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Question Number 207533 by mr W last updated on 18/May/24

Commented by mr W last updated on 18/May/24

find the radius of circumcircle

findtheradiusofcircumcircle

Answered by mr W last updated on 20/May/24

OA=p=R−a  OB=q=R−b  OC=r=R−c  X=q^2 +r^2 −u^2 =(R−b)^2 +(R−c)^2 −u^2 =2R^2 −2(b+c)R+b^2 +c^2 −u^2   Y=r^2 +p^2 −v^2 =(R−c)^2 +(R−a)^2 −v^2 =2R^2 −2(c+a)R+c^2 +a^2 −v^2   Z=p^2 +q^2 −w^2 =(R−a)^2 +(R−b)^2 −w^2 =2R^2 −2(a+b)R+a^2 +b^2 −w^2   volume of tetrahedron O−ABC  should be zero.  V=((√(4p^2 q^2 r^2 +XYZ−p^2 X^2 −q^2 Y^2 −r^2 Z^2 ))/(12))=0  4p^2 q^2 r^2 +XYZ−p^2 X^2 −q^2 Y^2 −r^2 Z^2 =0  4(R−a)^2 (R−b)^2 (R−c)^2 +XYZ−(R−a)^2 X^2 −(R−b)^2 Y^2 −(R−c)^2 Z^2 =0  after expansion we get a quadratic  equation for R:  {−(u^4 +v^4 +w^4 )+2(u^2 v^2 +v^2 w^2 +w^2 u^2 )+4[a(b+c)−bc−a^2 ]u^2 +4[b(c+a)−ca−b^2 ]v^2 +4[c(a+b)−ab−c^2 ]w^2 }R^2   +2{au^4 +bv^4 +cw^4 +2(a^3 u^2 +b^3 v^2 +c^3 w^2 )−(a+b)(u^2 v^2 +c^2 w^2 )−(b+c)(v^2 w^2 +a^2 u^2 )−(c+a)(w^2 u^2 +b^2 v^2 )+[bc(b+c)−a(b^2 +c^2 )]u^2 +[ca(c+a)−b(c^2 +a^2 )]v^2 +[ab(a+b)−c(a^2 +b^2 )]w^2 }R  −{a^2 u^2 (a^2 +u^2 )+b^2 v^2 (b^2 +v^2 )+c^2 w^2 (c^2 +w^2 )−[a^2 (b^2 +c^2 )−b^2 c^2 ]u^2 −[b^2 (c^2 +a^2 )−c^2 a^2 ]v^2 −[c^2 (a^2 +b^2 )−a^2 b^2 ]w^2 −(a^2 +b^2 )u^2 v^2 −(b^2 +c^2 )v^2 w^2 −(c^2 +a^2 )w^2 u^2 +u^2 v^2 w^2 }=0  with  A=−(u^4 +v^4 +w^4 )+2(u^2 v^2 +v^2 w^2 +w^2 u^2 )+4[a(b+c)−bc−a^2 ]u^2 +4[b(c+a)−ca−b^2 ]v^2 +4[c(a+b)−ab−c^2 ]w^2   B=au^4 +bv^4 +cw^4 +2(a^3 u^2 +b^3 v^2 +c^3 w^2 )−(a+b)(u^2 v^2 +c^2 w^2 )−(b+c)(v^2 w^2 +a^2 u^2 )−(c+a)(w^2 u^2 +b^2 v^2 )+[bc(b+c)−a(b^2 +c^2 )]u^2 +[ca(c+a)−b(c^2 +a^2 )]v^2 +[ab(a+b)−c(a^2 +b^2 )]w^2   C=a^2 u^2 (a^2 +u^2 )+b^2 v^2 (b^2 +v^2 )+c^2 w^2 (c^2 +w^2 )−[a^2 (b^2 +c^2 )−b^2 c^2 ]u^2 −[b^2 (c^2 +a^2 )−c^2 a^2 ]v^2 −[c^2 (a^2 +b^2 )−a^2 b^2 ]w^2 −(a^2 +b^2 )u^2 v^2 −(b^2 +c^2 )v^2 w^2 −(c^2 +a^2 )w^2 u^2 +u^2 v^2 w^2   ⇒AR^2 +2BR−C=0  ⇒R=((−B±(√(B^2 +AC)))/A)    example:  u=7, v=6, w=8  a=3, b=2, c=1  A=−(7^4 +6^4 +8^4 )+2(7^2 6^2 +6^2 8^2 +8^2 7^2 )+4[3(2+1)−2×1−3^2 ]7^2 +4[2(1+3)−1×3−2^2 ]6^2 +4[1(3+2)−3×2−1^2 ]8^2 =5855  B=3×7^4 +2×6^4 +1×8^4 +2(3^3 7^2 +2^3 6^2 +1^3 8^2 )−(3+2)(7^2 6^2 +1^2 8^2 )−(2+1)(6^2 8^2 +3^2 7^2 )−(1+3)(8^2 7^2 +2^2 6^2 )+[2×1(2+1)−3(2^2 +1^2 )]7^2 +[1×3(1+3)−2(1^2 +3^2 )]6^2 +[3×2(3+2)−1(3^2 +2^2 )]8^2 =−12895  C=3^2 7^2 (3^2 +7^2 )+2^2 6^2 (2^2 +6^2 )+1^2 8^2 (1^2 +8^2 )−[3^2 (2^2 +1^2 )−2^2 1^2 ]7^2 −[2^2 (1^2 +3^2 )−1^2 3^2 ]6^2 −[1^2 (3^2 +2^2 )−3^2 2^2 ]8^2 −(3^2 +2^2 )7^2 6^2 −(2^2 +1^2 )6^2 8^2 −(1^2 +3^2 )8^2 7^2 +7^2 6^2 8^2 =80929  R=((12895±(√(12895^2 +5855×80929)))/(5855))      =((2579)/(1171))±((3024(√(70)))/(5855))      ≈6.523586 / −2.118804

OA=p=RaOB=q=RbOC=r=RcX=q2+r2u2=(Rb)2+(Rc)2u2=2R22(b+c)R+b2+c2u2Y=r2+p2v2=(Rc)2+(Ra)2v2=2R22(c+a)R+c2+a2v2Z=p2+q2w2=(Ra)2+(Rb)2w2=2R22(a+b)R+a2+b2w2volumeoftetrahedronOABCshouldbezero.V=4p2q2r2+XYZp2X2q2Y2r2Z212=04p2q2r2+XYZp2X2q2Y2r2Z2=04(Ra)2(Rb)2(Rc)2+XYZ(Ra)2X2(Rb)2Y2(Rc)2Z2=0afterexpansionwegetaquadraticequationforR:{(u4+v4+w4)+2(u2v2+v2w2+w2u2)+4[a(b+c)bca2]u2+4[b(c+a)cab2]v2+4[c(a+b)abc2]w2}R2+2{au4+bv4+cw4+2(a3u2+b3v2+c3w2)(a+b)(u2v2+c2w2)(b+c)(v2w2+a2u2)(c+a)(w2u2+b2v2)+[bc(b+c)a(b2+c2)]u2+[ca(c+a)b(c2+a2)]v2+[ab(a+b)c(a2+b2)]w2}R{a2u2(a2+u2)+b2v2(b2+v2)+c2w2(c2+w2)[a2(b2+c2)b2c2]u2[b2(c2+a2)c2a2]v2[c2(a2+b2)a2b2]w2(a2+b2)u2v2(b2+c2)v2w2(c2+a2)w2u2+u2v2w2}=0withA=(u4+v4+w4)+2(u2v2+v2w2+w2u2)+4[a(b+c)bca2]u2+4[b(c+a)cab2]v2+4[c(a+b)abc2]w2B=au4+bv4+cw4+2(a3u2+b3v2+c3w2)(a+b)(u2v2+c2w2)(b+c)(v2w2+a2u2)(c+a)(w2u2+b2v2)+[bc(b+c)a(b2+c2)]u2+[ca(c+a)b(c2+a2)]v2+[ab(a+b)c(a2+b2)]w2C=a2u2(a2+u2)+b2v2(b2+v2)+c2w2(c2+w2)[a2(b2+c2)b2c2]u2[b2(c2+a2)c2a2]v2[c2(a2+b2)a2b2]w2(a2+b2)u2v2(b2+c2)v2w2(c2+a2)w2u2+u2v2w2AR2+2BRC=0R=B±B2+ACAexample:u=7,v=6,w=8a=3,b=2,c=1A=(74+64+84)+2(7262+6282+8272)+4[3(2+1)2×132]72+4[2(1+3)1×322]62+4[1(3+2)3×212]82=5855B=3×74+2×64+1×84+2(3372+2362+1382)(3+2)(7262+1282)(2+1)(6282+3272)(1+3)(8272+2262)+[2×1(2+1)3(22+12)]72+[1×3(1+3)2(12+32)]62+[3×2(3+2)1(32+22)]82=12895C=3272(32+72)+2262(22+62)+1282(12+82)[32(22+12)2212]72[22(12+32)1232]62[12(32+22)3222]82(32+22)7262(22+12)6282(12+32)8272+726282=80929R=12895±128952+5855×809295855=25791171±30247058556.523586/2.118804

Commented by mr W last updated on 19/May/24

Answered by ajfour last updated on 18/May/24

say A is origin. AO  y axis.  red circle x^2 +(y−R+a)^2 =R^2   x_B ^2 +y_B ^2 =w^2   x_C ^2 +y_C ^2 =v^2   x_B ^2 +(y_B −R+a)^2 =(R−b)^2   x_C ^2 +(y_C −R+a)^2 =(R−c)^2   ⇒w^2 −2(R−a)y_B =(R−b)^2   & v^2 −2(R−a)y_c =(R−c)^2   x_B =(√(w^2 −y_B ^2 ))  x_C =−(√(v^2 −y_C ^2 ))  (x_B −x_C )^2 +(y_B −y_C )^2 =u^2   ★

sayAisorigin.AOyaxis.redcirclex2+(yR+a)2=R2xB2+yB2=w2xC2+yC2=v2xB2+(yBR+a)2=(Rb)2xC2+(yCR+a)2=(Rc)2w22(Ra)yB=(Rb)2&v22(Ra)yc=(Rc)2xB=w2yB2xC=v2yC2(xBxC)2+(yByC)2=u2

Commented by mr W last updated on 18/May/24

welcome back sir!

welcomebacksir!

Commented by ajfour last updated on 18/May/24

thanks, but could u follow my solution  sir..

thanks,butcouldufollowmysolutionsir..

Commented by mr W last updated on 19/May/24

your path is right, thanks sir!

yourpathisright,thankssir!

Commented by ajfour last updated on 18/May/24

can this do?  Σcos^(−1) (((R−b)^2 +(R−c)^2 −u^2 )/(2(R−b)(R−c)))=2π

canthisdo?Σcos1(Rb)2+(Rc)2u22(Rb)(Rc)=2π

Commented by mr W last updated on 19/May/24

yes, this works either. but this  equation is not polynomial.  the distances from a point to the  vertices of a triangle fulfills  a polynomial equation.

yes,thisworkseither.butthisequationisnotpolynomial.thedistancesfromapointtotheverticesofatrianglefulfillsapolynomialequation.

Commented by mr W last updated on 19/May/24

the final equation can be simplied to  a quadratic equation, so we can get  the exact solution. see above.

thefinalequationcanbesimpliedtoaquadraticequation,sowecangettheexactsolution.seeabove.

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