Two Strange Constructions in the Euclidean Plane

Tw o Strange Constru ctions in the E uclidean Plane V OLKER TH ¨ UREY No v em b er 1, 20 18 MSC-class: 51N20 Abstract W e present t wo new c onstructions in the usual euclidean plane. W e only deal with ’Grecian Geometry’, with this phr ase we mean elemen tary geometry in the t wo- dimensional sp ace R 2 . W e describ e and pro v e t wo prop ositions ab out ’p ro jections’. The pro ofs need only elemen tary analytical kno wledge. The reader ma y find the foun dations and assumptions of the follo wing prop ositiones in many b o oks ab ou t plane geometry , for instance in [1], p.1-29 . Or y ou can lo ok in [2 ] , [3] , [4] , [5] , [6], [7] . See also [8], p.224-2 34 . Prop osition 1. Let us ta k e R 2 = { ( x | y ) | x , y ∈ R } , the t wo-dimensional euclidean p lane, with the horizont al x -axis and the vertica l y -axis. Assume tw o p arallel s tr aigh t lines G S and G T . Assume a third line L , not parallel to G S , G T , resp ectiv ely , with th e prop erty that L do es not meet the origin (0 | 0). The intersectio n of L with G S is calle d S = ( x S | y S ), and the in tersection of L w ith G T is calle d T = ( x T | y T ) . Note that, in the case that G S , G T are distinct, the three points (0 | 0 ), S, T are n ot collinear. W e can dra w t w o lines Z S and Z T , Z S connects the origin (0 | 0) and S , and Z T connects (0 | 0) a nd T . Z S and Z T are distinct if G S and G T are distinct. No w we distinguish t wo cases (A) and (B) , but note that they o v erlap. (A) : In th e case that G S and G T are not parallel to the horizon tal x -axis, w e ha v e tw o in tersections a S , a T of G S and G T , resp ectiv ely , with the x -axis. Th en th ere is an uniqu e p oint P hor = ( x hor | y hor ) on L , such that ( x hor − a S | y hor ) ∈ Z T , and ( x hor − a T | y hor ) ∈ Z S . (B) : In the case that G S and G T are not parallel to the vertic al y -axis, w e ha v e tw o in ter- sections b S , b T of G S and G T , resp ectiv ely , with th e y -axis. Th en there is an unique p oint P ver = ( x ver | y ver ) on L , suc h that ( x ver | y ver − b S ) ∈ Z T , and ( x ver | y ver − b T ) ∈ Z S . Before reading the pro of of the prop osition y ou should tak e a look on Picture 1 . 1 Picture 1 x y ✲ ✻ a S G T : y = 2 x + 2 G S : y = 2 x + 4 L : y = +1 a T ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ S T ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ ◗ Z S : y = − 2 3 x Z T : y = − 2 x × × P hor P ver b T b S Here: b S = 4 and b T = 2 , a S = − 2 and a T = − 1 , S = ( − 3 2 | 1) and T = ( − 1 2 | 1) . Hence P hor = ( − 5 2 | 1) and P ver = ( 3 2 | 1) . Pr o of. Note th at, if G S = G T , the prop osition is trivial. Hence we assume th at G S , G T are distinct. W e describ e the parallel straigh t lines G S , G T with equations G S := { ( x | y ) ∈ R 2 | e · y = m · x + b S } and G T := { ( x | y ) ∈ R 2 | e · y = m · x + b T } , with e, m, b S , b T ∈ R , ( m, e ) 6 = (0 , 0) . Without less of generalit y let either b e ( e = 0 and m = 1 ) or ( e = 1 ) . The straigh t line L can b e describ ed with t w o num b ers w 1 , w 2 ∈ R , ( w 1 , w 2 ) 6 = (0 , 0) . L := { ( x S | y S ) + t · ( w 1 | w 2 ) | t ∈ R } = { ( x T | y T ) + t · ( w 1 | w 2 ) | t ∈ R } , with w 1 · y T 6 = w 2 · x T , and w 1 · y S 6 = w 2 · x S , ( b ecause (0 | 0) / ∈ L ) , and with e · w 2 6 = m · w 1 , ( b ecause L is not parallel to G S and G T , resp ectiv ely) . Lemma 1. In the case of (A) , (that means that G S and G T are n ot parallel to the horizonta l x -axis), w e h a v e m 6 = 0 , and a S = − b S /m , a T = − b T /m . Then th ere are un iquely thr ee n um b ers , α, β ∈ R whic h solv e the system of four linear equations (1) x S +  · w 1 − a S = α · x T , (2) y S +  · w 2 = α · y T , (3) x S +  · w 1 − a T = β · x S , (4) y S +  · w 2 = β · y S . In the case of (B) , (that means that G S and G T are not p arallel to the v ertical y -axis), there are uniquely three n um b ers e , e α, e β ∈ R whic h solv e the system of four equations f (1) y S + e  · w 2 − b S = e α · y T , f (2) x S + e  · w 1 = e α · x T , f (3) y S + e  · w 2 − b T = e β · y S , f (4) x S + e  · w 1 = e β · x S . Pr o of. In ca se (A) the t wo equations (1),(2) yield  [1] , and the t w o equations (3),(4) yield  [2] ,  [1] = y S · x T − x S · y T + a s · y T w 1 · y T − w 2 · x T and  [2] = y S · a T w 1 · y S − w 2 · x S . 2 Because of ( x S | y S ) , ( x T | y T ) ∈ L , there is a ˘ t ∈ R suc h that ( x T | y T ) = ( x S | y S ) + ˘ t · ( w 1 , w 2 ), hence w 1 · y T − w 2 · x T = w 1 · y S − w 2 · x S . An d with ( x S | y S ) ∈ G S , ( x T | y T ) ∈ G T follo ws easily that y S · x T − x S · y T + a s · y T = y S · a T , hence  [1] =  [2] =:  . In the case of (B) the t w o equations f (1) f (2) yield e  [1] , and f (3) f (4) yield e  [2] , e  [1] = y T · x S − x T · y S + b s · x T w 2 · x T − w 1 · y T and e  [2] = x S · b T w 2 · x S − w 1 · y S . and with similar steps as only just follo ws e  [1] = e  [2] =: e  , and the lemma is prov ed. T o finish the pr o of of p rop osition 1 w e set in the case (A) : P hor = ( x hor | y hor ) := ( x S | y S ) +  · ( w 1 | w 2 ) , and in the case (B) : P ver = ( x ver | y ver ) := ( x S | y S ) + e  · ( w 1 | w 2 ) . The u niqueness of P hor and P ver is trivial, f or instance, for a non v ertical L , the horizon tal distance (with signs) from a p oint on L to Z S or Z T , resp ectiv ely , x 7− → the horizon tal distance (with sign) from a p oint ( x | y ) on L to Z S and x 7− → the horizon tal distance (with sign) from a p oint ( x | y ) on L to Z T , resp ectiv ely , are strictly monotone functions. This w as the last wh at we had to do to pro v e the prop osition. F or completeness, we wr ite d o w n other representa tions of P hor and P ver , resp ectiv ely . Note that if G S , G T are not parallel to the v ertical y -axis, they ha v e equations G S = { ( x | y ) ∈ R 2 | y = m · x + b S } and G T = { ( x | y ) ∈ R 2 | y = m · x + b T } , and if G S , G T are v ertical, they ha v e equations G S = { ( x | y ) ∈ R 2 | x = − b S =: a S } and G T = { ( x | y ) ∈ R 2 | x = − b T =: a T } . If L is not v ertical, we hav e w 1 6 = 0 , and L = { ( x S | y S ) + t · ( w 1 | w 2 ) | t ∈ R } = { ( x | y ) ∈ R 2 | y = m L · x + b L } , with m L := w 2 /w 1 and b L := y S − x S · m L , ( b L 6 = 0 , b ecause (0 | 0) / ∈ L ) . If L is v ertica l , w e s et a L := x S = x T , and L = { ( x | y ) ∈ R 2 | x = a L } . No w assume th at n either L n or G S , G T are parallel to one of the axes. Then P hor = ( x hor | y hor ) =  x T · b L + y T · a S b L | m L · x T · b L + y T · a S b L + b L  =  x S · b L + y S · a T b L | m L · x S · b L + y S · a T b L + b L  =  b L 2 · m + b S · b T · m L − m · b L · ( b S + b T ) b L · m · ( m − m L ) | b L 2 · m 2 + b S · b T · m L 2 − m · m L · b L · ( b S + b T ) b L · m · ( m − m L )  , P ver = ( x ver | y ver ) =  x T · ( b L − b S ) b L | m L · x T · ( b L − b S ) b L + b L  =  x S · ( b L − b T ) b L | m L · x S · ( b L − b T ) b L + b L  =  ( b L − b T ) · ( b L − b S ) b L · ( m − m L ) | m L · ( b S · b T − b L · b S − b L · b T ) + b 2 L · m b L · ( m − m L )  . 3 No w assume th at G S , G T are n ot parallel to one of the axes, and L is horizon tal. ( See the previous picture, to o.) Then w e ha ve an equation L : y = b L , and w e get P hor = ( x hor | y hor ) = ( x T + a S | b L ) = ( x S + a T | b L ) =  b L − b S − b T m | b L  , P ver =  x T · ( b L − b S ) b L | b L  =  x S · ( b L − b T ) b L | b L  =  ( b L − b S ) · ( b L − b T ) b L · m | b L  . Assume that G S , G T are not parallel to one of the axes, and L is v ertical. Then we h a v e an equation L : x = a L , and w e get P hor = ( x hor | y hor ) =  a L | m · a L + b S + b T + b S · b T a L · m  =  a L | ( m + b S a L ) · ( a L + b T m )  , P ver = ( x ver | y ver ) = ( a L | m · a L + b T + b S ) . No w assume that G S , G T are parallel to the horizon tal x -axis, and L is not parallel to the y -axis (and, of course, not parallel to the x -axis, to o) . Then w e get no P hor , and P ver = ( x ver | y ver ) =  x T · ( b L − b S ) b L | m L · x T · ( b L − b S ) b L + b L  =  x S · ( b L − b T ) b L | m L · x S · ( b L − b T ) b L + b L  =  ( b T − b L ) · ( b L − b S ) b L · m L | b L · b S + b L · b T − b S · b T b L  . If w e assume that G S , G T are p arallel to th e horizon tal x -axis, and L is p arallel to th e y -axis, then w e get no P hor , of course, and P ver = ( x ver | y ver ) = ( a L | b S + b T ) . No w assum e that G S , G T are p arallel to the vertic al y -axis, and L is not parallel to the x -axis (and, of course, not parallel to the y -axis, to o) . Then w e get no P ver , and P hor = ( x hor | y hor ) =  a T · b L + y T · a S b L | m L · a T · b L + y T · a S b L + b L  =  a S · b L + y S · a T b L | m L · a S · b L + y S · a T b L + b L  =  b L · ( a S + a T ) + m L · a S · a T b L | m L · x hor + b L  . And finally if we assume that G S , G T are parallel to th e y -axis, and L is p arallel to the x -axis, w e get P hor = ( x hor | y hor ) = ( a S + a T | b L ) . Remark 1. Note a few sp ecial trivial cases. Assume that G S , G T are not parallel to the h orizon tal x -axis (case (A) ) . If S = ( a S | 0) , then we ha ve Z S = x -axis and P hor = S = ( a S | 0) . If T = ( a T | 0) , then we ha ve Z T = x -axis and P hor = T = ( a T | 0) . Assume no w th at G S , G T are not parallel to the v ertical y -axis (case (B) ) . If S = (0 | b S ) , th en we h a v e Z S = y -axis and P ver = S = (0 | b S ) . If T = (0 | b T ) , th en we h a v e Z T = y -axis and P ver = T = (0 | b T ) . 4 No w w e describ e another p rop osition wh ich seems to b e more general, but indeed it is equiv alent, see lemma 2. Because w e pro v ed prop osition 1 , p r op osition 2 also is true. Prop osition 2. Let us tak e R 2 , the t w o-dimens ional euclidean p lane. Assume t wo p arallel straigh t lines G S and G T . Ass ume a thir d line L , not parallel to G S and G T , resp ectiv ely . The inte rsection of L with G S is called S , and the in tersection of L with G T is called T . Assume a fourth line Axis , Axis 6 = L , and Axis is not p arallel to G S and G T . The in tersection of Axis with G S is calle d S Axis , and the intersec tion of Axis w ith G T is calle d T Axis . F urther w e choose a p oint O r ig in on Axis \ L . W e can dra w t wo str aigh t lines Z S and Z T , Z S connects O r ig in and S , and Z T connects O r ig in and T . As ev er y line, Z S and Z T , resp ectiv ely , divide the plane in t w o halfplanes. Z S and Z T are distinct if G S and G T are d istinct. Then there is an unique p oint P ∈ L with the f ollo wing prop erties: W e dr aw the straigh t line Axis P whic h meets P and whic h is parallel to Axis . W e hav e to distinguish three cases, the main one (1) and t w o trivial ones (2),(3) . (1) S Axis 6 = S and T Axis 6 = T . Then Axis P 6 = Axis , th e inte rsection of Axis P with Z S is called S P , and the in tersection of Axis P with Z T is calle d T P . Then the distance of S Axis and O r ig in is equal to the d istance of P and T P , and the distance of T Axis and Or igin is equal to the distance of P and S P . F urthermore, S Axis and P are on the same sid e of Z T , and T Axis and P are on the same sid e of Z S , resp ectiv ely . ( S ee Pictur e 2 ) . (2) S Axis = S . ( Hence, if G S 6 = G T then T Axis 6 = T ) . Then P := S Axis = S , and Axis P = Axis = Z S . The in tersection of Axis P with Z T is T P := O r ig in . Then, b y trivialit y , the distance of S Axis and O r ig in is equal to the distance of P and T P , and furthermore, by trivialit y , S Axis and P are on the same side of Z T . (3) T Axis = T . ( Hence, if G S 6 = G T then S Axis 6 = S ) . Then P := T Axis = T , and Axis P = Axis = Z T . The in tersection of Axis P with Z S is S P := O r ig in . Then, by trivialit y , the d istance of T Axis and O r igin is equal to the distance of P and S P , and furtherm ore, by trivialit y , T Axis and P are on the same side of Z S . Axis O r ig in Axis P G T G S L ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ✏ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ ❤ S T ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✱ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ Z S Z T S Axis T Axis P × T P S P Picture 2 5 Lemma 2. W e ha v e that p rop osition 1 ⇐ ⇒ pr op osition 2 . Pr o of. prop osition 1 ⇐ = prop osition 2: Ob viously , the t w o situations whic h are d escrib ed in prop osition 1 are sp ecia l cases of the general situation in prop osition 2 . More detailed, we ha v e O r ig in := (0 | 0) and if we define Axis := x -axis we get P hor = P , and Axis := y -axis yields P ver = P , resp ectiv ely . prop osition 1 = ⇒ prop osition 2: With an easy transformation of co ordinates, w e get (0 | 0) = Or igin , and x -axis = Axis , hence P = P hor . No w follo ws another p iece of ’Grecian Geometry’. Prop osition 3. Let us agai n tak e R 2 = { ( x | y ) | x , y ∈ R } . with the horizon tal x -axis and the vertic al y -axis. Consider the tw o parallel lines G, P ( P m eans ’pro jection line’) , with the prop erty that G do es n ot meet (0 | 0). Assum e a fi xed ε ∈ R . Let us c ho ose a p oin t ( b x | b y ) on G , b y 6 = 0 , such that neither the line that conn ects (0 | 0) and S := ( b x − ε | b y ), nor the line that connects (0 | 0) and T := ( b x + ε | b y ) is parallel to G and P . W e call S the pro jection of S on the lin e P , and T the pro jection of T on th e line P . (That means that th e three p oin ts (0 | 0) , S, S , and the three p oin ts (0 | 0) , T , T , r esp ectiv ely , are collinear, S , T ∈ P .) The four p oints S , T , − S , − T a re the corners of a parallel ogram. W e call ′ ν ′ the in tersection of the line that connects T and − S with the horizon tal x -axis. F or the claim we d istin gu ish tw o disjoin t cases: (A) If P and G are parallel to th e v ertical y -axis, then ν dep end s only on ε and on the in tersections of the horizon tal x -axis with G and P , resp ectiv ely . (B) If P an d G are not parallel to the v ertical y -axis, then ν dep ends only on ε and on the in tersections of the v ertical y -axis with G and P , resp ectiv ely . ( See Picture 3 ) . In other w ords we claim that the v alue of ν do es not d ep end on the c hoice of ( b x | b y ) on G , and also, ( in case (B) ) , that ν do es not dep end on the slop e of G and P . x y ✲ ✻ S S − S T T ν ( b x | b y ) P : y = 2 · x + 2 G : y = 2 · x + 4 y = 1 2 · x − 1 ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅                           ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ W e fix t w o lines P , G (with slop e 2) , and ε := 4 . W e choose ( b x | b y ) := (0 | 4) . W e get S = ( − 2 3 | 2 3 ) , T = ( − 2 | − 2) . Hence w e get ν = 2 . Picture 3 6 Pr o of. First the trivial cases. If P meets the origin (0 | 0) , then (0 | 0) = T = S , and the parallelogram collapses into a single p oin t (0 | 0) = ν . If ε = 0 , th en T = S , and the parallelogram degenerates to a line b etw een T and − S , that meets (0 | 0 ) = ν . Hence we assume that P do es not meet th e origin (0 | 0) , and we tak e (without loss of generalit y) an ε > 0 . W e distinguish b et ween v ertical G, P and not v ertical G, P . Thus assume v ertical lines G and P with equations G : x = r a nd P : x = p . After choosing a p oin t ( b x | b y ) on G , b y 6 = 0 , we can compu te S and T . Because G do es not m eet (0 | 0) w e ha v e r 6 = 0 , and some easy calculations yield ν = p · ε/r . In the case that G, P are not v ertical th ey ha v e a s lop e m ∈ R , and there are equations G : y = m · x + b G and P : y = m · x + b P with m, b G , b P ∈ R , b G , b P 6 = 0 . After c ho osing a p oin t ( b x | b y ) on G , b y 6 = 0 , w e get with elemen tary calculatio ns S = b P b G + m · ε · ( b x − ε | b y ) and T = b P b G − m · ε · ( b x + ε | b y ) . Some more calculati ons yield the form u la y = m · b x + b G b x · b G + m · ε 2 · [ b G · x − b P · ε ] for a non vertic al straight line that intersect s T and − S , and finally w e get ν = b P · ε/b G . If the lin e th at connects T and − S is ve rtical we get the same f orm ula for ν , and the pro of of the prop osition is complete. Remark 2. The four p oints S , T , − S , − T form the corners of a parallelogram, and, corre- sp ond ing to th e the v alue of ν wh ic h is the intersecti on of the line through T and − S with th e horizon tal axis, the line through S and − T mee ts the same axis in − ν , hence in − p · ε/r , (if b oth lines G, P are v ertical ) , or in − b P · ε/b G (if b oth lines G, P are n ot vertica l ) . Corollary 1 . If w e reve rse the roles of th e x -axis and y -axis, w e are able to form ulate a corresp ondin g statemen t: Consider the t w o parallel lines G, P , with the prop erty that G do es not meet (0 | 0). Assume a fixed ε ∈ R . Let u s choose a p oint ( b x | b y ) on G , b x 6 = 0 , such that neither the line that connects (0 | 0 ) and S v := ( b x | b y − ε ) , n or the line that co nnects (0 | 0) and T v := ( b x | b y + ε ) is p arallel to G and P . W e call S v the pro jection of S v on the line P , and T v the pro jection of T v on the line P . (That means that th e three p oints (0 | 0) , S v , S v , and the th ree p oin ts (0 | 0) , T v , T v , resp ectiv ely , are col linear, S v , T v ∈ P .) Th e four p oints S v , T v , − S v , − T v are the corners of a paralle logram. W e call ′ µ ′ the inte rsection of the line that conn ects T v and − S v with the v ertical y -axis , and we claim that th e v alue of µ do es not dep end on the c h oice of ( b x | b y ) on G . Pr o of. T rivial with the previous prop osition. Again w e write do wn the last pr op osition in a seemingly more general form. Prop osition 4. Let us again tak e the eu clidean space R 2 . Consider the t w o parallel lines G, P ( P means ’pro jection line’) , and an arbitrary third line ( 6 = G ) that w e call Axis . W e fix a p oin t Or ig in on Axis \ G , an d an ε ≥ 0 . Let us c ho ose a p oin t ( b x | b y ) on G \ Axis , and dra w the straigh t line [ Axis , meeting ( b x | b y ) and p arallel to Axis . W e m ark tw o uniqu e p oints S, T on [ Axis , suc h that the distance of b oth to ( b x | b y ) is ε . W e assume the extra prop erty that the line that connects O r igin and S and the line that connects O r ig in and 7 T are not p arallel to G and P , resp ectiv ely . Th us we are ab le to ’pr o ject’ S and T onto P . W e call S the pr o jection of S on the line P , and T the p r o jection of T on the line P , b oth pro jections relativ ely to Or ig in . (That means th at the three p oin ts O r ig in, S, S , and the three p oin ts O r ig in, T , T , r esp ectiv ely , a re collinear, S , T ∈ P .) F urther we denote t w o p oint s − S , − T , s uc h that the f our p oin ts O r ig in, S, S , − S , and the four p oints O r ig in, T , T , − T , resp ectiv ely , are collinear, and the distance of O r igin and S is equal to the distance of O r ig in and − S , and the d istance of O r igin and T is equal to the distance of O r ig in an d − T , resp ectiv ely . The four p oin ts S , T , − S , − T form a parallelo gram with cen tre Or ig in . W e call ′ ν ′ the in tersection of the line that connects T an d − S with Axis , and we claim that ν do es not d ep end on the c hoice of ( b x | b y ) on G . (See Picture 4) . Axis S S − S T − T × T ν ( b x | b y ) P O r ig in G ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ [ Axis ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅                         ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ ✟ Picture 4 Lemma 3. W e hav e that prop ositio n 3 = ⇒ pr op osition 4 . Pr o of. With an easy transform ation of coord inates, we get (0 | 0) = Or ig in , and x -axis = Axis . Remark 3. F or all prop ositions it w ould b e desirable to h a v e a construction with compass and ruler, using the classical metho ds of the ’old Greeks’. Ac knowledgemen ts W e thank Dr. Genc ho S k ordev, V erena Th ¨ urey , Dr. Nils Th ¨ urey and sp ecial ly Prof. Dr. Eb erh ard O eljeklaus f or int eresting discussions, critical commen ts and some new ideas. V OLKER TH ¨ UREY Rheinstr. 91 28199 Bremen, German y T: 49 (0)421 /5917 77 E-Mail: v olk er@thuerey .de 8 References [1] W. G¨ otz, W. Sc h w eize r, W. F ranke , K. Sc h¨ onw ald, ’ Analytisc he Geometrie ’ , Ernst Klett V erlag , Stuttgart 1954 [2] Robin Hartsh orn e , ’ Geometry: Euclid and Bey ond ’ , Sp ringer 1997 [3] D. W. Hall , St. Szab o , ’ Plane Geometry ’ , Prentice- Hall, E nglew oo d Cliffs, New Jers ey , 1971 [4] H. K in der, U. 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