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Question-190340

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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=33⇒h=(a+b)32△ABCAEO2semblableshS+a=32⇒S=2h3−aS=(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,b−bcosϕ)P(asinθ,−a+acosθ)letM(h,k)bsinϕ=h+s2cosδb−bcosϕ=k+s2sinδasinθ=h−s2cosδ−a+acosθ=k−s2sinδ⇒2k=b−a+acosθ−bcosϕssinδ=b+a−acosθ−bcosϕ2h=bsinϕ+asinθscosδ=bsinϕ−asinθT(p,q)p=h+s32sinδq=k−s32cosδ(h+s32sinδ)2+(k−s32cosδ−b+a)2=(a+b)2θandϕarethenrelatedherein{bsinϕ+asinθ+3(b+a−acosθ−bcosϕ)}2+{(b−a+acosθ−bcosϕ)−3(bsinϕ−asinθ)−2(b−a)}2=4(a+b)2whiles2=(b+a−acosθ−bcosϕ)2+(bsinϕ−asinθ)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)(x−rsinθ)2+(y−rcosθ)2=s2pointA:(xA−rsinθ)2+(yA−rcosθ)2=s2(xA−b)2+yA2=a2(b−rsinθ)(2xA−b−rsinθ)−rcosθ(2yA−rcosθ)=s2−a22(b−rsinθ)xA−2rcosθyA=s2−a2+b2−r2?⇒yA=2(b−rsinθ)xA−s2+a2−b2+r22rcosθ(xA−b)2+[2(b−rsinθ)xA−s2+a2−b2+r22rcosθ]2=a2xA2−2bxA+b2+[(b−rsinθ)xArcosθ−s2−a2+b2−r22rcosθ]2=a2[1+(b−rsinθ)2r2cos2θ]xA2−2[b+(b−rsinθ)(s2−a2+b2−r2)2r2cos2θ]xA+(s2−a2+b2−r2)24r2cos2θ+b2−a2=0⇒xA=11+(b−rsinθ)2r2cos2θ{b+(b−rsinθ)(s2−a2+b2−r2)2r2cos2θ±[b+(b−rsinθ)(s2−a2+b2−r2)2r2cos2θ]2−[1+(b−rsinθ)2r2cos2θ][(s2−a2+b2−r2)24r2cos2θ+b2−a2]}pointB:(xB−rsinθ)2+(yB−rcosθ)2=s2(xB+a)2+yB2=b2−(a+rsinθ)(2xB+a−rsinθ)−rcosθ(2yB−rcosθ)=s2−b22(a+rsinθ)xB+2rcosθyB=−s2+b2−a2+r2⇒yB=−2(a+rsinθ)xB−s2+b2−a2+r22rcosθ(xB+a)2+[−2(a+rsinθ)xB−s2+b2−a2+r22rcosθ]2=b2(xB+a)2+[(a+rsinθ)xBrcosθ+s2−b2+a2−r22rcosθ]2=b2[1+(a+rsinθ)2r2cos2θ]xB2+2[a+(a+rsinθ)(s2−b2+a2−r2)2r2cos2θ]xB+(s2−b2+a2−r2)24r2cos2θ+a2−b2=0⇒xB=11+(a+rsinθ)2r2cos2θ{−a−(a+rsinθ)(s2−b2+a2−r2)2r2cos2θ±[a+(a+rsinθ)(s2−b2+a2−r2)2r2cos2θ]2−[1+(a+rsinθ)2r2cos2θ][(s2−b2+a2−r2)24r2cos2θ+a2−b2]}AB=s:(xA−xB)2+(yA−yB)2=s2xA2+yA2+xB2+yB2−2xAxB−2yAyB=s2⇒(a+xA)(b−xB)−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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