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Question Number 190340 by ajfour last updated on 01/Apr/23

Commented by a.lgnaoui last updated on 01/Apr/23

Commented by a.lgnaoui last updated on 01/Apr/23

Commented by mr W last updated on 01/Apr/23

Commented by mr W last updated on 01/Apr/23

there are infinite equilateral triangles.  one is the smallest (red one) and one  is the largest (green one).

thereareinfiniteequilateraltriangles.oneisthesmallest(redone)andoneisthelargest(greenone).

Answered by a.lgnaoui last updated on 01/Apr/23

soit MN Droites tengent aux cercles  MN∣∣ EA  et  EA∣∣BC  C_1 (O_1 ,b)  et cercleC_2 (O_2 ,a)  triangle EAO_2 (Equilaterale)  OA=R  point A tel que O_2 A=EA=R  R=a+b  tan (π/6)=(R/(2h))=((√3)/3)⇒h=(((a+b)(√3))/2)  △ABC   AEO_2   semblables   (h/(S+a))=((√3)/2)⇒S=((2h)/( (√3)))−a  S=(a+b)−a                  S=b

soitMNDroitestengentauxcerclesMN∣∣EAetEA∣∣BCC1(O1,b)etcercleC2(O2,a)triangleEAO2(Equilaterale)OA=RpointAtelqueO2A=EA=RR=a+btanπ6=R2h=33h=(a+b)32ABCAEO2semblableshS+a=32S=2h3aS=(a+b)aS=b

Answered by ajfour last updated on 01/Apr/23

Commented by ajfour last updated on 07/Apr/23

B(0,b)   A(0,−a)  Q(bsin φ, b−bcos φ)  P(asin θ,−a+acos θ)  let  M(h,k)  bsin φ=h+(s/2)cos δ  b−bcos φ=k+(s/2)sin δ  asin θ=h−(s/2)cos δ  −a+acos θ=k−(s/2)sin δ  ⇒  2k=b−a+acos θ−bcos φ  ssin δ=b+a−acos θ−bcos φ  2h=bsin φ+asin θ  scos δ=bsin φ−asin θ  T(p,q)  p=h+((s(√3))/2)sin δ  q=k−((s(√3))/2)cos δ  (h+((s(√3))/2)sin δ)^2 +(k−((s(√3))/2)cos δ−b+a)^2   =(a+b)^2   θ and φ  are then related herein  {bsin φ+asin θ+(√3)(b+a−acos θ−bcos φ)}^2   +{(b−a+acos θ−bcos φ)          −(√3)(bsin φ−asin θ)−2(b−a)}^2   =4(a+b)^2   while  s^2 =(b+a−acos θ−bcos φ)^2           +(bsin φ−asin θ)^2   .....

B(0,b)A(0,a)Q(bsinϕ,bbcosϕ)P(asinθ,a+acosθ)letM(h,k)bsinϕ=h+s2cosδbbcosϕ=k+s2sinδasinθ=hs2cosδa+acosθ=ks2sinδ2k=ba+acosθbcosϕssinδ=b+aacosθbcosϕ2h=bsinϕ+asinθscosδ=bsinϕasinθT(p,q)p=h+s32sinδq=ks32cosδ(h+s32sinδ)2+(ks32cosδb+a)2=(a+b)2θandϕarethenrelatedherein{bsinϕ+asinθ+3(b+aacosθbcosϕ)}2+{(ba+acosθbcosϕ)3(bsinϕasinθ)2(ba)}2=4(a+b)2whiles2=(b+aacosθbcosϕ)2+(bsinϕasinθ)2.....

Answered by mr W last updated on 02/Apr/23

Commented by mr W last updated on 04/Apr/23

r=a+b  C(r sin θ, r cos θ)  (x−r sin θ)^2 +(y−r cos θ)^2 =s^2   point A:  (x_A −r sin θ)^2 +(y_A −r cos θ)^2 =s^2   (x_A −b)^2 +y_A ^2 =a^2   (b−r sin θ)(2x_A −b−r sin θ)−r cos θ (2y_A −r cos θ)=s^2 −a^2   2(b−r sin θ)x_A −2r cos θ y_A =s^2 −a^2 +b^2 −r^2 ?  ⇒y_A =((2(b−r sin θ)x_A −s^2 +a^2 −b^2 +r^2 )/(2r cos θ))  (x_A −b)^2 +[((2(b−r sin θ)x_A −s^2 +a^2 −b^2 +r^2 )/(2r cos θ))]^2 =a^2   x_A ^2 −2bx_A +b^2 +[(((b−r sin θ)x_A )/(r cos θ))−((s^2 −a^2 +b^2 −r^2 )/(2r cos θ))]^2 =a^2   [1+(((b−r sin θ)^2 )/(r^2  cos^2  θ))]x_A ^2 −2[b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2  cos^2  θ))]x_A +(((s^2 −a^2 +b^2 −r^2 )^2 )/(4r^2  cos^2  θ))+b^2 −a^2 =0  ⇒x_A =(1/(1+(((b−r sin θ)^2 )/(r^2  cos^2  θ)))){b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2  cos^2  θ))±(√([b+(((b−r sin θ)(s^2 −a^2 +b^2 −r^2 ))/(2r^2  cos^2  θ))]^2 −[1+(((b−r sin θ)^2 )/(r^2  cos^2  θ))][(((s^2 −a^2 +b^2 −r^2 )^2 )/(4r^2  cos^2  θ))+b^2 −a^2 ]))}  point B:  (x_B −r sin θ)^2 +(y_B −r cos θ)^2 =s^2   (x_B +a)^2 +y_B ^2 =b^2   −(a+r sin θ)(2x_B +a−r sin θ)−r cos θ (2y_B −r cos θ)=s^2 −b^2   2(a+r sin θ) x_B +2r cos θ y_B =−s^2 +b^2 −a^2 +r^2   ⇒y_B =((−2(a+r sin θ)x_B −s^2 +b^2 −a^2 +r^2 )/(2r cos θ))  (x_B +a)^2 +[((−2(a+r sin θ)x_B −s^2 +b^2 −a^2 +r^2 )/(2r cos θ))]^2 =b^2   (x_B +a)^2 +[(((a+r sin θ)x_B )/(r cos θ))+((s^2 −b^2 +a^2 −r^2 )/(2r cos θ))]^2 =b^2   [1+(((a+r sin θ)^2 )/(r^2 cos^2  θ))]x_B ^2 +2[a+(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2  cos^2  θ))]x_B +(((s^2 −b^2 +a^2 −r^2 )^2 )/(4r^2  cos^2  θ))+a^2 −b^2 =0  ⇒x_B =(1/(1+(((a+r sin θ)^2 )/(r^2 cos^2  θ)))){−a−(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2  cos^2  θ))±(√([a+(((a+r sin θ)(s^2 −b^2 +a^2 −r^2 ))/(2r^2  cos^2  θ))]^2 −[1+(((a+r sin θ)^2 )/(r^2 cos^2  θ))][(((s^2 −b^2 +a^2 −r^2 )^2 )/(4r^2  cos^2  θ))+a^2 −b^2 ]))}    AB=s:  (x_A −x_B )^2 +(y_A −y_B )^2 =s^2   x_A ^2 +y_A ^2 +x_B ^2 +y_B ^2 −2x_A x_B −2y_A y_B =s^2   ⇒(a+x_A )(b−x_B )−y_A y_B =(s^2 /2)+ab  ......

r=a+bC(rsinθ,rcosθ)(xrsinθ)2+(yrcosθ)2=s2pointA:(xArsinθ)2+(yArcosθ)2=s2(xAb)2+yA2=a2(brsinθ)(2xAbrsinθ)rcosθ(2yArcosθ)=s2a22(brsinθ)xA2rcosθyA=s2a2+b2r2?yA=2(brsinθ)xAs2+a2b2+r22rcosθ(xAb)2+[2(brsinθ)xAs2+a2b2+r22rcosθ]2=a2xA22bxA+b2+[(brsinθ)xArcosθs2a2+b2r22rcosθ]2=a2[1+(brsinθ)2r2cos2θ]xA22[b+(brsinθ)(s2a2+b2r2)2r2cos2θ]xA+(s2a2+b2r2)24r2cos2θ+b2a2=0xA=11+(brsinθ)2r2cos2θ{b+(brsinθ)(s2a2+b2r2)2r2cos2θ±[b+(brsinθ)(s2a2+b2r2)2r2cos2θ]2[1+(brsinθ)2r2cos2θ][(s2a2+b2r2)24r2cos2θ+b2a2]}pointB:(xBrsinθ)2+(yBrcosθ)2=s2(xB+a)2+yB2=b2(a+rsinθ)(2xB+arsinθ)rcosθ(2yBrcosθ)=s2b22(a+rsinθ)xB+2rcosθyB=s2+b2a2+r2yB=2(a+rsinθ)xBs2+b2a2+r22rcosθ(xB+a)2+[2(a+rsinθ)xBs2+b2a2+r22rcosθ]2=b2(xB+a)2+[(a+rsinθ)xBrcosθ+s2b2+a2r22rcosθ]2=b2[1+(a+rsinθ)2r2cos2θ]xB2+2[a+(a+rsinθ)(s2b2+a2r2)2r2cos2θ]xB+(s2b2+a2r2)24r2cos2θ+a2b2=0xB=11+(a+rsinθ)2r2cos2θ{a(a+rsinθ)(s2b2+a2r2)2r2cos2θ±[a+(a+rsinθ)(s2b2+a2r2)2r2cos2θ]2[1+(a+rsinθ)2r2cos2θ][(s2b2+a2r2)24r2cos2θ+a2b2]}AB=s:(xAxB)2+(yAyB)2=s2xA2+yA2+xB2+yB22xAxB2yAyB=s2(a+xA)(bxB)yAyB=s22+ab......

Commented by ajfour last updated on 06/Apr/23

lengthy but you solved it, sir. Very  commendable! thank you.

lengthybutyousolvedit,sir.Verycommendable!thankyou.

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