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new-idea-and-solution-to-questions-35178-amp-35195-triangle-ABC-a-BC-b-CA-c-AB-CAB-ABC-BCA-d-a-b-c-a-b-c-a-b-c-a-b-c-put-it-as-this-A-0-0-B-c-0-C-

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Question Number 35871 by MJS last updated on 25/May/18
new idea (and solution) to questions 35178 & 35195 triangle: ABC; a=BC, b=CA, c=AB; α=∠CAB, β=∠ABC, γ=∠BCA d=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) put it as this: A= ((0),(0) ), B= ((c),(0) ), C= ((((−a^2 +b^2 +c^2 )/(2c))),((d/(2c))) ) circumcircle: center=M_1 = (((c/2)),((((a^2 +b^2 −c^2 )c)/(2d))) ) radius=R=((abc)/d) (calculated by intersection of circles with centers A, B, C or of symmetry−axes of AB and AC) 2 circles touching b, c and circumcircle, one from inside, the other from outside: center=M_2 lies on y=kx with k=tan (α/2) M_2 = ((x),((xtan (α/2))) ) radius=r_1 =R−∣M_1 M_2 ∣=xtan (α/2) (inside) r_2 =∣M_1 M_2 ∣−R=xtan (α/2) (outside) (obviously any circle with center M_2 (x) and touching the x−axis has radius xtan (α/2) and also obviously the touching point of 2 circles is located on the line connecting their centers) 1. ∣M_1 M_2 ∣=R−xtan (α/2) M_1 M_2 =(R−xtan (α/2))^2 2. ∣M_1 M_2 ∣=R+xtan (α/2) M_1 M_2 =(R+xtan (α/2))^2 tan (α/2)=((sin (α/2))/(cos (α/2)))=((√((1−cos α)/2))/( (√((1+cos α)/2))))=(√((1−cos α)/(1+cos α)))= [cos α=((−a^2 +b^2 +c^2 )/(2bc))] =(√((a^2 −b^2 +2bc−c^2 )/(−a^2 +b^2 +2bc+c^2 )))=(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))) M_1 M_2 =(m_1 −m_2 )^2 +(n_1 −n_2 )^2 = =((c/2)−x)^2 +((((a^2 +b^2 −c^2 )c)/(2d))−xtan (α/2))^2 = [after some transformation work] =((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x+((a^2 b^2 c^2 )/d^2 ) [((a^2 b^2 c^2 )/d^2 )=R^2 ] (R±xtan (α/2))^2 =x^2 tan^2 (α/2)±2Rxtan (α/2)+R^2 = =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x+R^2 so we have ((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x= =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x which leads to x_3 =0 (as I explained before, the point A can be seen as a circle with radius 0 still meeting the requirements) x_1 =((2bc)/(a+b+c)) ⇒ r_1 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)^3 (−a+b+c)))) x_2 =((2bc)/(−a+b+c)) ⇒ r_2 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)^3 ))) for the circles corresponding with the points B and C just interchange {a, b, c} with {b, c, a} and {c, a, b}
newidea(andsolution)toquestions35178&35195triangle:ABC;a=BC,b=CA,c=AB;α=∠CAB,β=∠ABC,γ=∠BCAd=(a+b+c)(a+b−c)(a−b+c)(−a+b+c)putitasthis:A=(00),B=(c0),C=(−a2+b2+c22cd2c)circumcircle:center=M1=(c2(a2+b2−c2)c2d)radius=R=abcd(calculatedbyintersectionofcircleswithcentersA,B,Corofsymmetry−axesofABandAC)2circlestouchingb,candcircumcircle,onefrominside,theotherfromoutside:center=M2liesony=kxwithk=tanα2M2=(xxtanα2)radius=r1=R−∣M1M2∣=xtanα2(inside)r2=∣M1M2∣−R=xtanα2(outside)(obviouslyanycirclewithcenterM2(x)andtouchingthex−axishasradiusxtanα2andalsoobviouslythetouchingpointof2circlesislocatedonthelineconnectingtheircenters)1.∣M1M2∣=R−xtanα2M1M2=(R−xtanα2)22.∣M1M2∣=R+xtanα2M1M2=(R+xtanα2)2tanα2=sinα2cosα2=1−cosα21+cosα2=1−cosα1+cosα=[cosα=−a2+b2+c22bc]=a2−b2+2bc−c2−a2+b2+2bc+c2=(a+b−c)(a−b+c)(a+b+c)(−a+b+c)M1M2=(m1−m2)2+(n1−n2)2==(c2−x)2+((a2+b2−c2)c2d−xtanα2)2=[aftersometransformationwork]=4bc(a+b+c)(−a+b+c)x2−2bc(b+c)(a+b+c)(−a+b+c)x+a2b2c2d2[a2b2c2d2=R2](R±xtanα2)2=x2tan2α2±2Rxtanα2+R2==(a+b−c)(a−b+c)(a+b+c)(−a+b+c)x2±2abc(a+b+c)(−a+b+c)x+R2sowehave4bc(a+b+c)(−a+b+c)x2−2bc(b+c)(a+b+c)(−a+b+c)x==(a+b−c)(a−b+c)(a+b+c)(−a+b+c)x2±2abc(a+b+c)(−a+b+c)xwhichleadstox3=0(asIexplainedbefore,thepointAcanbeseenasacirclewithradius0stillmeetingtherequirements)x1=2bca+b+c⇒r1=2bc(a+b−c)(a−b+c)(a+b+c)3(−a+b+c)x2=2bc−a+b+c⇒r2=2bc(a+b−c)(a−b+c)(a+b+c)(−a+b+c)3forthecirclescorrespondingwiththepointsBandCjustinterchange{a,b,c}with{b,c,a}and{c,a,b}

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