PROVING SOME GEOMETRIC IDENTITIES BY USING THE DETERMINANTS

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1 Journal of cience and Arts Year No () pp OIGINAL PAPE POVING OME GEOMETIC IDENTITIE BY UING THE DETEMINANT DAM VAN NHI Manuscript received: 7; Accepted paper: 79; Published online: Abstract In this note we will give proofs about some identities by using determinants The method of determinants in Geometry is a powerful technique Keywords: Matrix determinant tetrahedron volume Mathematical ubject Classification: 6D99 97D5 INTODUCTION ome results in geometry have been proved by the traditional methods In this note we want to use the identities to reprove and find out the new results uccessfully we have gained the below achievements: Proposition Given the tetrahedron A A A 3 A with the lengths of 6 sides a = l ; b = l 3 ; c = l ; x = l 3 ; y = l ; z = l 3 Put = ax + by + cz Then the radius of circumscribed sphere of the tetrahedron will be defined by the formula: ax by cz a a b z a c a b z b b c x a c y b c x c y Proposition Given the tetrahedron ABCD Let A B C D be the barycenters of the surfaces BCD; CDA; DAB; ABC; correspondingly Let's prove that with any point O and uvt for choosing properly VOB CD uv OCDA vv ODA B tv OAB C V ABCD 7 Pedagogical University Ha Noi Hanoi Vietnam damvannhi@yahoocom IN: 8 958

2 386 Proving some geometric Dam Van Nhi POVING OME IDENTITIE BY UING THE DETEMINANT Lemma Decomposing the vectors x y z into the orthogonal basis we have x x x y xz x y z y x y y y z zx zy zz Proof: uppose that x abc y a' b' c' and z a" b" c" a b c a b c a b c x y z a' b' c' a' b' c' a' b' c' a" b" c" a" b" c" a" b" c" x x x y x z x y z yx yy yz x z y z z z therefore we have ince Proposition The volume of a triangular pyramid with A = a B = b C = c BC = x CA = y AB = z is a a b z a c y V a b z b b c x a c y b c x c Proof Denote x A y B z C and V V We have x y a b z x x x y xz x z a c y and y z b c x Using 36 V y x y y y z by Lemma we x z y z z z have a a b 88V a b z b b c x a c y b c x c z a c y Lemma 3 [] Given the tetrahedron ABCD with A (x y z ) A (x y z ) A 3 (x 3 y 3 z 3 ) and A (x y z ) Denoting is the radius of circumscribed sphere of the tetrahedron ABCD and l ij = l ji is the length of side A i A j = A j A i i j Then we always have l3ll3l lll3l3 ll3ll3 ll3 l3l ll3 V Proof: It was well-known that the volume of a tetrahedron is not changed under a translation Therefore without loss of generality we can suppose that the center of the circumscribed sphere of the tetrahedron A A A 3 A is the orgin of coordinates Then we have wwwjosaro

3 Proving some geometric Dam Van Nhi 387 x y z x y z x3 y3 z3 x y z xi xj yi yj zi zj From this we deduce that xx i j yy i j zz i j with any i j = 3 i j Let s t xx y y z z Then t ij i j i j i j ij ji ij i j i j i j x x y y z z l l tji Consider the following decomposition x y z x y z P x y z x y z x y z x y z x y z x y z x y z x x x x 3 x y z y y y y 3 x y z z z z z x y z t t t t 3 3 t t t t t t t 3 t with t ij = t ji Because so x y z x y z x y z 36V and l l l 3 x y z x y z x y z l 3 l3 l l l3 l l3 l3 l 6 l 3 3 l3 l3 l l l3 l l l3 l l l all equals with 6V From the last determinant and taking the root we have the formula calculating l3ll3l lll3l3 ll3ll3 ll3 l3l ll3 V From the Proposition and the Lemma 3 we can deduce the formula calculating the radius of circumscribed sphere through the lengths of 6 sides of the quadrilateral Proposition Given the tetrahedron A A A 3 A with the lengths of 6 sides a = l b = l 3 c = l x = l 3 y = l z = l 3 Put = ax + by + cz Then the radius of circumscribed sphere of the tetrahedron will be defined by the formula: IN: 8 958

4 388 Proving some geometric Dam Van Nhi ax by cz a a b z a c a b z b b c x a c y b c x c y Proof: By the Ptolemy's Inequality we have ax by cz by cz ax and cz ax by We can easily check the formula 6 ax by cz l l l l l l l l l l l l l l l l l l Therefore we have ax by cz 6V Hence we obtain the ax by cz relationship a a b z a c y a b z b b c x a c y b c x c Example With the tetrahedron A A A 3 A Let's prove that with any points O we always have the inequality: VAA AA 3 OAi 6 i olution: Building the coordinates Oxyz uppose A i (x i y i z i ) i = 3 We have x y z x y z 36V x y z AAAA 3 i i i i x y z following the Hadamard's Inequality Therefore V AA AA 3 6 OAi i Example Given the trihedral angle xyz uppose that the point N is in the trihedron and the plane (P) through N cuts x y z at A B C correspondingly Then the volume ratio V is constant VNBC VNCA VNAB olution: Calling the distance from N to (BC) (CA) (AB) is a b c correspondingly and setting A=x B=y C=z Then a b c is constant and we have 3 3 V 6 V 6 V V V V ayz bzx cxy abcx y z NBC NCA NAB wwwjosaro

5 Proving some geometric Dam Van Nhi 389 On the other hand we also have 6 V x y z cos cos cos coscos cos following the Lemma in which BC CA AB Therefore the volume ratio which is calculated by V 6 cos cos cos coscos cos VNBC VNCA VNAB abc is constant Example Given the point O in the tetrahedron ABCD A plane (P) through O cuts ABACAD igning the distance from A B C D to (P) is t a t b t c t d correspondingly Prove that tv a OBCD tv b OCDA tv c ODAB tv d OABC olution: Building the coordinates Oxyz so that O( ) (P) : z = and A(x y z ) B(x y z ) C(x 3 y 3 z 3 ) D(x y z ) with t a = z > t b = -z > ; t c = -z 3 > ; t d = -z > Conventing that the volume of the tetrahedron is calculated by the determinants following the right hand rule: x y z x y z tv a OBCD z z x3 y3 z3 x y z x y z x y z x y z tv b OCDA z z x3 y3 z3 x y z x y z x y z x y z tv c ODAB z3 z3 x y z x y z x y z x y z x y z x y z tv d OABC z z x y z x3 y3 z3 x y z Therefore the difference T tavobcd tv b OCDA tv c ODAB tdvoabc exactly equals x y z z x y z x y z x y z x y z x y z z T z z z3 x y z z x y z z3 x y z x y z x y z x y z z ummarizing tv tv tv tv a OBCD b OCDA c ODAB d OABC IN: 8 958

6 39 Proving some geometric Dam Van Nhi Proposition 5 Given the tetrahedron ABCD Let A B C D be the barycenters of the surfaces BCD CDA DAB ABC correspondingly Let's prove that with any point O and u v t for choosing properly VOB CD uv OCD A vv ODA B V OAB C V ABCD 7 Proof: Builiding the coordinates Oxyz so that O( ) A(x y z ) B(x y z ) C(x 3 y 3 z 3 ) and D(x y z ) The coordinates of the barycenters of the surfaces of the tetrahedron ABCD x x3 x y y3 y z z3 z A x x3 x y y3 y z z3 z B x x x y y y z z z C x x3 x y y3 y z z3 z D Let's X x Y y i i i i and Z zi setting Va 6VOBC D We perform x x3 x y y3 y z z3 z Va x x x y y y z z z x x3 x y y3 y z z3 z X Y Z X x Y y Z z x y z X x3 Y y3 Z z3 7 7 X x Y y Z z x y z ignal X Y Z X Y Z X Y Z X Y Z x y z x y z x y z x y z T x y z x y z x y z x y z x y z X Y Z x y z with the proper choice of + - From x y z deducing x y z i wwwjosaro

7 Proving some geometric Dam Van Nhi 39 x y z X Y Z X Y Z X Y Z x y z x y z x y z x y z x y z x y z x y z x y z x y z x y z x y z x y z Therefore V 7V 7V 7V 7V ABCD OBC D OCDA ODA B OAB C deduce the result: Existing a proper choice u v t so that X Y Z x y z x y z x y z From this we VOB CD uv OCD A vv ODA B V OAB C V ABCD 7 3 OME EXAMPLE ABOUT AEA Proposition 3 Given ΔABC Taking the points M N P belonging to the edges BC MB NC PA CA AB correspondingly Let u v t Then we have the area ratio MC NA PB MNP uvt uvt Proof: uppose A(x y ) B(x y ) C(x 3 y 3 ) Because BM umc x ux 3 3 y M uy u u 3 3 imilarly we also have x vx y N vy and x tx y ty P v v t t Let uvt and change T x ux x vx x tx u v t y uy y vy y ty u v t u v t x ux x vx x tx uvt 3 3 y uy y vy y ty 3 3 v uvt x x x3 t x x x3 y y y u y y y 3 3 IN: 8 958

8 39 Proving some geometric Dam Van Nhi Therefore MNP ABC gtt x ux3 x3 vx xtx u v t y uy3 y3 vy yty u v t uvt gtt x x x 3 y y y 3 u v t or MNP uvt u v ABC t Example [The VN 6 th Olympic Mathematics] Given ΔABC with the area equals to Taking the points M N P belonging to the edges BC CA AB correspondingly so that MB NC PA k k k3 with k k k 3 < Determining the area of the triangle with MC NA PB the vertices are the intersections among AM BN CP olution: uppose A(x y ) B(x y ) C(x 3 y 3 ) Because BM k MC x kx 3 y ky 3 Mxm ym k k imilarly we also have x3 kx y3 ky Nxn yn k k and x kx 3 3 p y ky Px yp k3 k3 AE BM NC Lets E AM BN Following the Menlaut's theorem we have EM BC NA o k kk xk xm kk yk ym AE EM and Exe ye From this we kk kkk kkk deduce that kk x x kx 3 3 kk y y ky Exe ye kkk kkk Let's F BNCP G CP AM imilarly as above we receive kkx 3 xkx kky 3 yky Fxf yf k kk3 k kk3 Gx g kkx 3 3 xkx 3 kky 3 3 yky 3 ye k3 k3k k3 k3k k kk k k k k k k and implementing the changes etting wwwjosaro

9 Proving some geometric Dam Van Nhi 393 T kk x x kx kkx x k x kkx x kx k kk k k k k k k kk y y ky kk y y k y kky y k y k kk k k k k k k kkk k kk3 k3 k3k kk x xkx 3 kkx 3 xkx k3ky3 yk3y kk y y ky kk y y k y kky y k y kk k kkk 3 x x x3 kk3 k3 x x x3 y y y k k k y y y From this we deduce EFG kkk 3 k kk k k k k k k Example Given ΔABC is acute triangle with the radiuses of the circumscribed and inscribed circles r uppose that the roots of the hights are A B C Prove that 3 ABC r 7 AB ccos B BC acosc CA bcos A olution: Because and AC bcosc B A ccos A CB acos B ABC abccos Acos BcosC cos Acos BcosC ABC abc 3 r r Because of cos Acos BcosC we have following the 7 Cauchy's Inequality Example Given the triangle ABC Let M N P be the midpoints of BC CA and AB Prove that with any point O outside the ΔABC having one of the areas OAM OBN OCP equals the sum of the two remaining areas olution: Building the coordinates Oxy so that O( ) and A(x y ) B(x y ) C(x y 3 ) The coordinates of the midpoints x x y M y x 3 x 3 y N y x x y P y Let's consider the case O belonging to AGN (G is the forcus of ΔABC) Conventing the area of triangle is calculated by the determinants following the right hand rule we have: IN: 8 958

10 39 Proving some geometric Dam Van Nhi x y OAM x x y y 3 3 x y x x y y 3 3 to x y OBN x y x3 x y3 y x3 x y3 y x x y y x x y y OCP Nh vy x3 y3 x y 3 3 x y x y x y x y because all are equal OAM OBN OCP x y x3 y3 x y x3 y3 x y x y imilarly for the other situations x y x y EFEENCE [] Pogorelov A Geometry Mir Publishers Moscow 987 [] Proskuryakov I V bornik zadatr po liheinoi algebre Nauka 978 wwwjosaro