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Question Number 45730 by ajfour last updated on 16/Oct/18

Commented by ajfour last updated on 16/Oct/18

If base BC of △ABC has slipped to the base edges of box and its sides AB and AC touch (rests against) poimts P and Q respectively of the two upper edges of the box, find A(x,y,z) in terms of sides a,b, c of △ and dimensions l, w, h of the open box.

IfbaseBCof△ABChasslippedtothebaseedgesofboxanditssidesABandACtouch(restsagainst)poimtsPandQrespectivelyofthetwoupperedgesofthebox,findA(x,y,z)intermsofsidesa,b,cof△anddimensionsl,w,hoftheopenbox.

Commented by ajfour last updated on 17/Oct/18

MrW Sir please solve this.. what a Question !

MrWSirpleasesolvethis..whataQuestion!

Commented by MrW3 last updated on 17/Oct/18

I′ll give a try using classic method, but without knowing how far one can go.

I′llgiveatryusingclassicmethod,butwithoutknowinghowfaronecango.

Answered by ajfour last updated on 17/Oct/18

let y_P = p , x_Q =q r_A ^� = xi^� +yj^� +zk^� = x_B i^� +λ[(l−x_B )i^� +p j^� +hk^� ] = y_C j^� +μ[qi^� +(w−y_C )j^� +hk^� ] ⇒ λ=μ = ε (say) q=l+(((1−ε)/ε))x_B ....(i) p=w+(((1−ε)/ε))y_C ....(ii) (l−x_B )^2 +p^2 +h^2 = c^2 /ε^2 ....(iii) q^2 +(w−y_C )^2 +h^2 = b^2 /ε^2 ...(iv) x_B ^2 +y_C ^2 =a^2 ....(v) let ((1−ε)/ε)= f ⇒ ε = (1/(1+f)) using (ii) in (iii) (l−x_B )^2 +(w+fy_C )^2 +h^2 =(1+f)^2 c^2 similarly using (i) in (iv) (l+fx_B )^2 +(w−y_C )^2 +h^2 =(1+f)^2 b^2 let l^2 +w^2 +h^2 = s^2 ⇒ −2lx_B +x_B ^2 +2fwy_C +f^2 y_C ^2 = (1+f)^2 c^2 −s^2 ...(I) −2wy_C +y_C ^2 +2flx_B +f^2 x_B ^2 = (1+f)^2 b^2 −s^2 ..(II) adding the two −2(1−f)(lx_B +wy_C )+(1+f^2 )a^2 = (1+f)^2 (b^2 +c^2 )−2s^2 let lx_B +wy_C =(a(√(l^2 +w^2 )) )sin φ sin φ = (((1+f)^2 (b^2 +c^2 )−(1+f^2 )a^2 −2s^2 )/(2(f−1)a(√(l^2 +w^2 )))) and x_B ^2 +y_C ^2 = a^2 ; ⇒ x_B =acos θ , y_C = asin θ ⇒ a(√(l^2 +w^2 )) sin(θ+tan^(−1) (l/w))= (a(√(l^2 +w^2 )) )sin φ ⇒ θ = φ−β ; β = tan^(−1) (l/w) hence x_B = acos (φ−β) y_C = asin (φ−β) since φ(f) function of f we can now use either of eqs. (I) and (II) to obtain f ; then φ, x_B , y_B , p, q, and finally r_A = xi^� +yj^� +zk^� ; A(x,y,z) .

letyP=p,xQ=qr¯A=xi^+yj^+zk^=xBi^+λ[(l−xB)i^+pj^+hk^]=yCj^+μ[qi^+(w−yC)j^+hk^]⇒λ=μ=ϵ(say)q=l+(1−ϵϵ)xB....(i)p=w+(1−ϵϵ)yC....(ii)(l−xB)2+p2+h2=c2/ϵ2....(iii)q2+(w−yC)2+h2=b2/ϵ2...(iv)xB2+yC2=a2....(v)let1−ϵϵ=f⇒ϵ=11+fusing(ii)in(iii)(l−xB)2+(w+fyC)2+h2=(1+f)2c2similarlyusing(i)in(iv)(l+fxB)2+(w−yC)2+h2=(1+f)2b2letl2+w2+h2=s2⇒−2lxB+xB2+2fwyC+f2yC2=(1+f)2c2−s2...(I)−2wyC+yC2+2flxB+f2xB2=(1+f)2b2−s2..(II)addingthetwo−2(1−f)(lxB+wyC)+(1+f2)a2=(1+f)2(b2+c2)−2s2letlxB+wyC=(al2+w2)sinϕsinϕ=(1+f)2(b2+c2)−(1+f2)a2−2s22(f−1)al2+w2andxB2+yC2=a2;⇒xB=acosθ,yC=asinθ⇒al2+w2sin(θ+tan−1lw)=(al2+w2)sinϕ⇒θ=ϕ−β;β=tan−1lwhencexB=acos(ϕ−β)yC=asin(ϕ−β)sinceϕ(f)functionoffwecannowuseeitherofeqs.(I)and(II)toobtainf;thenϕ,xB,yB,p,q,andfinallyrA=xi^+yj^+zk^;A(x,y,z).

Commented by MrW3 last updated on 17/Oct/18

very powerful sir!

verypowerfulsir!

Answered by MrW3 last updated on 17/Oct/18

Commented by MrW3 last updated on 17/Oct/18

Commented by MrW3 last updated on 17/Oct/18

Commented by MrW3 last updated on 17/Oct/18

the position of triangle ABC is given by parameters θ and φ. the altitude of ΔABC over BC is d: Area ΔABC=(1/2)ad=(√(s(s−a)(s−b)(s−c))) ⇒d=(2/a)(√(s(s−a)(s−b)(s−c))) with s=((a+b+c)/2) z_A =d sin φ=b sin φ_1 =c sin φ_2 ⇒sin φ_1 =(d/b) sin φ ⇒sin φ_2 =(d/c) sin φ MR=(h/(sin φ)) PQ=((MR)/d)a=((ah)/(d sin φ)) BP′=(h/(tan φ_2 ))=((h(√(1−(d^2 /c^2 )sin^2 φ)))/((d/c)sin φ))=(h/(sin φ))(√((c^2 /d^2 )−sin^2 φ)) CQ′=(h/(tan φ_1 ))=((h(√(1−(d^2 /b^2 )sin^2 φ)))/((d/b)sin φ))=(h/(sin φ))(√((b^2 /d^2 )−sin^2 φ)) BP′^2 =(l−a cos θ)^2 +(w−PQ sin θ)^2 =(h^2 /(sin^2 φ))((c^2 /d^2 )−sin^2 φ) ⇒(l−a cos θ)^2 +(w−((ah)/(d sin φ)) sin θ)^2 =h^2 ((c^2 /d^2 ) sin^2 φ−1) ⇒l^2 −2al cos θ+a^2 cos^2 θ+w^2 −((2ahw)/(d sin φ)) sin θ+((a^2 h^2 sin^2 θ)/(d^2 sin^2 φ))=((h^2 c^2 )/d^2 ) sin^2 φ−h^2 ⇒2al cos θ−a^2 cos^2 θ+((2ahw sin θ)/(d sin φ)) −((a^2 h^2 sin^2 θ)/(d^2 sin^2 φ))+((h^2 c^2 sin^2 φ)/d^2 )−(l^2 +w^2 +h^2 )=0 ⇒2ad^2 l cos θ sin^2 φ−a^2 d^2 cos^2 θ sin^2 φ+2adhw sin θ sin φ −a^2 h^2 sin^2 θ+c^2 h^2 sin^4 φ−d^2 (l^2 +w^2 +h^2 ) sin^2 φ=0 ...(i) CQ′^2 =(w−a sin θ)^2 +(l−PQ cos θ)^2 =(h^2 /(sin^2 φ))((b^2 /d^2 )−sin^2 φ) ⇒(w−a sin θ)^2 +(l−((ah)/(d sin φ)) cos θ)^2 =(h^2 /(sin^2 φ))((b^2 /d^2 )−sin^2 φ) ⇒w^2 −2aw sin θ+a^2 sin^2 θ+l^2 −((2ahl)/(d sin φ)) cos θ+((a^2 h^2 cos^2 θ)/(d^2 sin^2 φ))=((h^2 b^2 )/d^2 ) sin^2 φ−h^2 ⇒2aw sin θ−a^2 sin^2 θ+((2ahl)/(d sin φ)) cos θ−((a^2 h^2 cos^2 θ)/(d^2 sin^2 φ))+((h^2 b^2 )/d^2 ) sin^2 φ=l^2 +w^2 +h^2 ⇒2ad^2 w sin θ sin^2 φ−a^2 d^2 sin^2 θ sin^2 φ+2adhl cos θ sin φ−a^2 h^2 cos^2 θ+b^2 h^2 sin^4 φ−d^2 (l^2 +w^2 +h^2 ) sin^2 φ=0 ...(ii) .... .... two unknowns θ and φ in two eqn.

thepositionoftriangleABCisgivenbyparametersθandϕ.thealtitudeofΔABCoverBCisd:AreaΔABC=12ad=s(s−a)(s−b)(s−c)⇒d=2as(s−a)(s−b)(s−c)withs=a+b+c2zA=dsinϕ=bsinϕ1=csinϕ2⇒sinϕ1=dbsinϕ⇒sinϕ2=dcsinϕMR=hsinϕPQ=MRda=ahdsinϕBP′=htanϕ2=h1−d2c2sin2ϕdcsinϕ=hsinϕc2d2−sin2ϕCQ′=htanϕ1=h1−d2b2sin2ϕdbsinϕ=hsinϕb2d2−sin2ϕBP′2=(l−acosθ)2+(w−PQsinθ)2=h2sin2ϕ(c2d2−sin2ϕ)⇒(l−acosθ)2+(w−ahdsinϕsinθ)2=h2(c2d2sin2ϕ−1)⇒l2−2alcosθ+a2cos2θ+w2−2ahwdsinϕsinθ+a2h2sin2θd2sin2ϕ=h2c2d2sin2ϕ−h2⇒2alcosθ−a2cos2θ+2ahwsinθdsinϕ−a2h2sin2θd2sin2ϕ+h2c2sin2ϕd2−(l2+w2+h2)=0⇒2ad2lcosθsin2ϕ−a2d2cos2θsin2ϕ+2adhwsinθsinϕ−a2h2sin2θ+c2h2sin4ϕ−d2(l2+w2+h2)sin2ϕ=0...(i)CQ′2=(w−asinθ)2+(l−PQcosθ)2=h2sin2ϕ(b2d2−sin2ϕ)⇒(w−asinθ)2+(l−ahdsinϕcosθ)2=h2sin2ϕ(b2d2−sin2ϕ)⇒w2−2awsinθ+a2sin2θ+l2−2ahldsinϕcosθ+a2h2cos2θd2sin2ϕ=h2b2d2sin2ϕ−h2⇒2awsinθ−a2sin2θ+2ahldsinϕcosθ−a2h2cos2θd2sin2ϕ+h2b2d2sin2ϕ=l2+w2+h2⇒2ad2wsinθsin2ϕ−a2d2sin2θsin2ϕ+2adhlcosθsinϕ−a2h2cos2θ+b2h2sin4ϕ−d2(l2+w2+h2)sin2ϕ=0...(ii)........twounknownsθandϕintwoeqn.

Commented by MrW3 last updated on 17/Oct/18

Commented by ajfour last updated on 17/Oct/18

i have corrected my solution only now.. thank you Sir, for yours.. Excellent diagrams, let me have some time comprehending your solution.

ihavecorrectedmysolutiononlynow..thankyouSir,foryours..Excellentdiagrams,letmehavesometimecomprehendingyoursolution.

Commented by MrW3 last updated on 17/Oct/18

let p=tan (θ/2), q=sin φ ⇒2ad^2 l (((1−p^2 )/(1+p^2 )))q^2 −a^2 d^2 (((1−p^2 )/(1+p^2 )))^2 q^2 +2adhw(((2p)/(1+p^2 )))q−a^2 h^2 (((2p)/(1+p^2 )))^2 +c^2 h^2 q^4 −d^2 (l^2 +w^2 +h^2 )q^2 =0 ⇒(1/((l^2 +w^2 +h^2 )))[2l−a+(2l+a)p^2 ](1−p^2 )q^2 +((4ahw)/(d(l^2 +w^2 +h^2 )))p(1+p^2 )q−((4a^2 h^2 )/(d^2 (l^2 +w^2 +h^2 )))p^2 +[((c^2 h^2 )/(d^2 (l^2 +w^2 +h^2 )))q^2 −1](1+p^2 )^2 q^2 =0 ...(i) ⇒4ad^2 wp(1+p^2 )^2 q^2 −4a^2 d^2 p^2 q^2 +2adhl(1−p^4 )q−a^2 h^2 (1−p^2 )^2 +b^2 h^2 (1+p^2 )^2 q^4 −d^2 (l^2 +w^2 +h^2 )(1+p^2 )q^2 =0 ⇒((4a)/((l^2 +w^2 +h^2 )))[w(1+p^2 )^2 −ap]pq^2 +((ah)/(d^2 (l^2 +w^2 +h^2 )))[2dl(1+p^2 )q−ah(1−p^2 )](1−p^2 )+[((b^2 h^2 )/(d^2 (l^2 +w^2 +h^2 )))q^2 −1](1+p^2 )^2 q^2 =0 ....(ii)

letp=tanθ2,q=sinϕ⇒2ad2l(1−p21+p2)q2−a2d2(1−p21+p2)2q2+2adhw(2p1+p2)q−a2h2(2p1+p2)2+c2h2q4−d2(l2+w2+h2)q2=0⇒1(l2+w2+h2)[2l−a+(2l+a)p2](1−p2)q2+4ahwd(l2+w2+h2)p(1+p2)q−4a2h2d2(l2+w2+h2)p2+[c2h2d2(l2+w2+h2)q2−1](1+p2)2q2=0...(i)⇒4ad2wp(1+p2)2q2−4a2d2p2q2+2adhl(1−p4)q−a2h2(1−p2)2+b2h2(1+p2)2q4−d2(l2+w2+h2)(1+p2)q2=0⇒4a(l2+w2+h2)[w(1+p2)2−ap]pq2+ahd2(l2+w2+h2)[2dl(1+p2)q−ah(1−p2)](1−p2)+[b2h2d2(l2+w2+h2)q2−1](1+p2)2q2=0....(ii)

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