Q In fct if such well centered qudrngle exists then by Corollry 1 it hs to be the intersection of two orthogonl moons L 1 nd L of ngles θ 1 (0, π) nd θ (0, π), respectively

6 544 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles With the Figure 6 nnottion, one hs s we hve seen before And Figure 6 cos α = sin θ 1 sin θ, cos = cos θ 1 + cos (π α) sin (π α) = cos θ 1 + cos α sin α It is strightforwrd exercise to show tht the system of equtions (1) hs unique solution nd tht L1 L is congruent to Q Observe tht θ 1 (0, π), where the cosine function is injective nd θ (0, π ), where the sine function is lso injective Remrk 1 Let α : (0, π) (0, π) ( π, π) nd : (0, π) (0, π) (0, π) be such tht α(θ 1, θ ) = rccos( sin θ 1 sin θ ) nd (θ 1, θ ) = rccos cos θ 1 + sin θ1 sin θ 1 sin θ 1 sin θ The contour levels of α nd re illustrted in Figure 7 (done by Mthemtic) We my observe tht the intersection of ny two contour levels of α, nd determine unique pir of ngles (θ 1, θ ) (0, π) (0, π), which mens tht sphericl qudrngle in the conditions of the lst proposition is congruent to well centered sphericl qudrngle (the intersection of two orthogonl well centered sphericl moons of ngles θ 1 nd θ ) Figure 7

7 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles 545 Proposition 7 Let Q be sphericl qudrngle with ll congruent sides, sy (0, π) nd with congruent opposite ngles, sy α 1, α, α 1 α Then α 1 rccos(1 ) nd 1+cos nyone of the prmeters, α 1 nd α is completely determined by the other two Proof Suppose tht Q is in the bove conditions 1 If α 1 = α = α then s seen in proposition 3, cos = 1+cos α 1 cos α, tht is, cos α = 1 1+cos ; If α 1 > α then continuity rgument llows us to conclude tht α 1 > rccos(1 1+cos ) > α This cn be seen drgging two opposite vertices of Q long the digonl of Q contining them, see Figure Figure 8 Now, given nd α 1, α is completely determined by the system of equtions, { cos α = cos α 1 + sin α 1 cos l cos l = cos + sin cos α where l denotes the digonl of Q bisecting α 1, see Figure 9 Thus, Figure 9 cos α = tn α 1 cos nd cos = cot α 1 cot α Proposition 8 Let Q be sphericl qudrngle with ll congruent sides, sy (0, π) nd with congruent opposite pirs of ngles, α 1, α, α 1 > α, with α = α (α 1, ) nd α 1 > rccos(1 ) Then Q is congruent to WCSQ 1+cos

8 546 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles Proof Let Q be sphericl qudrngle s indicted bove Let us show first tht when two well centered sphericl moons with congruent ngles, sy θ, nd π x, x (0, π ) s the ngle mesure between them, then the following system of equtions { cos α1 = sin θ cos θ sin x cos = cos θ+cos α 1 cos α sin α 1 sin α hs unique solution which defines WCSQ congruent to Q As seen in Corollry 1, if such well centered sphericl qudrngle exists then it hs to be the intersection of two well centered sphericl moons with congruent ngles θ (0, π), nd such tht the ngle mesure between them is π x, x (0, π ), see Figure 10 With the lbelling of Figure 10 one hs, Figure 10 cos α 1 = cos π θ nd on the other hnd, + sin π θ cos( π + x) = sin θ cos θ sin x cos = cos θ + cos(π α ) cos(π α 1 ) sin(π α ) sin(π α 1 ) = cos θ + cos α 1 cos α sin α 1 sin α Using similr rgument to the one used in proposition 6 it cn be seen tht the solution is unique nd tht Q is congruent to WCSQ Remrk Let α 1 : (0, π) (0, π) ( π, π) nd : (0, π) (0, π) (0, π ) be such tht nd α 1 (θ, x) = rccos( sin θ θ cos sin x) cos θ + ( sin θ (θ, x) = rccos cos θ sin x)( sin θ + cos θ sin x) 1 ( sin θ cos θ sin x) 1 ( sin θ + cos θ sin x) The contour levels of α 1 nd re represented in Figure 11 (done by Mthemtic)

9 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles 547 Observe tht if α 1 ( π, π) nd (0, π) such tht α 1 rccos(1 ) then the 1+cos intersection of ny two contour levels of α 1 nd is unique point (θ, x) (0, π) (0, π) In other words ny sphericl qudrngle in the conditions of the previous proposition is congruent to the intersection of two well centered sphericl moons with the sme ngle mesure, θ, nd being π x the ngle mesure between them Figure 11 Proposition 9 Let Q be sphericl qudrngle with congruent opposite sides, sy nd b nd with congruent opposite ngles, sy α 1 nd α with α 1 α Then, i) + b < π; ii) α 1 rccos( tn tn b ); iii) nyone of the prmeters α 1, α, or b is completely determined by the other three Proof If Q is qudrngle s described bove then it follows tht 0 < + b < π nd lso α 1 + α π > 0, with α 1 (0, π) nd α (0, π) Tht is, 0 < + b < π, α 1 + α > π, α (0, π) nd α 1 ( π, π), since α 1 α Assume, in first plce, tht α 1 = α = α Then, by lemm 1 we hve cos α = tn tn b nd so α = α 1 = α = rccos( tn tn b ) As before continuity rgument llows us to conclude tht if α 1 > α, then α 1 > rccos( tn tn b ) > α Now, we show how to determine α s function of, b nd α 1 The digonl l of Q through α gives rise to two ngles, x nd y, (α = x + y) s illustrted in Figure 1 Figure 1

10 548 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles One hs, Besides, cos x = cos l = cos cos b + sin sin b cos α 1 cos cos b cos l sin b sin l nd cos y = Since α = x + y, then α is function of, b nd α 1 cos b cos cos l sin sin l We cn lso determine b s function of, α 1 nd α s follows Let b 1, b nd θ be, respectively, the sides nd the internl ngle mesure (to be determined) of the tringle obtined by lengthening the b sides of Q, see Figure 13 One hs, Figure 13 cos θ = cos α 1 cos α + sin α 1 sin α cos nd cos b 1 = cos α 1 + cos α cos θ sin α sin θ cos b = cos α + cos α 1 cos θ sin α 1 sin θ Finlly, b = π (b 1 + b ) is function of, α 1 nd α Proposition 10 Let Q be sphericl qudrngle with congruent opposite sides, sy nd b such tht + b < π nd with congruent internl ngles α 1, α, α 1 > α Let us suppose lso tht α 1 > rccos( tn tn b ) nd α = α (, b, α 1 ) Then, Q is congruent to WCSQ Proof Let Q be sphericl qudrngle in the bove conditions We shll show tht when we hve two well centered sphericl moons with ngle mesure θ 1 nd θ nd such tht π x, x (0, π ) is the ngle mesure between them, see Figure 14, then the unique solution of the system of equtions cos α 1 = sin θ 1 sin θ cos θ 1 cos θ sin x cos = cos θ 1+cos α 1 cos α sin α 1 sin α cos b = cos θ +cos α 1 cos α sin α 1 sin α

11 A M d Azevedo Bred, A F Sntos: Well Centered Sphericl Qudrngles 549 defines well centered sphericl qudrngle congruent to Q As seen in Corollry 1, if such WCSQ exists it should be the intersection of two well centered sphericl moons (not orthogonl) with ngles mesure θ 1 nd θ, 0 < θ i < π, i = 1, nd with π x, 0 < x < π s the ngle mesure between them, see Figure 14 q 1 p x q b x 1 q b 1 With the nottion used in Figure 14 one hs, On the other hnd, cos α 1 = cos π θ 1 cos π θ q 1 Figure 14 + sin π θ 1 = sin θ 1 sin θ cos θ 1 cos θ sin x sin π θ cos( π + x) cos = cos θ 1 + cos(π α 1 ) cos(π α ) sin(π α 1 ) sin(π α ) = cos θ 1 + cos α 1 cos α sin α 1 sin α nd cos b = cos θ + cos(π α 1 ) cos(π α ) sin(π α 1 ) sin(π α ) = cos θ + cos α 1 cos α sin α 1 sin α As before, it is strightforwrd exercise to stte the uniqueness of the solution References [1] Berger, Mrcel: Geometry, Volume II Springer-Verlg, New York 1996 cf Geometry I, II Trnsl from the French by M Cole nd S Levi, Springer 1987 Zbl [] d Azevedo Bred, An M: Isometric foldings PhD Thesis, University of Southmpton, UK, 1989 Received Februry 13, 00