Miscellaneous Results on Tetrahedron

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1 Miscellaneous Results on Tetraheron vni llana We start with eva's theorem for tetraheron in barycentric coorinates 1, D f b a c e a b c Fig. 1 a(d)*b()*c()*() : a()*b()*c(d)*() : a()*b(d)*c()*() = 1 : 1 : 1, (1) where is an arbitrary point an a = ( 0 : a() : a() : a(d) ), b = ( b() : 0 : b() : b(d) ), c = ( c() : c() : 0 : c(d) ), = ( () : () : () : 0 ) are its traces on the respective faces of tetraheron D, see Fig. 1. From relation (1) it follows immeiately that cevians to the points ai, bi, ci, i built as isotomic conjugates of traces a, b, c,, are also concurrent. Now let us consier the insphere an exspheres of tetraheron. Let Ge an Na be respectively the points where the insphere an the respective exsphere meet the tetraheron face. oint Ge is isogonal conjugate to Na. Let α, β, γ be the angles between the tetraheron faces sharing the sies a, b, c respectively. The barycentric coorinates of Na an Ge are as follows Na = ( a*tan( α/) : b*tan( β/) : c*tan(γ /) : 0 ), () Ge = ( a/tan( α/) : b/tan( β/) : c/tan(γ /) : 0 ), (3) 1 Generalization of eva's Theorem for Tetraheron: 1

2 or Ge = ( a^/(a*tan( α/)) : b^/(b*tan( β/)) : c^/(c*tan(γ /)) : 0 ). (4) omparing (4) an () follows that Na an Ge are isogonal conjugate to each other. From the spherical trigonometry we have the nice formula tan( α/) = sqrt(sin(s-bb)*sin(s-cc)/(sin(s)*sin(s-aa))), (5) where aa is the angle between sies c, e; angle bb is between sies e, a; angle cc is between sies a, c, an s = (aa+bb+cc)/. Similarly we fin tan( β/) an tan(γ /). The analogy to Danelin spheres is obvious, so the points Ge an Na are foci of the ellipse which passes through the reflection points of the rays from Ge to Na on the sies of triangle. Motivate by this property we efine the following mapping. Let = ( u : v : w ) be an arbitrary point an Q its isogonal conjugate with respect to triangle. Let Qa, Qb, Qc be the mirror points of Q with respect to the sies a, b, c of triangle. Let Ra, Rb, Rc be the intersection points of lines Qa, Qb, Qc with triangle sies,, respectively. Now lines Ra, Rb, Rc are concurrent at point T with barycentric coorinates (Tx:Ty:Tz), Tx = 1/(u*(w*(b^*(w+v)+v*(c^-a^))+c^*v^)), Ty = 1/(v*(u*(c^*(u+w)+w*(a^-b^))+a^*w^)), Tz = 1/(w*(v*(a^*(v+u)+u*(b^-c^))+b^*u^)). Qb Rb T Ra Qa Rc Q Qc Fig. Danelin Spheres:

3 For = ( a : b : c ), that is = X(1) the incenter of triangle, center T represents the Gergonne point X(7). For = ( 1 : 1 : 1 ), that is = X() the centroi of triangle, T represents X(598). The following relation also hols for arbitrary. Ra+RaQ = Rb+RbQ = Rc+RcQ, (6) Let = (x : y : z : w) be an arbitrary point, then Q = (Sa^/x : Sb^/y : Sc^/z : S^/w) is the isogonal conjugate of, where Sa, Sb, Sc, S are the areas of tetraheron faces. Let Ra, Rb, Rc, R be the reflection points of the rays from to Q on the respective faces of tetraheron, then a relation similar to (6) hols Ra+RaQ = Rb+RbQ = Rc+RcQ = R+RQ. Let = (px : py : pz : pw) an Q = (qx : qy : qz : qw) be arbitrary points, then we call point S = (px^/qx : py^/qy : pz^/qz : pw^/qw) the iso- conjugate of Q. This mapping represents a generalization of the isotomic an isogonal conjugate mapping. The geometric interpretation of this generalize mapping will be explaine by means of the respective cevian traces on face of tetraheron, see Fig. 3. Ra S S1 1 Q1 Q Fig.3 In this planar case for the traces of the points on face, for example, instea of (px : py : pz : 0) we use the notation (px : py : pz). Let 1 be an arbitrary point on the extension of line. Let Q1 an S1 be the intersection points of lines an S with the lines 1 an 1 respectively. Then the line Q1S1 intersects the line at the point Ra = (0 : py : -pz), which epens only on the coorinates of. This gives us the possibility to 3

4 construct point S1 for given an Q. It is obvious that line S1 coincies with the cevian of S from. Similarly we obtain points Rb = (-px : 0 : pz) an Rc = (px : -py : 0) for the respective eges b an c of triangle. We note that Ra, Rb, Rc are collinear, that is et([0 py pz ; -px 0 pz ; px py 0 ]) = 0. Let us now turn to the incenter It of tetraheron. enter It is the intersection point of planes that bisect the angles between the tetraheron faces. Let Sa, Sb, Sc, S be the areas of tetraheron faces, then It has barycentric coorinates The trace of It in the face has barycentric coorinates It = ( Sa : Sb : Sc : S ). (7) It = ( Sa : Sb : Sc : 0 ). (8) Relation (8) represents a generalization of angle bisector theorem for tetraheron. Let Et, Et, Et, EtD be the excenters of tetraheron corresponing to the tetraheron vertices,,, D respectively. We show that,, Et, Et are complanar. Since = (1 : 0 : 0 : 0), =(0 : 1 : 0 : 0), Et = (-Sa : Sb : Sc : S), Et = (Sa : Sb : Sc : S), it follows et([ ; ; -Sa Sb Sc S ; Sa -Sb Sc S]) = 0. Olso, D, Et, Et are complanar. Since = (0 : 0 : 1 : 0), D=(0 : 0 : 0 : 1), it follows The same hols for other eges of tetraheron. et([ ; ; -Sa Sb Sc S ; Sa -Sb Sc S]) = 0. Let r, ra, rb, rc, r be raii of insphere an exspheres of tetraheron D, then we have following relations 3*V/r = Sa+Sb+Sc+S, (9) 3*V/ra = -Sa+Sb+Sc+S, (10) 3*V/rb = Sa -Sb+Sc+S, (11) 3*V/rc = Sa+Sb -Sc+S, (1) 3*V/r = Sa+Sb+Sc -S, (13) where V is the volume of tetraheron. From (9), (10), (11), (1), (13) we obtain /r = 1/ra + 1/rb + 1/rc + 1/r. (14) The circumcenter Ot or the center of the circumsphere of tetraheron is the intersection point of planes through mipoints of sies an orthogonal to those sies of tetraheron. Let a, b, c,, e, f be the sie lengths of tetraheron, then Ot has barycentric coorinates ( Ot : Ot : Ot : OtD ), 4

5 Ot = ^*a^*(f^+e^-a^) + b^*e^*(a^+f^-e^) + c^*f^*(e^+a^-f^) - *a^*e^*f^, (15) Ot = e^*b^*(f^+^-b^) + c^*f^*(^+b^-f^) + a^*^*(b^+f^-^) - *b^*^*f^, (16) Ot = f^*c^*(e^+^-c^) + b^*e^*(^+c^-e^) + a^*^*(c^+e^-^) - *c^*e^*^, (17) OtD = ^*a^*(b^+c^-a^) + e^*b^*(c^+a^-b^) + f^*c^*(a^+b^-c^) - *a^*b^*c^. (18) From (15), (16), (17), (18) we obtain Ot + Ot + Ot + OtD = 88*V^, (19) where V is the volume of tetraheron. Relation (19) is similar to the sum of barycentric coorinates of the circumcenter of triangle : a^*(b^+c^-a^) + b^*(c^+a^-b^) + c^*(a^+b^-c^) = 16*S^, (0) where S is the area of triangle. We recall the ayley-menger eterminant 3 for the volume of tetraheron 88*V^ = K, (1) where K , () an 3 a, b, 1 1 c, 14 41, 4 4 e, omparing (19) an (1) we obtain where Ot = K 1, Ot = K 13, Ot = K 14, OtD = K 15, (3) K ij are the respective cofactors of matrix K. f. 3 ayley-menger eterminant: 5

Collinearity/Concurrence

Collinearity/Concurrence Ray Li (rayyli@stanford.edu) June 29, 2017 1 Introduction/Facts you should know 1. (Cevian Triangle) Let ABC be a triangle and P be a point. Let lines AP, BP, CP meet lines BC,