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Question-88188

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Question Number 88188 by mr W last updated on 08/Apr/20
Commented by mr W last updated on 08/Apr/20
Given: triangle ABC with side lengthes a, b, c. P is midpoint of BC. Find: perimeter of inscribed triangle PQR with the smallest perimeter.
$${Given}:\:{triangle}\:{ABC}\:{with}\:{side}\:{lengthes} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a},\:{b},\:{c}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\:{is}\:{midpoint}\:{of}\:{BC}. \\ $$$${Find}:\:\:\:{perimeter}\:{of}\:{inscribed}\:{triangle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{PQR}\:{with}\:{the}\:{smallest}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{perimeter}. \\ $$
Commented by ajfour last updated on 09/Apr/20
Commented by ajfour last updated on 09/Apr/20
let ∠A=α , ∠B=β , ∠C=γ , ∠QPC=θ, p(slope of AB′)=tan (π−2γ−β) tan θ=m, tan β=q, A(h,k) AR=BR′ = l eq. of PQ: y=m(x−a/2) eq. of A ′B : y=−qx eq. of AB′ : y−k=p(x−h) for R′_(−) m(x−(a/2))=−qx x_(R′) =((m(a/2))/(q+m)) , y_(R′) =−((qm(a/2))/(q+m)) for R_(−) m(x−a/2)=k+p(x−h) x_R =((k−ph+m(a/2))/(m−p)) y_R =m(((k−ph+p(a/2))/(m−p))) l^2 =(BR′)^2 =[((m(a/2))/(q+m))]^2 (1+q^2 ) =(AR)^2 = (((k−mh+m(a/2))/(m−p)))^2 +{((m[−ph+p(a/2)]+pk)/(m−p))}^2 from above eq. we solve for m. Then min( perimeter △PQR ) = R′R .
$${let}\:\angle{A}=\alpha\:,\:\angle{B}=\beta\:,\:\angle{C}=\gamma\:, \\ $$$$\angle{QPC}=\theta,\: \\ $$$${p}\left({slope}\:{of}\:{AB}'\right)=\mathrm{tan}\:\left(\pi−\mathrm{2}\gamma−\beta\right) \\ $$$$\mathrm{tan}\:\theta={m},\:\mathrm{tan}\:\beta={q},\:\:{A}\left({h},{k}\right) \\ $$$${AR}={BR}'\:=\:{l} \\ $$$${eq}.\:{of}\:{PQ}:\:\:\:{y}={m}\left({x}−{a}/\mathrm{2}\right) \\ $$$${eq}.\:{of}\:{A}\:'{B}\::\:\:\:{y}=−{qx} \\ $$$${eq}.\:{of}\:{AB}'\::\:\:{y}−{k}={p}\left({x}−{h}\right) \\ $$$$\underset{−} {{for}\:{R}'} \\ $$$${m}\left({x}−\frac{{a}}{\mathrm{2}}\right)=−{qx} \\ $$$${x}_{{R}'} =\frac{{m}\left({a}/\mathrm{2}\right)}{{q}+{m}}\:,\:\:{y}_{{R}'} =−\frac{{qm}\left({a}/\mathrm{2}\right)}{{q}+{m}} \\ $$$$\underset{−} {{for}\:{R}} \\ $$$${m}\left({x}−{a}/\mathrm{2}\right)={k}+{p}\left({x}−{h}\right) \\ $$$${x}_{{R}} =\frac{{k}−{ph}+{m}\left({a}/\mathrm{2}\right)}{{m}−{p}} \\ $$$${y}_{{R}} ={m}\left(\frac{{k}−{ph}+{p}\left({a}/\mathrm{2}\right)}{{m}−{p}}\right) \\ $$$${l}^{\mathrm{2}} =\left({BR}'\right)^{\mathrm{2}} =\left[\frac{{m}\left({a}/\mathrm{2}\right)}{{q}+{m}}\right]^{\mathrm{2}} \left(\mathrm{1}+{q}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\left({AR}\right)^{\mathrm{2}} =\:\left(\frac{{k}−{mh}+{m}\left({a}/\mathrm{2}\right)}{{m}−{p}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\left\{\frac{{m}\left[−{ph}+{p}\left({a}/\mathrm{2}\right)\right]+{pk}}{{m}−{p}}\right\}^{\mathrm{2}} \\ $$$${from}\:{above}\:{eq}.\:{we}\:{solve}\:{for}\:{m}. \\ $$$${Then}\:{min}\left(\:{perimeter}\:\bigtriangleup{PQR}\:\right) \\ $$$$\:\:\:\:\:=\:{R}'{R}\:. \\ $$
Commented by ajfour last updated on 09/Apr/20
couldn′t compress it Sir!
$${couldn}'{t}\:{compress}\:{it}\:{Sir}! \\ $$
Commented by mr W last updated on 09/Apr/20
thank you sir! it is the geometric way which i prefer. i′ll try in a similar way.
$${thank}\:{you}\:{sir}!\:{it}\:{is}\:{the}\:{geometric}\:{way} \\ $$$${which}\:{i}\:{prefer}.\:{i}'{ll}\:{try}\:{in}\:{a}\:{similar}\:{way}. \\ $$
Answered by mr W last updated on 09/Apr/20
Commented by mr W last updated on 09/Apr/20
let ∠A=α, ∠B=β, ∠C=γ we construct the image of side BC about AB and about AC. P^ ′ and P ′′ are the corresponding image of P. BP ′=BP=(a/2), CP ′′=CP=(a/2) P ′R=PR, QP ′′=QP perimeter of ΔPQR=PQ+QR+PR =P ′R+RQ+QP ′′ we see it is minimum, if P ′,R,Q,P ′′ are collinear, then perimeter of ΔPQR=P ′P ′′ ϕ=π−2β φ=π−2γ θ=π−ϕ−φ=2(β+γ)−π=π−2α ((DB)/(sin φ))=((DC)/(sin ϕ))=((BC)/(sin θ)) ⇒DB=((sin φ)/(sin θ))×BC=((sin 2γ)/(sin 2α))×a ⇒DC=((sin ϕ)/(sin θ))×BC=((sin 2β)/(sin 2α))×a DP ′=DB+BP ′=((sin 2γ)/(sin 2α))×a+(a/2)=((1/2)+((sin 2γ)/(sin 2α)))a DP ′′=DC+CP ′′=((sin 2β)/(sin 2α))×a+(a/2)=((1/2)+((sin 2β)/(sin 2α)))a let l=P ′P ′′ l^2 =((1/2)+((sin 2β)/(sin 2α)))^2 a^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 a^2 −2((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α)))a^2 cos θ l^2 =a^2 {((1/2)+((sin 2β)/(sin 2α)))^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 +2((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α))) cos 2α} minimum perimeter of ΔPQR=P ′P ′′=l =a(√(((1/2)+((sin 2β)/(sin 2α)))^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 +2 cos 2α ((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α)))))
$${let}\:\angle{A}=\alpha,\:\angle{B}=\beta,\:\angle{C}=\gamma \\ $$$${we}\:{construct}\:{the}\:{image}\:{of}\:{side}\:{BC} \\ $$$${about}\:{AB}\:{and}\:{about}\:{AC}.\:{P}^{\:} '\:{and}\:{P}\:'' \\ $$$${are}\:{the}\:{corresponding}\:{image}\:{of}\:{P}. \\ $$$${BP}\:'={BP}=\frac{{a}}{\mathrm{2}},\:{CP}\:''={CP}=\frac{{a}}{\mathrm{2}} \\ $$$${P}\:'{R}={PR},\:{QP}\:''={QP} \\ $$$${perimeter}\:{of}\:\Delta{PQR}={PQ}+{QR}+{PR} \\ $$$$={P}\:'{R}+{RQ}+{QP}\:'' \\ $$$${we}\:{see}\:{it}\:{is}\:{minimum},\:{if}\:{P}\:',{R},{Q},{P}\:'' \\ $$$${are}\:{collinear},\:{then} \\ $$$${perimeter}\:{of}\:\Delta{PQR}={P}\:'{P}\:'' \\ $$$$ \\ $$$$\varphi=\pi−\mathrm{2}\beta \\ $$$$\phi=\pi−\mathrm{2}\gamma \\ $$$$\theta=\pi−\varphi−\phi=\mathrm{2}\left(\beta+\gamma\right)−\pi=\pi−\mathrm{2}\alpha \\ $$$$\frac{{DB}}{\mathrm{sin}\:\phi}=\frac{{DC}}{\mathrm{sin}\:\varphi}=\frac{{BC}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{DB}=\frac{\mathrm{sin}\:\phi}{\mathrm{sin}\:\theta}×{BC}=\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}×{a} \\ $$$$\Rightarrow{DC}=\frac{\mathrm{sin}\:\varphi}{\mathrm{sin}\:\theta}×{BC}=\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}×{a} \\ $$$${DP}\:'={DB}+{BP}\:'=\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}×{a}+\frac{{a}}{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right){a} \\ $$$${DP}\:''={DC}+{CP}\:''=\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}×{a}+\frac{{a}}{\mathrm{2}}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right){a} \\ $$$$ \\ $$$${let}\:{l}={P}\:'{P}\:'' \\ $$$${l}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} {a}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right){a}^{\mathrm{2}} \:\mathrm{cos}\:\theta \\ $$$${l}^{\mathrm{2}} ={a}^{\mathrm{2}} \left\{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right)\:\mathrm{cos}\:\mathrm{2}\alpha\right\} \\ $$$$ \\ $$$${minimum}\:{perimeter}\:{of}\:\Delta{PQR}={P}\:'{P}\:''={l} \\ $$$$={a}\sqrt{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right)^{\mathrm{2}} +\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\alpha\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{sin}\:\mathrm{2}\alpha}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{sin}\:\mathrm{2}\alpha}\right)} \\ $$
Commented by ajfour last updated on 09/Apr/20
Superb! Sir. Its smooth and very nice solution, i′d delayed following it thinking′ it wd be tough for me to comprehend; i was wrong, thank you Sir.
$${Superb}!\:{Sir}.\:{Its}\:{smooth}\:{and}\: \\ $$$${very}\:{nice}\:{solution},\:{i}'{d}\:{delayed} \\ $$$${following}\:{it}\:{thinking}'\:{it}\:{wd}\:{be} \\ $$$${tough}\:{for}\:{me}\:{to}\:{comprehend}; \\ $$$${i}\:{was}\:{wrong},\:{thank}\:{you}\:{Sir}. \\ $$

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