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Question Number 50352 by mr W last updated on 16/Dec/18

The distances from a point to the sides  of a triangle are p,q,r. Find the   maximum (or minimum) area of the  triangle, if it exists.  Assume r≤q≤p.

Thedistancesfromapointtothesidesofatrianglearep,q,r.Findthemaximum(orminimum)areaofthetriangle,ifitexists.Assumerqp.

Commented by ajfour last updated on 16/Dec/18

Commented by ajfour last updated on 17/Dec/18

D[rcos θ−pcos (θ+φ)]=Nsin φsin (2θ+φ)  xcos θ−ysin θ = q  eq. of BC  xcos φ+ysin φ = r  y_C  = ((r−pcos φ)/(sin φ))    ,   y_A  = ((pcos θ−q)/(sin θ))  x_B  = ((qsin φ+rsin θ)/(sin (θ+φ)))  △ = (((x_B −p)/2))(y_C −y_A )      = [((qsin φ+rsin θ−psin (θ+φ))/(2sin (θ+φ)))]             ×[((qsin φ+rsin θ−psin (θ+φ))/(sin θsin φ))]  ___________________________  △ = (([qsin φ+rsin θ−psin (θ+φ)]^2 )/(2sin (θ+φ)sin θsin φ))  ___________________________

D[rcosθpcos(θ+ϕ)]=Nsinϕsin(2θ+ϕ)xcosθysinθ=qeq.ofBCxcosϕ+ysinϕ=ryC=rpcosϕsinϕ,yA=pcosθqsinθxB=qsinϕ+rsinθsin(θ+ϕ)=(xBp2)(yCyA)=[qsinϕ+rsinθpsin(θ+ϕ)2sin(θ+ϕ)]×[qsinϕ+rsinθpsin(θ+ϕ)sinθsinϕ]___________________________=[qsinϕ+rsinθpsin(θ+ϕ)]22sin(θ+ϕ)sinθsinϕ___________________________

Commented by ajfour last updated on 16/Dec/18

yes Sir, but still not an easy going..

yesSir,butstillnotaneasygoing..

Commented by mr W last updated on 16/Dec/18

it is indeed not easy.

itisindeednoteasy.

Commented by mr W last updated on 17/Dec/18

thank you sir for the working!  i′m still thinking about this question.  we can see in this diagram:  when θ and φ →π/2, Δ→∞  and when θ and φ →0, Δ→∞  does it not mean that a mininmum  for Δ should exist?

thankyousirfortheworking!imstillthinkingaboutthisquestion.wecanseeinthisdiagram:whenθandϕπ/2,Δandwhenθandϕ0,ΔdoesitnotmeanthatamininmumforΔshouldexist?

Answered by mr W last updated on 16/Dec/18

Commented by mr W last updated on 16/Dec/18

draw three circles from the same center  with radii p,q,r.  it is to find a triangle whose sides  tangent these three circles and has  maximum area (like ΔABC) or  minimum area (like ΔA′B′C′).  A(−a,0) and B(b,0) and C(h,k).  Eqn. of AC:  (y/k)=((x+a)/(h+a))  ⇒(k/(h+a))x−y+((ak)/(h+a))=0  Eqn. of circle with radius r:  x^2 +(y−p)^2 =r^2   AC is tangent,  ⇒((k/(h+a)))^2 r^2 +r^2 =(−p+((ak)/(h+a)))^2 =p^2 −2pa((k/(h+a)))+a^2 ((k/(h+a)))^2   ⇒(a^2 −r^2 )((k/(h+a)))^2 −2pa((k/(h+a)))+(p^2 −r^2 )=0  ⇒(k/(h+a))=((pa±(√(p^2 a^2 −(a^2 −r^2 )(p^2 −r^2 ))))/(a^2 −r^2 ))=((pa±r(√(p^2 −r^2 +a^2 )))/(a^2 −r^2 ))  we consider at first the big triangle,  ⇒(k/(h+a))=((pa+r(√(p^2 −r^2 +a^2 )))/(a^2 −r^2 ))  ⇒h=((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))k−a    Eqn. of BC:  (y/k)=((x−b)/(h−b))  ⇒(k/(h−b))x−y−((bk)/(h−b))=0  Eqn. of circle with radius q:  x^2 +(y−p)^2 =q^2   BC is tangent,  ⇒((k/(h−b)))^2 q^2 +q^2 =(−p−((bk)/(h−b)))^2 =p^2 +2pb((k/(h−b)))+b^2 ((k/(h−b)))^2   ⇒(b^2 −q^2 )((k/(h−b)))^2 +2pb((k/(h−b)))+(p^2 −q^2 )=0  ⇒(k/(h−b))=((−pb+q(√(p^2 −q^2 +b^2 )))/(b^2 −q^2 ))  ⇒h=((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))k+b  ⇒((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))k−a=((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))k+b  ⇒[((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))]k=a+b  ⇒k=((a+b)/(((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))))    area of ΔABC  Δ=(1/2)(a+b)k  ⇒Δ=(1/2)×(((a+b)^2 )/(((a^2 −r^2 )/(pa+r(√(p^2 −r^2 +a^2 ))))−((b^2 −q^2 )/(−pb+q(√(p^2 −q^2 +b^2 ))))))  (∂Δ/∂a)=(∂Δ/∂b)=0  ......

drawthreecirclesfromthesamecenterwithradiip,q,r.itistofindatrianglewhosesidestangentthesethreecirclesandhasmaximumarea(likeΔABC)orminimumarea(likeΔABC).A(a,0)andB(b,0)andC(h,k).Eqn.ofAC:yk=x+ah+akh+axy+akh+a=0Eqn.ofcirclewithradiusr:x2+(yp)2=r2ACistangent,(kh+a)2r2+r2=(p+akh+a)2=p22pa(kh+a)+a2(kh+a)2(a2r2)(kh+a)22pa(kh+a)+(p2r2)=0kh+a=pa±p2a2(a2r2)(p2r2)a2r2=pa±rp2r2+a2a2r2weconsideratfirstthebigtriangle,kh+a=pa+rp2r2+a2a2r2h=a2r2pa+rp2r2+a2kaEqn.ofBC:yk=xbhbkhbxybkhb=0Eqn.ofcirclewithradiusq:x2+(yp)2=q2BCistangent,(khb)2q2+q2=(pbkhb)2=p2+2pb(khb)+b2(khb)2(b2q2)(khb)2+2pb(khb)+(p2q2)=0khb=pb+qp2q2+b2b2q2h=b2q2pb+qp2q2+b2k+ba2r2pa+rp2r2+a2ka=b2q2pb+qp2q2+b2k+b[a2r2pa+rp2r2+a2b2q2pb+qp2q2+b2]k=a+bk=a+ba2r2pa+rp2r2+a2b2q2pb+qp2q2+b2areaofΔABCΔ=12(a+b)kΔ=12×(a+b)2a2r2pa+rp2r2+a2b2q2pb+qp2q2+b2Δa=Δb=0......

Commented by mr W last updated on 17/Dec/18

you′re right.  x→0,y→0 ⇒Δ→0  x→∞,y→∞⇒Δ→∞  I thought there is a minimum for  ΔABC and a maximum for ΔA′B′C′.  but this is not true. so this is not a  good question. thanks for trying and  for the correct conclusion.

youreright.x0,y0Δ0x,yΔIthoughtthereisaminimumforΔABCandamaximumforΔABC.butthisisnottrue.sothisisnotagoodquestion.thanksfortryingandforthecorrectconclusion.

Commented by ajfour last updated on 17/Dec/18

but i think minimum is zero  and maximum → ∞.

butithinkminimumiszeroandmaximum.

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