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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

$G i v e n : t r i a n g l e A B C w i t h s i d e l e n g t h e s$ $a, b, c .$ $P i s m i d p o i n t o f B C .$ $F i n d : p e r i m e t e r o f i n s c r i b e d t r i a n g l e$ $P Q R w i t h t h e s m a l l e s t$ $p e r i m e t e r .$
Commented by ajfour last updated on 09/Apr/20

Commented by ajfour last updated on 09/Apr/20
$let ∠A=α , ∠B=β , ∠C=γ , ∠QPC=θ, p(slope of AB′)=tan (π−2γ−β) tan θ=m, tan β=q, A(h,k) AR=BR′ = l eq. of PQ: y=m(x−a/2) eq. of A ′B : y=−qx eq. of AB′ : y−k=p(x−h) for R′_(−) m(x−(a/2))=−qx x_(R′) =((m(a/2))/(q+m)) , y_(R′) =−((qm(a/2))/(q+m)) for R_(−) m(x−a/2)=k+p(x−h) x_R =((k−ph+m(a/2))/(m−p)) y_R =m(((k−ph+p(a/2))/(m−p))) l^2 =(BR′)^2 =[((m(a/2))/(q+m))]^2 (1+q^2 ) =(AR)^2 = (((k−mh+m(a/2))/(m−p)))^2 +{((m[−ph+p(a/2)]+pk)/(m−p))}^2 from above eq. we solve for m. Then min( perimeter △PQR ) = R′R .$
$l e t ∠ A = α, ∠ B = β, ∠ C = γ,$ $∠ Q P C = θ,$ $p (s l o p e o f {A B}^{'}) = \tan (π - 2 γ - β)$ $\tan θ = m, \tan β = q, A (h, k)$ $A R = {B R}^{'} = l$ $e q . o f P Q : y = m (x - a / 2)$ $e q . o f A^{'} B : y = - q x$ $e q . o f {A B}^{'} : y - k = p (x - h)$ $\underline{f o r R^{'}}$ $m (x - \frac{a}{2}) = - q x$ $x_{R^{'}} = \frac{m (a / 2)}{q + m}, y_{R^{'}} = - \frac{q m (a / 2)}{q + m}$ $\underline{f o r R}$ $m (x - a / 2) = k + p (x - h)$ $x_{R} = \frac{k - p h + m (a / 2)}{m - p}$ $y_{R} = m (\frac{k - p h + p (a / 2)}{m - p})$ $l^{2} = {({B R}^{'})}^{2} = {[\frac{m (a / 2)}{q + m}]}^{2} (1 + q^{2})$ $= {(A R)}^{2} = {(\frac{k - m h + m (a / 2)}{m - p})}^{2}$ $+ {\frac{m [- p h + p (a / 2)] + p k}{m - p}}^{2}$ $f r o m a b o v e e q . w e s o l v e f o r m .$ $T h e n m i n (p e r i m e t e r △ P Q R)$ $= R^{'} R .$
Commented by ajfour last updated on 09/Apr/20

${c o u l d n}^{'} t c o m p r e s s i t S i r!$
Commented by mr W last updated on 09/Apr/20

$t h a n k y o u s i r! i t i s t h e g e o m e t r i c w a y$ $w h i c h i p r e f e r . i^{'} l l t r y i n a s i m i l a r w a y .$
	Answered by mr W last updated on 09/Apr/20

	Commented by mr W last updated on 09/Apr/20
	$let ∠A=α, ∠B=β, ∠C=γ we construct the image of side BC about AB and about AC. P^ ′ and P ′′ are the corresponding image of P. BP ′=BP=(a/2), CP ′′=CP=(a/2) P ′R=PR, QP ′′=QP perimeter of ΔPQR=PQ+QR+PR =P ′R+RQ+QP ′′ we see it is minimum, if P ′,R,Q,P ′′ are collinear, then perimeter of ΔPQR=P ′P ′′ ϕ=π−2β φ=π−2γ θ=π−ϕ−φ=2(β+γ)−π=π−2α ((DB)/(sin φ))=((DC)/(sin ϕ))=((BC)/(sin θ)) ⇒DB=((sin φ)/(sin θ))×BC=((sin 2γ)/(sin 2α))×a ⇒DC=((sin ϕ)/(sin θ))×BC=((sin 2β)/(sin 2α))×a DP ′=DB+BP ′=((sin 2γ)/(sin 2α))×a+(a/2)=((1/2)+((sin 2γ)/(sin 2α)))a DP ′′=DC+CP ′′=((sin 2β)/(sin 2α))×a+(a/2)=((1/2)+((sin 2β)/(sin 2α)))a let l=P ′P ′′ l^2 =((1/2)+((sin 2β)/(sin 2α)))^2 a^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 a^2 −2((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α)))a^2 cos θ l^2 =a^2 {((1/2)+((sin 2β)/(sin 2α)))^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 +2((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α))) cos 2α} minimum perimeter of ΔPQR=P ′P ′′=l =a(√(((1/2)+((sin 2β)/(sin 2α)))^2 +((1/2)+((sin 2γ)/(sin 2α)))^2 +2 cos 2α ((1/2)+((sin 2β)/(sin 2α)))((1/2)+((sin 2γ)/(sin 2α)))))$
	$l e t ∠ A = α, ∠ B = β, ∠ C = γ$ $w e c o n s t r u c t t h e i m a g e o f s i d e B C$ $Prime causes double exponent: use braces to clarify$ $a r e t h e c o r r e s p o n d i n g i m a g e o f P .$ $B P^{'} = B P = \frac{a}{2}, C P^{″} = C P = \frac{a}{2}$ $P^{'} R = P R, Q P^{″} = Q P$ $p e r i m e t e r o f Δ P Q R = P Q + Q R + P R$ $= P^{'} R + R Q + Q P^{″}$ $w e s e e i t i s m i n i m u m, i f P^{'}, R, Q, P^{″}$ $a r e c o l l i n e a r, t h e n$ $p e r i m e t e r o f Δ P Q R = P^{'} P^{″}$ $φ = π - 2 β$ $ϕ = π - 2 γ$ $θ = π - φ - ϕ = 2 (β + γ) - π = π - 2 α$ $\frac{D B}{\sin ϕ} = \frac{D C}{\sin φ} = \frac{B C}{\sin θ}$ $\Rightarrow D B = \frac{\sin ϕ}{\sin θ} \times B C = \frac{\sin 2 γ}{\sin 2 α} \times a$ $\Rightarrow D C = \frac{\sin φ}{\sin θ} \times B C = \frac{\sin 2 β}{\sin 2 α} \times a$ $D P^{'} = D B + B P^{'} = \frac{\sin 2 γ}{\sin 2 α} \times a + \frac{a}{2} = (\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α}) a$ $D P^{″} = D C + C P^{″} = \frac{\sin 2 β}{\sin 2 α} \times a + \frac{a}{2} = (\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α}) a$ $l e t l = P^{'} P^{″}$ $l^{2} = {(\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α})}^{2} a^{2} + {(\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α})}^{2} a^{2} - 2 (\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α}) (\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α}) a^{2} \cos θ$ $l^{2} = a^{2} {{(\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α})}^{2} + {(\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α})}^{2} + 2 (\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α}) (\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α}) \cos 2 α}$ $m i n i m u m p e r i m e t e r o f Δ P Q R = P^{'} P^{″} = l$ $= a \sqrt{{(\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α})}^{2} + {(\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α})}^{2} + 2 \cos 2 α (\frac{1}{2} + \frac{\sin 2 β}{\sin 2 α}) (\frac{1}{2} + \frac{\sin 2 γ}{\sin 2 α})}$
	Commented by ajfour last updated on 09/Apr/20

	$S u p e r b! S i r . I t s s m o o t h a n d$ $v e r y n i c e s o l u t i o n, i^{'} d d e l a y e d$ $f o l l o w i n g i t {t h i n k i n g}^{'} i t w d b e$ $t o u g h f o r m e t o c o m p r e h e n d;$ $i w a s w r o n g, t h a n k y o u S i r .$
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