Geometry Beyond Algebra

Transcription

1 r Q a 6 4R Q R 4 4R Q 4R 4Q R 0 r = R h Geoetry Beyond lgebra Hard Hard = Easy The Theore of Overlapped Polynoials (TOP) and its pplication to the Sawa Masayoshi s Sangaku Proble The dventure of Solving a Matheatical Challenge Stated in 8 Jesús Álvarez Lobo lobo@ateaticas.net Spain - 0

2 In eoria of Tere. The solution presented here to the geoetric Sangaku proble proposed by Masayoshi in 8 and for which no solutions were known, don't belong to e, because I was inspired by an angel naed Tere (Maria Teresa Suárez Fernández). I offer this work as a tribute to the woan who decided to live beside e until the last day of her life, and now fro Heaven she guides e with her Light and protects e with her Love. I needed a way to tell her how uch she eans to e, until I can do it by yself, when at last she and I unite in Eternity. Jesús Álvarez Lobo. Oviedo, June 4, 0.

3 BSTRCT. This work presents for the first tie a solution to the 8 unsolved Sawa Masayoshi s proble, giving an explicit and algebraically exact solution for the syetric case (particular case b = c, i.e., BC right-angled isosceles triangle), see (.60) and (.6). Despite the isosceles triangle restriction is not necessary, in view of the coplexity of the explicit algebraic solution for the syetric case, one can guessing the ipossibility of achieving an explicit relationship for the asyetric case (the ore general case: BC right-angled scalene triangle). For this case is given a proof of existence and uniqueness of solution and a proof of the ipossibility of getting such a relationship, even iplicitly, if the sextic equation (.54) it isn t solvable. Nevertheless, in (.56) (.58) it is shown the way to solve the asyetric case under the condition that (.54) be solvable. Furtherore, it is proved that with a slight odification in the final set of variables (F ), it is still possible to establish a relation between the, see (.59) and (.6), which provides a bridge that connects the priitive relationship by eans of nuerical ethods, for every given right-angled triangle BC. nd as the attept to solve Ferat's conjecture (or Ferat's last theore), culinated ore than three centuries later by ndrew Wiles, led to the developent of powerful theories of ore general scope, the attept to solve the Masayoshi s proble has led to the developent of the Theory of Overlapping Polynoials (TOP), whose application to this proble reveals a great potential that ight be extrapolated to other fraeworks. CKNOWLEDGEMENTS The first part of this work was presented at the University of Nagoya (Japan) on 5 ugust of 0, by Hidetoshi Fukagawa, worldwide leading expert in Sangaku Geoetry. It was also overseen by Tony Rothan (Princeton University), Kirsty Klafton (University of Edinburgh), Eanuele Delucchi (Breen University, Gerany), Peter Wong (Bates College, US) and Miguel engual Covas (eber of the coission of the European Matheatical Olypiad). I want to express to all of the y sincere appreciation. I wish to express y gratitude in the ost ephatic for to lan Horwitz (Penn State University, US) to assue the unpleasant task of checking y work, as an author authorized in arxiv to endorse e to subit y article.

4 Introdution. Proposed by the Nippon atheatician Sawa Masayoshi in 8, this proble reained unsolved. s shown in the picture below, an ellipse E is inscribed in BC, right-angled triangle at, with its ajor axis parallel to the hypotenuse a of BC. Within the ellipse are inscribed two circuferences of radius r, C and C, that are tangent to the ellipse and each other at the center of the ellipse. Inside-tangent to the sides b and c of the right angle and outsidetangent to the ellipse, there is another circuference, C, with the sae radius r. Challenge: find a relationship between a, b, c and r. B a C r C r r C c C b The Travel Diary of Matheatician Yaaguchi Kanzan. I have arrived at Tatsuno, city near Hieji, and visited the Syosya teple to record a sangaku proposed by Sawa in 8. In the evening, visited e and he showed e an unsolved proble and two sangaku probles of the Syosya teple. I have written the down in y diary as follows:... Fro the book Sacred Matheatics. Japanese Teple Geoetry, by Fukagawa Hidetoshi and Tony Rothan. ccording to the brilliant atheatician Paul Erdös, ath proble that takes ore than a century to be solved, it is a nuber theory proble. nd I would add: If a seeingly siple atheatical proble, because it hides an underlying extree coplexity, takes ore than a century to be solved, ost likely it is that it will be a nuber theory proble... or a geoetric Sangaku proble, and only an exhaustive analysis could lead to its solution. Jesús Álvarez Lobo. The Masayoshi s proble kept a surprise: the explicit relationship between the length of one side of BC and the radius of the circuferences or is so extensive and intricate that it cannot be written without fragentation, or both the path to prove the ipossibility of getting a relationship for the asyetric case and the equations for the alternative case are huge and of an bewildering coplexity, despite the valuable aid provided by the TOP. Everybody can believe in the existence of ore siply relationships, (that s easy!), but the reality is that, up today, the ones presented here are the unique known. Soe approaches, cut-off paths and other collateral interesting results found through the surveying, has been oved to final appendixes, in order to lighten the reading. In spite of the proofs are carefully developed, soe steps have been oitted for brevity. However, it isn t hard to follow the exhaustive evolving of the work.

5 0. Theory of Overlapped Polynoials. 0.. Definitions. Definition. Class On ( x r ) of Overlapped Polynoials: Set of all polynoials of degree n in the variable x which have the root x = r with ultiplicity not lower than. Definition. Monic Transfored of a Polynoial: M = n n i aix i i= 0 an i= 0 a x i (0.0) 0.. Lea of Overlapped Polynoials (LOP). The difference of the onic transfors of two overlapped polynoials of degree n in the variable x, that share the root r with ultiplicity not lower than, it is a polynoial of degree < n that keeps the root r with ultiplicity at least : S( x), T( x) ( x r) O M{ S( x)} M{ T( x)} O ( x r), n n (0.0) Proof. M{ S( x)} = ( x r) s( x) ( )} ( )} ( ) ( ) ( ) M{ S x M{ T x = x r s x t x M{ T( x)} = ( x r) t( x) (0.0) where s and t are onic polynoials of degree n, n n n i n i i = s( x) = x + a x n n n i n i i = t( x) = x + b x (0.04) n s( x) t( x) = ( a b ) x i = n i n i n i (0.05) Therefore, n n i n i n i = n i M{ S( x)} M{ T( x)} = ( x r) ( a b ) x O ( x r) (0.06) Be noted that, by Viète-Girard forula, the degree of S(x) - T(x) is less than n - iff the su of the roots of s(x) is equal to the su of the roots of t(x), the degree of S(x) - T(x) is less than n - iff also the su of the binary products of the roots of s(x) is equal to the su of the binary products of the roots of t(x), and so on... (0.07)

6 0.. Lea of Overlapped Polynoials (LOP). M S T x r S T x r k n ( (, On ( ) ST M Ok ( ), (0.08) ( ( denoting S, T the derivatives of order - of S and T, respectively. Proof. S(x) and T(x) can be factored as in (0.0), ( ) ( - ) S x = x r s( x), T ( x) = ( x - r) t( x), where s and t are the onic polynoials defined in (0.0). pplying the Leibniz s forula for the - th derivative of the product to S(x), S( x) = {( x r) s} = {( x r) } s i = 0 i ( ( ( i ( i (0.09) and siilarly for T(x), T( x) = {( x r) t } = {( x r) } t i = 0 i ( ( ( i ( i (0.0) It is straightforward that the ters of the sus (0.0) and (0.) are all of degree n + and contain the powers of x r with integer exponents between and, so, S( x) ( = ( x r) p( x) ( T( x) = ( x r) q( x) (0.) where p and q are polynoials of degree n, Hence, ( ) = n k p x p x ( ) = n k q x q x k k = 0 k= 0 n ( M S = M ( x r) p( x) = ( x r) M p( x) = ( x r) pix p k n i = 0 i (0.) (0.) n ( M T = M ( x r) q( x) = ( x r) M q( x) = ( x r) qix q n i = 0 i (0.4) n n ( ( i i MS MT = ( x r) pix qix pn i= 0 q n i= 0 (0.5) But the suations of (0.5) are onic polynoials of degree n -, therefore, applying the LOP, n n ( ( i i ST MS MT = ( x r) pi x qi x k ( x r), k n O pn i= 0 q n i= 0 (0.6)

7 0.4. Theore of Overlapped Polynoials (TOP). This theore is an extension of LOP, reoving the restriction of equal degree of the overlapped polynoials involved. Therefore, the TOP represents the widest generalization of the Theory of Overlapped Polynoials. The theore is stated as follows: S On ( x r) T O ( x r) ( ( ST MS M( x T) Ok ( x r), k n n in, (0.7) Proof. S On ( x r) T O ( ) n( ) n( ) x r S O x r O x r n x T O+ ( x r) = On ( x r) On ( x r) in, (0.8) where the inclusion relationship are due to the nested characterization stated in the definition of the overlapped classes of polynoials. Hence, applying the LOP, M S x T x r S x T x r k n ( (, On ( ) ST M ( ) Ok ( ), (0.9) 0.4. pplications of the TOP. In the Masayoshi s proble, to relate the radius of the circuferences inscribed in the ellipse with the radius of the outside tangent circle to the ellipse, is necessary to calculate the coordinates of the tangent point T as a function only of b, c and r, but the equations that deterine it are coplete quartics of a extree coplexity. On the other hand, as should be shown below, it is possible to define three independent quartics in the abscise x T of T. t first glance, it could look like superfluous to state three different condition for the sae purpose, but aazingly it is possible to get advantage of this circustance, what can be crucial in certain conditions. In general, there ay be several possibilities, depending on the degree of ultiplicity of the root shared and on the depth (order of the derivative) at which the theore be applied. The tangent point T of two curves is twofold and this reasoning led to the developent of the Theory of Overlapping Polynoials in the generalized for in which it is presented, i.e., considering ultiplicities of the roots. However, we ust consider that ultiplicity cannot be inferred fro the ere consideration of the geoetry of the proble (lgebra is generous and often gives ore than is asked!), since in the transforation of rationalization (squaring) strange solutions can be introduced. t this proble the TOP has been ipleented in its siplest for (considering siple roots), but even so a unit reduction in the degree of an equation can be crucial to achieve a solution, and will always represent a substantial siplification of the calculations and the solution.

8 Moreover, the TOP can be applied repeatedly, thereby achieving further reductions in the degree of the equations. For instance, fro three polynoial equations of any degree bigger than that share a root x = r, by successive applications of the TOP, is possible to reduce the degree of the polynoial equation of least degree in two units. For exaple, suppose you have two equations of 6 th degree and one of 5 th degree for the deterination of a certain agnitude s. t first, the proble should be algebraically unsolvable, but it ight go beyond the liits of lgebra by applying the TOP: with a first application, it is we obtained a quintic fro the two sextics, and then, applied to the two quintics gives a quartic which is already solvable algebraically. Suppose now that you have three equations of 6 th grade for the deterination of a certain agnitude s. Since the sextics are, in general, algebraically unsolvable, it looks like that the proble hasn t algebraic solution, but it possible trespass the liits of lgebra by applying the TOP: applied at a pair of sextics (for instance, to the first a second) we get a quintic, and applied to other pair of sextics (for exaple, first and third) we get another quintic. Finally, the application of the TOP to the pair of quintics derived fro the sextics gives us a quartic that is always solvable by radicals. It is not particularly unusual to find different equations for the deterination of a value of a variable under certain conditions, as probles often lets different approaches leading to different equations. Usually the equation will be of the sae degree as they have to contain all the necessary inforation to define the proble, but even if they have different degrees it is possible to apply the TOP. Let s see a concrete exaple to clarify the way the TOP is applied. Suppose you have three polynoials, u x x x x x x x ( ) v x x x x x x 5 4 ( ) w x x x x x 5 4 ( ) (0.0) (0.) (0.) that share certain unknown root, x = r. pplying the TOP to u and v, u x r O6 ( ) O5 ( ) 0 uv v x r Mu Mxv Ok ( x r), k 6 = = (0.) i.e., u( xw) = u xv = x + x x + x x (0.4) Now we have three polynoials of 5 th degree that share the root x = r, and applying the TOP to the three pairs of quintic equations we will obtain three quartic equations that preserves the sae coon root. Let us denote with u w the quintic inherited by eans of the TOP fro the sextic u and the quintic w.

9 ( u w) v Mu w Mv = 6u w v = 9x x + x x ( u w) w Mu w Mw = 6u w w = x x + 6x x v w 0 4 vw = M v M w = v w = x x + x + x 6 6 (0.5) (0.6) (0.7) Now we have three polynoials of 4 th degree that share the root x = r, and applying the TOP to the three pairs of quartic equations we will obtain three cubic equations that preserves the root x = r, u w v u w w u w v u w w = 46 x x 56 x ( ) ( ) M( ) M( ) u w v v w = u w v v w = 46 x x 56 x ( ) M( ) M u w w v w = u w w v w = 87 x x 50 x ( ) M( ) M (0.8) (0.9) (0.0) If we would try to ipleent now the TOP to any pair of the above cubic equations, for instance, M u w v u w w M u w vv w ( ) ( ) ( ) 0 (0.) we would get the null polynoial because the three equations are equivalent. So, it is ipossible to further reduce the degree of the equations. Solving any of the cubic equations, for instance, get, x=, x= 8 8 (0.) one of whose roots is that which is shared by the three initial equations. To deterine which of the three roots is what is seeking is necessary to use soe criteria inferred fro the conditions of the proble. In this case, the root shared is x = /.

10 . Syetric case: b = c. In the orthonoral Cartesian reference syste provided in the for shown in the figure below, the equation of the ellipse is reduced to its canonical for and geoetric configuration has axial syetry on the coordinate axes. y C T C O C x B M C Equation of the ellipse in canonical for (center at the origin and ajor-axis of the ellipse coincident with the x-axis): x y E + = (.0) Let t be the straight line defined by and C: t y = x + y (.0) Fro the intersection of t and the ellipse E : x ( x+ y ) t E + =, that can be written as, + 0 x yx + y = (.0) However, the necessary and sufficient condition for t to be tangent to E at T is that the last equation has twofold roots, or, equivalently, discriinant null: or, t E = T = ( y) 4 + ( y ) 0, = y + ( y ) 0 0, = y + + y = = +

11 and for the positive sei-ellipse, y = + + (.04) Furtherore, straight by siple observation of the picture are obtained the coordinates of, as C is the diagonal of a square of side r. Hence, (, ) = ( 0, + + ) x y r r y = ( + ) r+ (.05) The pair of equations (.04) and (.05), that arises in a fortunate choice of the reference syste, will be crucial to the resolution of the proble. Equating these two expressions of y, we have: that can be expressed as + = + ( + ) r + = ( ) r ( ) r, ( ) = r r (.06) In order to copact the expression, let us denote p = + (.07) Then, (.06) is written as, p = r+ r, p (.08) and fro this one, = + r p p r (.09) Intersection of C and E : x y E + = x r ( x r) E C, + = C ( x r) + y = r quadratic equation in x that can be written as, x rx + = 0 (.0)

12 Now, taking into account that, by definition, the eccentricity of the ellipse is the equation (.0) can be expressed as, = = = x rx + = 0 (.) (.) Moreover, the condition of tangency between the circle with center at C and the ellipse is that the contact occurs only between two points of equal abscissa, what will happen iff the discriinant of (.), as a quadratic equation in x, is null, and this condition is translated in the following relationship: that iplies, = = 0 ( r) 4 0, r = (.) However, should be noted that the axiu radius of the circles inscribed in the sei-ellipse of sei-ajor axis is /. Whereas as a constant, r can be expressed as function of as the only variable: but in the context of the proble, r =, r = ( ) = = Let us study the increasing / decreasing of this function: r r( ) = r ( ) = = 0 =, r ( ) 0, 0 r ( ) 0, (.4) (.5) 0 as can be checked trivially. So, r reaches a relative and absolute axiu for i.e., Hence, the doain of r( ) is,, = of value r ( ) ( ) r( ) ax = r = Do r( ) =, = =, (.6) (.7)

13 The left end of the doain of r ( ) coincides with the iniu value of the eccentricity for which the radius of the circles inscribed in the sei-ellipse reaches a axiu. This liit value can also be deterined as follows: i.e., r = = = =, = 0 (.8) biquadratic equation whose unique real and positive root is =. The circle with center at C inscribed in the right sei-ellipse has axiu radius r = for, as it only has contact with the ellipse at the right apex (point of tangency). If = the ellipse is degenerated in a segent. In short, if, r = if 0, (.9) But r = cannot be a solution of the proble. Y T C O X B M C Indeed, referring to the figure above, according to (.04) and (.05), y = O = + = ( + ) r +. Hence, + = ( + ) r + ( + ) r +, ( + ) r + ( + ) r = 0 (.0) quadratic equation in r, whose solutions are, ( )( ) r = +.

14 Discarding the negative root, that has no sense in this context, ( )( ) r = + (.) and fro (.), we obtain: ( ) ( ) ( ) ( ) r = ( + ) = ( + ) = = r = r. Consequently, the solution of the proble, unique due the onotony of r ( ), is confined to the doain of r ( ), established in (.7). Now, eliinating r in (.09) through the relationship (.), which links the radius of the inside the ellipse circunferences with the sei-ajor axis of the ellipse and its eccentricity, we obtain: p p r = + p = p p + = p + fro which it is obtained the coplete quartic equation in, r r = ( ) ( ), 4 + p p + p = 0 (.) By the lineal Tschirnhausen transforation, p (.) is eliinated the ter of third grade in (.) and, after replacing p by its value (.07) and siplifying, we get the depressed quartic: ( ) = (.4) nd denoting the coefficients of this equation by 5 C, D 4 6, E 6 4 (.5) the equation (.4) can be written as, 4 + C + D + E = 0 (.6) Copleting a square on the left side and transposing the other ters to the right side, we have, C + D + E = 0 + C + C = C D + C E, ( + C) = C D + C E (.7)

15 Let us transfor (.7) by introducing the variable thus, ( + C) + = C D + C E + ( + C) + ( + C) (.8) This is the crucial trick, since for soe the right side of (.8), which is quadratic in, will be a perfect square, and the necessary and sufficient condition for this is that the discriinant of is null, cubic equation in : ( C + ) D + C E + C + = 0 = = 0 ( D) 4( C ) C E C 0 5 C ( ) ( ) C C E D = 0 8 C E (.9) (.0) (.) Denoting i.e., 5 C C( C E) D c d C E e 8,, 5(6 ) c, d, e (.) (.) we can write (.) in a ore copact for + c + d + e = 0 (.4) and applying again a Tschirnhausen transforation, c z (.5) is eliinated the quadratic ter and we get the depressed cubic: Denoting its coefficients by we can write (.6) siply as 7 z + 4 z+ = , 08 z + z+ = 0 (.6) (.7) (.8)

16 To solve (.8), let us ake the change, z u + v (.9) Then, z z u v u v u v uv u v u v + + = 0 ( + ) + ( + ) + = ( + ) + ( + ) + = 0, i.e., u v uv u v + + ( + )( + ) + = 0 (.40) If we choose u, v so that + uv = 0, we get two equations: u + v = uv = (.4) (.4) Equations that give us the su and product, respectively, of the cubes of u and v, which allows us to construct a quadratic equation whose roots are these cubes: The solutions of (.4) are, With no loss of generality, we can write: t + t = 0 t = + (.4) (.44) u = + +, v = + (.45) Thus, by (.9), a solution of (.8) is given by (Cardano - Tartaglia forula): z = (.46) Note: this root is real for being 0. ccording to (.7), the value of the discriinant of (.8) is, 69 + = 08 8 (.47) Let us write (.46) in a ore copact for, z = + + (.48)

17 Undoing now the variable change (.5) and the value of c given by (,), 5( 6 ) = (.49) For the value of given by (.49) the quadratic equation (.9) has a double root (since its discriinant vanishes). Therefore, (.8) can be expressed as, D ( + C) + = C+ ( + ) C (.50) Extracting the square root of both sides of (.50) and siplifying the right side, D + + = + + C C C (.5) Fro (.5) we get two quadratic equations in, D = 0 C + C C D = 0 C + C C (.5) (.5) and each of these ones provides a pair of solutions of the depressed quartic (.6): D = + C+ C C + D = C+ + C C + (.54) (.55) nd finally, undoing the change of variable (.) and the value of D, given by (,5), 4( ) = + C+ + C C ( ) = C + C i C + (.56) (.57) where the values of C,, and are given by (.5), (.7), (.47) and (.49), respectively. Obviously, in the context of the proble the last two roots has no sense, and the second, by the discussion ade above, it is not valid. So, the only acceptable solution is the first, i.e., r = r = r =

18 Relationship between b and r. Y C R H T C O C R V X B M C Straight fro the above picture, we get the height M of the triangle CB : M = + r + r = + r( + ) (.58) Since MC is an isosceles right triangle, the Pythagorean Theore gives: i.e., r b = C = M = r( ) + + = + r( + ) = r + +, b r = + + (.59) Therefore, an explicit and algebraically exact relation between b and r is b 4 = + + r 4( ) + C+ + C C + (.60) where, 08 5 k C ( ) 69 + k = 08 9 (.6)

19 Graphic representation of r. For a geoetrically exactly representation of the circle inscribed in the right sei-ellipse, its radius can be deterined as shown in the above drawing, where R is the intersection point of the circle with center at the idpoint of OV and diaeter and the circle with center at V and diaeter. Then, RR OV RR = r (.6) Indeed, VRO is rectangle in R, and verified: VO = VR = RR VR RO = RR = VR = = = r RR = r RO = RO VO VO RR O VRO (.6)

20 b r pproxiate value of the ratio b with 000 decial digits: r

21 . syetric case: b c... Existence and uniqueness of solution. By observation of the below picture can be seen straightforward that the solution is unique in each quadrant, since the radius of the circles inscribed in the ellipse decreases as the eccentricity of the ellipse increases, according to (.5) and (.7), while the radius of the outside circle grows. Therefore, can only be a value of the eccentricity for which the radius of the three circles are equal. Hence, there is a bijection r. On the other hand, the eccentricity increases when decreasing the angle and, consequently, the radius r increases as increases. There is, therefore, also a bijection r, that applies (90, 0) (r ax, 0). (0, + ) B O C The radius for the three circles inscribed reaches a axiu when the value of eccentricity is the one obtained for the particular case studied before ( = 90 BC is a right-angled isosceles triangle). But now the BC has no axial syetry, so it is expected a significant coplication in the calculations. The outline of several approaches (shown in appendixes B, C, D and F) fro which outcrops an intractable coplexity, allows glipsing the ipossibility of obtaining an explicit algebraic solution. Despite that the extreely coplexity of the relationship gotten for the syetric case, doesn t represents a atheatical proof of the ipossibility of getting relationships for the asyetric case, indeed it is a proof for the intuition and coon sense: the asyetry increases the coplication, both in the outlines as in the difficult of the equations. Nevertheless, the aforeentioned ipossibility will be rigorously translated to an algebraic condition: the solubility of a sextic, (.).

22 .. Outlining the path to the solution: What we need for defining algebraically the geoetric conditions?. First of all, we ust see the proble in a bird view, looking for the necessary equations for translate the geoetric picture into an algebraic description. In the study of the syetric case we had yet deterined the equation (.), that describe the condition of tangency iposed to the circuferences C and C, and this relationship is also valid for the asyetric case, since it doesn t depend on BC. Other condition is that the ellipse E has to be tangent to the three sides of BC. The condition of tangency aong E and the hypotenuse a is trivial for being parallel to a the ajor axis of E. The condition of tangency aong E and the other two sides, b and c, will be deterine cutting E with the lines that contains b and c, and iposing that the intersection verifies in a single point (the intersection of a line and an ellipse is deterined by a quadratic equation, and the condition of tangency iplies that its discriinant is null). The condition of tangency between C and b and c is given siply iposing that the center C of C belongs to the bisector line of the right-angle. The condition of tangency between C and E can be given iposing that, at an undefined point T of C and E, both curves share the tangent line, and this will occur iff the derivatives of C and E take the sae value at T (i.e., tangents equal slope). The equation of E involves two paraeters, and, and the relationship (.) a third paraeter,. Therefore, we need, at least, four independent equations for reoving these three paraeters. But by the definition of eccentricity of the ellipse we have an additional relationship between the three paraeters involved, and by ean of this one and (.) the reaining equation can be easily reduced to a ono - paraetric for, for instance, to an - paraetric for. It is necessary to find a third relationship, an - paraetric equation independent of the coordinates of T, for expressing in the set of the final variables, F = { a, b, c, r }. Let s denote this hypothetic relationship as F. Substituting now F into the equation derived of the condition of tangency between C and E, we ll find an expression for one coordinate of T, (for instance, the abscise x T) in ters of the FVs, and using the equation of C we could find the other coordinate of T (for instance, the ordinate y T) in ters of the FVs. Finally, equaling to r the distance between C and T, we would find the searched relationship aong the variables of F. Nevertheless, we ll see that the hardest difficulty is to find an explicit relationship F, that would be the key for the door to the solution, but, as it will be shown later, F is, in fact, the key for proving that it is ipossible to find a relation in the set F = { a, b, c, r }, neither explicit nor iplicit, since the equation relating r to is of 6th grade in, without any especial relation of syetry in its coefficients, so unsolvable by radicals. However, taking as final set of variables F = { a, b, c, }, it is indeed possible establish a relationship, because the equation relating r to is of 4th grade in r, therefore solvable algebraically. Getting a relationship in F = { a, b, c, } allows to get another one in F = { a, b, c, r } for a given right-angled triangle BC, by eans of nueric ethods.

23 .. Switching the reference syste. The previous reference syste provides the siplest expression for the equation of the ellipse (reduced to its canonical for), but instead, the equations for the reaining eleents result too coplex (in appendix B we show its expressions). Let us take now the Cartesian reference syste showed in the figure below, obtained fro the previous one by a rotation and a translation of axes: rotation of 80º (keeping the parallelis of the coordinate s axes with respect to the axis of the ellipse) and translation of the origin to the vertex. Y C M B T r C O X.. Equations in the new reference syste. To relate the new coordinates with the old, just switch x by x + h, and the priitives coordinates of C with the new ones. The slope is invariant under the transforation perfored (80 degree turn and translation). In this new scenery, the equation of the bisector of, and the equations for the coordinates of C, are significantly siplify. The slope of the bisector of is, tan + tan 4 + tan tan + = =, 4 tan tan tan 4 where tan is the slope of the side B. Setting, tan (.0) the equation of the bisector of is as siply as + y= x (.0) The new coordinates of C can be obtained as intersection of a circuference with center at the origin and radius equal to the diagonal of a square of side r. ssuing > 0, + y = x + + x = r x + y = ( r ) x = r + ( + r ) y = + (.0)

24 and expressing as a function of the sides of BC, x c b r = y a ( c + b) r = a (.04).4. Centre of the ellipse: analytic and geoetric deterination. The equation of the ellipse with center at E(x 0, y 0) is, ( x x0) ( y y0) E + = (.05) and the equation of the line deterined by the vertex of and B, y = x (.06) The intersection of this line and the ellipse is given by, or, ( x x0) ( x y0) + = ( x x 0) + ( x y0) = 0, ( + ) x ( x + y ) x + x + y = (.07) but the line (.06) will be tangent to the ellipse (.05) iff the discriinant of the quadratic equation in x (.07) is null, and siplifying, x0 y 0 x0 y0 = ( + ) 4( + )( + ) = 0 ( x ) x y + y = (.08) (.09) Substituting in (.09) by -/ we obtain a siilar expression for the condition of tangency of the line deterined by the vertices and C with the ellipse (.05): x + x y + ( y ) = (.0) dding equations (.09) and (.0) side by side, and factoring, we have, ( + )( x + y ) = (.) that iplies, + = x + y 0 0 (.) This condition is also obtained considering that the vertex belongs to the director circuference (see ppendix B).

25 Fro (.), = x + y 0 0 (.) and substituting in (.09) or (.0) by the right side of (.), ( y )( ) = x y (.4) Into this equation hides an interesting geoetric relationship, which coes to light when expressed thus, since, y = x y = tan (.5) (.6) Using (.6) and writing right now (.4) in the shape, x0 ( y0 + )cot = y y 0 0 (.7) follows a useful geoetrical construction to deterine the center of the ellipse in ters of : Y C L M B L G E H M O X GH L M LM x0 ( y0 + )cot = = = OG OL OL y y 0 0 (.8) But OBM and OCM are isosceles because both triangles have two equals angles; hence, M is the iddle point of the hypotenuse of BC, and the vertex H of the ellipse is siply the intersection of M with the parallel to BC by G, where GL = and M is the projection of M over the ajor axis of the ellipse.

26 Fro (.7), taking into account that y 0 + is the height of BC relative to its hypotenuse, and can be expressed as, since is the slope of c in BC and = tan B = b. c y 0 bc + = a (.9) Solving for x 0 in (.7) and replacing y 0 + by the value given by (.9), since tan = b/c, y0 x0 = ( y0 + )cot = y 0 bc a c b bc, y 0 c b x0 = a y 0 (.0) Note that, b c + x + x = = =, that is, the abscise of the iddle point of BC. a c b a a C B xm Let s denote, c b k a bc h a (.) (.) Theore. The locus of the centers of the ellipses with inor sei-axis of length inscribed in a right-angled triangle BC, with ajor-axis parallel to the hypotenuse, in a Cartesian reference syste with origin at the vertex of the rightangle and abscissa axis below the hypotenuse and parallel to it, belong to the hyperbole, y = x k (.) Reoving y 0 in (.0) by eans of (.9), we obtain the - paraetric equations of the center of the ellipse: h x0 = k y0 = h (.) and by eliinating through (.) we get the - paraetric equations of the center of the ellipse: Notice the siplified relationship, r x0 = k h r r y0 = h (.4) a h + k = (.5)

27 .5. Coordinates of the tangent point T between C and E. For deterining the point of tangency between the C and E, as shown below, it is possible to raise three independent equations. t first glance, defining in three different ways the sae object (unknown xt) ight see unnecessary work, and therefore superfluous. However, is quite the opposite, since the ipleentation of the TOP (which displays here soe of its potential) reduces the three quartic to a siple quadratic equation which preserves the root of interest (the abscissa of T) shared by the three quartics. The TOP acts synergistically on the equations, so that the joint action of the equations is uch ore effective than just the su of their individual actions. Below are defined the three quartics..5.. Quartic for x T derived fro the intersection of C and E. ( x x0 ) ( y y0) ( x x0) + = y= y 0 ( x x0 ) y0 = y r ( x x ). ( x x ) ( y y ) r y y r ( x x ) + = = Taking into account that the contact take place between the lower sei-ellipse and the upper sei-circuference, the signs of the radicals ust be in the equation of the ellipse and + in the equation of the circuference: ( x x ) y y r x x 0 0 = + ( ) (.6) Strealining, developing and collecting ters (used the DERIVE coputer algebra syste), there arises a quartic equation, that we denote by u(x) = 0, showing the following dissuasive-looking: (.7).5.. Quartic for x T which is obtained by equating the derivatives of C and E. in its explicit for. Derivatives of the upper canonic sei-ellipse E and the lower sei-circuference C : x x ( x x ) ( y y ) ( x x ), = y = y 0 y = ( x x0 ) y = x0 x (.8)

28 x x ( x x ) + ( y y ) = r y = y r ( x x ) y =, r ( x x ) y = r x x (.9) the derivatives of both curves at the tangent point T represent the slope of the coon tangent line; hence, they have to be equals: r = = r x x x0 x x0 x x x r = x x x0 x (.0) where can be expressed in ters of r and, r = ( ) (.) Fro (.0) and (.), or, r r r x0 x x x ( )( x0 x) ( x x ) ( ) = = ( )( x0 x) ( x x ) + r ( ) r ( x x ) = 0 (.) Developing and collecting ters in x we obtain the following quartic denoted v(x) = 0: (.).5.. Quartic for x T got by equating the derivatives of C and E. in its iplicit for. Deriving iplicitly the equations of both curves, ( x x ) ( y y ) ( x x ) ( + y y ) 0 x y= y= x y y E + = C ( x x ) + ( y y ) = r x x 0 0 x x ( ) + ( y y) y = 0 y = y y 0 (.4) (.5)

29 But C and E share the tangent line at T, hence, their derivatives (slope of the tangent line) take the sae value at T: x x0 x x = y y y y 0 (.6) Paraeterizing (.6) in, eliinating denoinators and expressing it in explicit for, ( ) ( x x0 )( y y) = ( x x )( y y0) (.7) y = y ( )( x x ) y ( x x ) 0 0 ( ) ( x x0) ( x x) (.8) Replacing by eans of (.8) the variable y in the equation of C, C y ( )( x x0 ) y0( x x ) ( x x ) + y = r ( ) ( x x0) ( x x) 0 0 ( x x ) r x + x ( ) x ) + ( y y ) ( x x ) = 0 (.9) (.40) after a long chain of operations and siplifications, we get the following quartic that we denote by w(x) = 0, (.4) or, ore copact, (.4) The quartic equations (.7), (.) and (.4) have to share the real root x = x T at the tangent point T. This property can be exploited applying the Theore of Overlapped Polynoials (TOP), very useful for dealing with the enorous coplexity of these equations. By applying the TOP three ties, we get a quadratic fro three quartics. First, applying the TOP to the quartics u an v it is obtained a cubic, denoted uv, then, applying it to the quartics v an w it is obtained other cubic, denoted vw. Finally, the application of the TOP to this pair of cubic equations it is obtained a quadratic equation, denoted uvw. The algorith can be scheatized as follow: (.4) u, v ( x x ) u v ( x x ) O4 T uv M M O T uvw M uv M vw O x xt O4 T vw M M O T v, w ( x x ) v w ( x x ) ( )

30 where uvw is a quadratic equation one of whose roots is the abscissa x = x T of the tangent point T, inherited fro the three quartics that shared it, by eans of three overlapping onic polynoial transforations. Developing the algorith, after doing heavy calculations derived fro the enorous coplexity of the polynoials involved, fro the overwheling ocean of nubers and letters eerges the following expression for the quadratic equation uvw that gives the abscissa of the tangent point T: (.44) or in a ore copact for, (.45) But the equation (.7) ust have a double root, as the tangency point is the liit to which tend both points of intersection of the circle and the ellipse. Therefore, we can replace the quartic (.7) by its derivative, which is a cubic: (.46)

31 Using (.46) instead of (.7), the algorith to obtain the quadratic equation by iteration of the TOP would be: (.47) u x x u x x O4( T) O( T) u vw Mu M vw O x xt O4 T vw M M O T v, w ( x x ) v w ( x x ) ( ) Thus, we have the abscise x T as a function of a, b, c, r,. Replacing x T in the equation of C it is obtained the corresponding ordinate y T, also as a function of a, b, c, r,..6. Defining the proble. Let s gather now a coplete set of independent equations to define the proble, translating geoetrical eanings to algebraic descriptions. Condition of tangency of C and C with each other and with E. Fusing (.) and (.7): r =,, (.48) Condition of tangency aong the ellipse E and the sides a, b and c of BC. Unifying (.) and (.4): r r + = k + h h r (.49) where h and k are constants that depend only of a, b, c according to (.) and (.). Condition of tangency aong the outside-circuference C and the ellipse E. There are three ways of setting this condition, iposing that: C and E share the tangent point T, or the noral line at T contains the center C of C, or C and E share the tangent line at T. It will be use (.44) and the equation of C. Condition of tangency between the outside-circuference C and the sides b and c of BC. x c b r = y a ( c + b) r = a (.04) Condition of that the distance between the center C of C and the tangent point T is r. ( x x ) + ( y y ) = r T T (.50) Relationship aong the paraeters involved in the proble (lengths of the sei-axis of E and its eccentricity): = (.) The condition of equal radius for the three circuferences it is yet included iplicitly in the above equations since are used the sae variable r for all the. This is the siplest syste of equations for describing the Masayoshi s proble, but subjacent to its apparent siplicity is hidden an extreely coplexity that outcrops when trying to reoving the involved paraeters, as shown below.

32 .7. Reoving the paraeters. Fro (.) and (.48), r ( ) =,, (.5) and fro (.48), r =,, (.5) Now only reains to eliinate the paraeter, and there is just a way to try it: by eans of (.49), (.5) and (.5). Reoving and in (.) by eans of (.49) and (.5), and applying (.4) to the right side of (.), r r k r h r + = + ( ) h r (.5) and after a tedious series of operations, expanding and siplifying by eans of (.), (.) and (.5), it is obtained, r Q a 6 4R Q R 4 4R Q 4R 4Q R 0 r = R h (.54) Here it is, eureka!: the asyetric case is solvable iff the sextic (.54) is solvable algebraically. There is no further way to eliinate, since there are no ore independent equations that interrelate these paraeters than those that have been used yet. It is, therefore, likely ipossible to eliinate the paraeter and, consequently, therefore, it is ost likely that nor be possible to get a relationship between the variables a, b, c and r that doesn t involve any of the paraeters,, or. See appendix C for the analysis of solubility of this equation..8. Final relationship between a, b, c and r. If would be possible to solve the sextic (.54), denoting the right root (that corresponding to the eccentricity of E ) by ( a, b, c, r) r (.55) the coordinates of T would be also functions of a, b, c, r, and (.50) would state an iplicit relationship aong r and a, b, c: ( ) ( ) ( ) ( ) T T x r x r + y r y r = r (.56) where x T is one of the two roots of (.44), y T is the function obtained substituting x T in the equation of C, x and y the functions defined in (.04).

33 .9. pproxiate solution by iteration. Writing (.56) in the for, r = x ( r) x ( r) + y ( r) y ( r) ( a, b, c,, r) T T r (.57) the algorith for n iterations would be, r = ( r0) r = ( r) rn = ( rn ) (.58) where r 0 is an initial value, arbitrarily chosen, but in the range of convergence of the algorith. The value of n depends on the desired accuracy and the algorith terinates when the values of r becoe stable for a certain nuber of digits. Obviously, if is known the value of r it is trivial to define one (or any nuber) relationships between the sides of the triangle and r.

34 .9. n alternative way. However, by rearranging the equation (.54) according to the powers of r, we get a coplete quartic in r, 4 p = 0 bc 4 a0 ( ) a abc ( ) p ap = 0 a ( a + h ) a a h a4 (.59) equation which always has algebraic solution. It would, therefore, be necessary to slightly odify the conditions of the stateent so that the proble had a solution, siply by changing the variable r by the paraeter. pplying the forula (.0) given in the appendix, its roots have the following explicit algebraic expression: a ( a a a ) ( a a a ) a B a + ( 4a0 aa ) C + 4B a C a a C 4 r = + E D F + + B D C a a a C a a C 4 E D+ r = + E D F 4 a aa a F 4 C + E Substituting in (.0) the generic coefficients of the quartic by the specifics ones given by (.59), a ( a a a ) ( a a a ) a B a + ( a a a ) 4 0 bc 4 a0 ( ) C + B 4 a abc ( ) B D C a ( a + h ) a a a a h E D+ 4 a a 4 aa a F 4 C + E (.60)

35 nd after soe boring siplifications, we obtain as value of the constants, B,..., F, 4 6 ( a + h ) a + 9b c ( ) ( a a h ) + ( + ) ( a + h ) a 4 B ( a + h ) a b c ( )( ) C + 4B B D C ( a + h ) a 6 E D + h 6 h h ( a + h ) a abc( ) F C + E (.6).0. nueric exaple. Given the following lengths for the sides of BC, a.8994 BC b c.6950 (.6) the values for the reaining paraeters involved in the proble, with a 4-decial precision, are: = = = 0.4 r = 0.58 x0 = 0.75 y0 = x = 0. y = xt = yt = (.6) (.64)

36 ppendix. n alternative representation for the explicit relationship given by (.60) and (.6). Expressing (.) as, 4 + a + a + a + a = 0 0 (.0) its roots have the following explicit algebraic expression, (.0): a ( a a a ) ( a a a ) a B a + ( 4a0 aa ) C + 4B a C a a C 4 = + E D F + + B D C a a a C a a C 4 E D+ = + E D F 4 a aa a F 4 C + E Substituting the generic coefficients of the quartic by the specifics ones of our equation, we obtain the value of the constants, B,..., as a function of the constant p defined by (.07): 4 08 p + 44 p B 4 p + C + 4B a = p = B a D 4 + p p + p = 0 C a = p = + + a0 p E D p p( p ) F C + E (.0) B = 7 48 B D = C C = ( 0 7) + ( ) p = E = D + ( ) 4( ) F C + E (.04)

37 Substituting the values given by (.6) into the explicit expressions for the solutions of the quartic, (.59), we obtain: C C = + + E 4( ) D + F C C = + E 4( ) D + F i (.05) (.06) and, b 4 = + + r C C + + E + 4( ) D + F B = 7 48 B D = C C = ( 0 7) + ( ) E = D+ ( ) (.07) (.08) F 4( ) C + E To shorten the writing of this forula (whose length exceeds the width of this page for a font size readable), have been split into, B,..., F parts. This not only represents a greater copactness of the expression but also an econoy of calculations because soe of these parts is repeated within the expression. Otherwise, writing this steps fully requires too uch typing, so it is skipped, but it can be easily verified.

38 ppendix B. Developing the equations for the asyetric case in the reference syste used in the syetric case. Let s consider the sae reference syste used in the solution of the syetric case, as shown in the pictures below. Y T C T T c T T b B C C O C T X B T a C Y T C T c T B C C O C T T T b X B T a C

39 B.. Explicit equation of the tangent to the ellipse with slope and ordinate at the origin n. Let t y = x + n a line that intersects the ellipse e, (.0): i.e., x y e + = x ( x + n) t y = x + n t e + =, ( + ) x + n x + ( n ) = 0 (B.0) The necessary and sufficient condition for t to be tangent to the ellipse is that there is only one point of contact between the line and the curve, which is equivalent to equation (B.0) has only a root (double), ie, its discriinant ust be zero: = + = ( n ) 4( ) ( n ) 0 (B.0) Operating and siplifying (B.0), we obtain: n= + (B.0) Thus, the equation of the tangent of slope to the ellipse (.0), for y > 0 (positive sei-ellipse), is t y = x + + (B.04) B.. Locus of intersection points of pairs of orthogonal tangents: director circle. Let us consider a couple of tangents to the positive sei-ellipse in canonical for, generic and orthogonal each other, (B.04) and s y = x + + (B.05) Intersection point of (B.04) and (B.05): s t x y = x + + y = x = + (B.06) Let ( x, y ) s t. Multiplying by the nuerator and denoinator of (B.06), we have: x = (B.07)

40 + + x y t y x n y + (, ) = +, = + +, i.e., y = + (B.08) Let us copact the notation through the variable changes, u +, v +, of the squares of the coordinates of, given by (B.07) and (B.08): and calculate the su and undoing the variable changes, i.e., x x u v u + v + y = + = + + ( ) ( ) ( + ) ( u + v ) ( + ) = u + v u v ( + ) ( + ) + y = = =, + + +, x + y = + (B.09) The locus of the points fro which can be drawn pairs of orthogonal tangents to the canonical ellipse is the circle centered at the origin of coordinates and radius R = +. This circle is called the director circle. B.. Coordinates of the tangent point T C of the line with slope and the ellipse. Solving the equation (B.0), under the tangency condition established in (B.0) and the value of n according to (B.04), we get the abscissa x of the tangent point T C : or, x T = By replacing T = + = 0 xt =, ( + ) n ( ) x n x ( n ) 0 xt into t y = x + n, we obtain, = +. Gathering the above results, we have the following expression for the coordinates of T C : y T = + + = + + T, + + (B.0) nd fro (.), as a function of the eccentricity of the ellipse, x T r r =, y T = + + (B.)

41 B.4. Equation of the bisector of the right angle. x y x + y c y = ( x x ) + y = 0 + x y x + y = x + y x y x + y x ( ) y = + = 0 b y x x y + ( ) x y x + y = + ( x + y x y ) ( ) x + ( + ) y ( ) x ( + ) y = 0 x y x + y = ( x + y x y ) ( + ) x ( ) y ( + ) x + ( ) y = 0 For in the first quadrant its bisector has positive slope, so its iplicit equation is, and explicitly, ( + ) x ( ) y ( + ) x + ( ) y = y = x + y x (B.) nother way to deduce the equation of the bisector. The slope of the bisector is Hence, the equation of the bisector is: Substituting tgb + tg 4 + = tg B + = =. 4 tgb tg y = ( x x) + y = x x + y. x by the expression (B.07) and y by the right side of (B.04), taking into account that t, i.e., y x x x n x x = + + = = = x + + +, y= x+ (B.) or, y= x+ r (B.4)

42 B.5. Coordinates of C. C is deterined by the intersection of the bisector of, (.), and the circunference with center and radius r : ( x x) + ( y y) = r y = y + r ( x x) + + r ( x x) = x x + + y = x+ y x or, hence, + = r ( x x ) ( x x ), + + r = + ( x x) r = ( x x), if 0 ( ) fro which, x = x r + (B.5) and substituting this expression for x into (B.), + y = y r + (B.6) Replacing in (B.5) and (.6) the coordinates of given by (B.07) and (B.08), respectively, ( x, y) = r, r x = r y = r + + (B.7) or, as a function of the eccentricity of the ellipse: r x = + r r y = ( ) + r ( + ) + (B.8)

43 ppendix C. nalysis of the solubility of the sextic (.54). s is well known, [] and [4]. by eans of a Tschirnhausen transforation the general sextic, x a x a x a x a x a x a = 0 (C.0) can be reduced to the for, 6 z + z + az + a = 0 (C.0) Thus, via Tschirnhausen transforation, the sextic (.54) can be reduced to the for, 5 r r + z z + z = 0; b, c, r ; r b c b c (C.0) ll equations of degree 4 are algebraically solvable over the coplex nuber field, but fro the work of Ruffini, bel and Galois it is now well-known that all quintic equations do not []. In [] are stated the necessary and sufficient conditions for the quintic factor of (C.0), a Bring-Jerrard for or principal quintic for the general quintic, be solvable by radicals: the existence of three rational nubers, =, p > 0 and q 0, such that, applied to our specific values, q ( 4 p ) 4 q ( p ) r r + =, = + p + p + b c (C.04) fro where, 4 q( + p) r r = + 5( 4 p) b c (C.05) that can be written as, 4 q ( + p ) 4 a = r 5( 4 p) bc (C.06) or, 4 q( + p) r = 5( 4 p) h (C.07) where r is the radius coon to the three circuferences and h the height relative to the hypotenuse of BC. In [4] are given explicit forulas to get the roots for this irreducible particular for of quintic when it is algebraically solvable: 4 z = e u, j = [0, 4] j k= i jk 5 k (C.08)

44 with, v v v v v v v v u u u u p ,,, 4, + v +, v +, v + +, v 4 (C.09) (C.0) Furtherore, any irreducible equation with rational coefficients is solvable by radicals if and only if its Galois group G is a subgroup of one of the two transitive subgroups of S 6: J = S # S and K = S # Z, denoting the wreath product of two groups by #, [5]. t first glance, the lack of the first-degree ter in (.54) ight see to facilitate the analysis of solvability of the equation, but is just the opposite because the resolvent equation of (C.0) is of the for [], z 6 a z (4a + ) a z + + (44 a a a ) a a = (C.) and (C.0) is solvable by radicals and its Galois group is a subgroup of J iff (G.) has a rational root, but it is straightforward that this happens if the constant ter vanishes because then z = 0, i.e., if 4 = 44a a a + (C.) and taking, for instance, a = ½, we find that the Galois group of the irreducible polynoial 6z 6 + 6z + 8z + 5 is a subgroup of J and solvable. But in our case, a = 0, and so (C.0) is reducible, therefore this theore is not applicable. Moreover, the second condition of (C.04) involves a relationship between b, c and r, which is the final goal of the proble, so this condition neither is ipleentable. But, according to Duit s analysis [], solvable irreducible quintic equations are rare, hence, it is reasonable to expect that solvable irreducible sextics are also rare. Through the exhaustive analysis carried out we have seen that the insurountable barrier to solving the asyetric case is the sextic in (.54), the only way to eliinate the paraeter.

45 ppendix D. Warnings about the polynoial 6 th degree equation (.54). One of the roots of the sextic (.54), expressed in its ost copact for, 6 q = 0 r c0 bc c 0 ar c bc q cq = 0 c abc 4 r( a r ) a c4 r 4 + bc a 4ar c5 bc c6 (D.0) is the eccentricity of the ellipse, in ters of a, b, c and r, but this equation could be unsolvable by radicals. s a warning to sailors, it can be useful to reproduce the following chain of thoughts: we know that (D.0) has to have, at least, one real solution, and taking into account that not real roots appear only in pairs of conjugate coplex, necessarily, the nuber of real roots have to be even, and by the uniqueness of the solution discussed before, we d can think (wrongly) that the unique real solution of this equation has to have ultiplicity, 4 or 6, and, therefore, the derivative of this polynoial should preserve the sae real root, allowing the substitution of the before equation by the following: 6 k = k kc = 0 k (D.0) But c = 0, therefore, (D.0) can be written as, 6 k = k kc = 0 k (F.0) t this point, we would think that the lack of the first grade ter in in the equation (D.0) should be the key for the door to the solution, since it allow to get down one grade ore by eliination of the coon factor in (D.0) (it has been proved yet that the solution = 0 it isn t adissible), getting finally the following quartic: 6 4 k k k = p= 0 p kc = ( p + ) c = 0 p+ (D.04) Unfortunately, the hypothesis that the equation (F.0) has one unique anifold real root can t be proved, hence, it isn t warranted to reach the solution trough this path. Indeed, for the nuerical exaple presented in., the six roots of (.54) are all real and different.

46 ppendix E. Exploring another ways. The transforation by inversion, the ost widely-used ethod in Sangaku challenges, applied to the ellipse, transfor its canonical equation into a quartic equation which is uch ore coplicated. For instance, the curves on the right picture arise fro the apping by inversion on the unit circle, that transfors the ellipse into the quartic x (, ) (, ) x y x y + x y x y y + = + = ( x + y ). It is neither possible to siplify the proble trough affine transforations, since an affinity that transfor the ellipse into a circuference, it also will transfor, at the sae tie, the circuferences into ellipses.