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Question Number 162071 by mr W last updated on 26/Dec/21

Commented by mr W last updated on 27/Dec/21

$i f t h e d i s t a n c e s f r o m a p o i n t t o t h e$ $v e r t i c e s o f a t r i a n g l e a r e p, q, r$ $r e s p e c t i v e l y, f i n d t h e m a x i m u m a r e a$ $o f t h e t r i a n g l e a n d i t s s i d e l e n g t h e s .$
Commented by mr W last updated on 26/Dec/21

Commented by mr W last updated on 27/Dec/21

$i t i s t o f i n d t h e t r i a n g l e w h o s e v e r t i c e s$ $l i e o n t h e t h r e e c i r c l e s r e s p e c t i v e l y$ $a n d w h i c h h a s t h e m a x i m u m a r e a .$
Commented by Tawa11 last updated on 27/Dec/21

$Wow, great sir .$
Commented by mr W last updated on 29/Dec/21

$t h a n k s f o r t h i s v e c t o r w a y s i r!$ $i c a m e t o t h e s a m e c o n c l u s i o n a s$ $s t a t e d b e l o w .$
Commented by aleks041103 last updated on 29/Dec/21

$L e t p, q, r r e p r e s e n t t h e v e c t o r s .$ $\vec{S} = S \vec{e_{z}} = \frac{1}{2} (\vec{p} \times \vec{q} + \vec{q} \times \vec{r} + \vec{r} \times \vec{p})$ $S i n c e S i s e x t r e m a l, t h e n w e m a y$ $v a r i a t e t h e v e c t o r s a s$ $\vec{p} \to \vec{p} + δ \vec{p}$ $\vec{q} \to \vec{q} + δ \vec{q}$ $\vec{r} \to \vec{r} + δ \vec{r}$ $A n d s e a r c h f o r \vec{p}, \vec{q}, \vec{r} w h i c h m a k e δ S = 0 a n d$ $s o δ \vec{S} = 0 .$ $A l s o t h e v a r i a t i o n s m u s t b e s u c h t h a t$ $t h e y p r e s e r v e ∣ p ∣, ∣ q ∣ a n d ∣ r ∣ .$ $δ (\vec{p} . \vec{p}) = 2 \vec{p} . δ \vec{p} = 0 \Rightarrow \vec{p} ⊥ δ \vec{p} .$ $W i t h o u t r e s t r i c t i o n w e m a y c h o o s e$ $δ \vec{q} = δ \vec{r} = 0 s i n c e t h e v a r i a t i o n s a r e a r b i t r a r y$ $\Rightarrow δ S \vec{e_{z}} = \frac{1}{2} (δ \vec{p} \times \vec{q} + \vec{r} \times δ \vec{p}) = 0$ $\Rightarrow δ \vec{p} \times (\vec{q} - \vec{r}) = 0$ $\Rightarrow \vec{p} ⊥ δ \vec{p} ∥ \vec{q} - \vec{r} \Rightarrow \vec{p} ⊥ (\vec{q} - \vec{r})$ $B u t (\vec{q} - \vec{r}) i s t h e s i d e o f t h e t r i a n g l e$ $o p p o s i t e t h e v e r t e x d e f i n e d b y \vec{p} .$ $\Rightarrow \vec{p} m u s t b e c o l i n e a r t o t h e h e i g h t$ $T h i s m u s t b e t r u e f o r \vec{q} a n d \vec{r} t o o .$ $\Rightarrow T h e s t a r t i n g p o i n t i s t h e o r t h o c e n t e r .$
Commented by papa last updated on 04/May/22

$h i p l z t e l l m e h o w t o m a k e p v e c t o r$
Commented by mr W last updated on 04/May/22

Commented by mr W last updated on 04/May/22

	Answered by mr W last updated on 27/Dec/21

	Commented by mr W last updated on 27/Dec/21

	$t h a n k s f o r r e v i e w i n g s i r! y o u s e e m$ $t o h a v e b e e n a b s e n t f o r a l o n g t i m e,$ $w e l c o m e b a c k!$
	Commented by mr W last updated on 27/Dec/21
	$the area of ABC is Δ=(1/2)(pq sin α+qr sin β+rp sin γ) α+β+γ=2π Δ=(1/2)[pq sin α+qr sin β−rp sin (α+β)] such that Δ is maximum, (∂Δ/∂α)=(1/2)[pq cos α−rp cos (α+β)]=0 ⇒pq cos α=rp cos (α+β)=rp cos γ (∂Δ/∂β)=(1/2)[qr cos β−rp cos (α+β)]=0 ⇒qr cos β=rp cos (α+β)=rp cos γ so we have pq cos α=qr cos β=rp cos γ=k say cos α=(k/(pq)) ⇒sin α=(√(1−(k^2 /(p^2 q^2 )))) cos β=(k/(qr)) ⇒sin β=(√(1−(k^2 /(q^2 r^2 )))) cos γ=(k/(rp)) ⇒sin γ=(√(1−(k^2 /(r^2 p^2 )))) k−rp [(k^2 /(pq^2 r))−(√((1−(k^2 /(p^2 q^2 )))(1−(k^2 /(q^2 r^2 )))))]=0 (√((p^2 q^2 −k^2 )(q^2 r^2 −k^2 )))=k^2 −q^2 k p^2 q^4 r^2 −(p^2 q^2 +q^2 r^2 )k^2 +k^4 =k^4 +q^4 k^2 −2q^2 k^3 p^2 q^2 r^2 −(p^2 +q^2 +r^2 )k^2 +2k^3 =0 determinant ((((1/k^3 )−(((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )))(1/k)+(2/(p^2 q^2 r^2 ))=0))) p^2 q^2 r^2 −(1/(27))(p^2 +q^2 +r^2 )^3 ≤0 ⇒three roots (1/k)=(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {((2nπ)/3)+(1/3) sin^(−1) pqr((3/(p^2 +q^2 +r^2 )))^(3/2) } (n=0,1,2) but suitable for maximum Δ is only the root with n=2: determinant ((((1/k)=−(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(π/3)+(1/3) sin^(−1) pqr((3/(p^2 +q^2 +r^2 )))^(3/2) }))) Δ_(max) =(1/2)((√(p^2 q^2 −k^2 ))+(√(q^2 r^2 −k^2 ))+(√(r^2 p^2 −k^2 ))) a^2 =q^2 +r^2 −2qr cos β=q^2 +r^2 −2k a=(√(q^2 +r^2 −2k)) b=(√(r^2 +p^2 −2k)) c=(√(p^2 +q^2 −2k)) example: p=10, q=8, r=4 Δ_(max) ≈67.303409 a≈11.084677 b≈12.604367 c≈14.382978 coresponding equilateral triangle: s≈11.943609, Δ≈61.769169$
	$t h e a r e a o f A B C i s$ $Δ = \frac{1}{2} (p q \sin α + q r \sin β + r p \sin γ)$ $α + β + γ = 2 π$ $Δ = \frac{1}{2} [p q \sin α + q r \sin β - r p \sin (α + β)]$ $s u c h t h a t Δ i s m a x i m u m,$ $\frac{\partial Δ}{\partial α} = \frac{1}{2} [p q \cos α - r p \cos (α + β)] = 0$ $\Rightarrow p q \cos α = r p \cos (α + β) = r p \cos γ$ $\frac{\partial Δ}{\partial β} = \frac{1}{2} [q r \cos β - r p \cos (α + β)] = 0$ $\Rightarrow q r \cos β = r p \cos (α + β) = r p \cos γ$ $s o w e h a v e$ $p q \cos α = q r \cos β = r p \cos γ = k s a y$ $\cos α = \frac{k}{p q} \Rightarrow \sin α = \sqrt{1 - \frac{k^{2}}{p^{2} q^{2}}}$ $\cos β = \frac{k}{q r} \Rightarrow \sin β = \sqrt{1 - \frac{k^{2}}{q^{2} r^{2}}}$ $\cos γ = \frac{k}{r p} \Rightarrow \sin γ = \sqrt{1 - \frac{k^{2}}{r^{2} p^{2}}}$ $k - r p [\frac{k^{2}}{{p q}^{2} r} - \sqrt{(1 - \frac{k^{2}}{p^{2} q^{2}}) (1 - \frac{k^{2}}{q^{2} r^{2}})}] = 0$ $\sqrt{(p^{2} q^{2} - k^{2}) (q^{2} r^{2} - k^{2})} = k^{2} - q^{2} k$ $p^{2} q^{4} r^{2} - (p^{2} q^{2} + q^{2} r^{2}) k^{2} + k^{4} = k^{4} + q^{4} k^{2} - 2 q^{2} k^{3}$ $p^{2} q^{2} r^{2} - (p^{2} + q^{2} + r^{2}) k^{2} + 2 k^{3} = 0$ $\begin{matrix} \frac{1}{k^{3}} - (\frac{p^{2} + q^{2} + r^{2}}{p^{2} q^{2} r^{2}}) \frac{1}{k} + \frac{2}{p^{2} q^{2} r^{2}} = 0 \end{matrix}$ $p^{2} q^{2} r^{2} - \frac{1}{27} {(p^{2} + q^{2} + r^{2})}^{3} ⩽ 0 \Rightarrow t h r e e r o o t s$ $\frac{1}{k} = \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{2 n π}{3} + \frac{1}{3} \sin^{- 1} p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}} (n = 0, 1, 2)$ $b u t s u i t a b l e f o r m a x i m u m Δ i s o n l y$ $t h e r o o t w i t h n = 2 :$ $\begin{matrix} \frac{1}{k} = - \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}} \end{matrix}$ $Δ_{m a x} = \frac{1}{2} (\sqrt{p^{2} q^{2} - k^{2}} + \sqrt{q^{2} r^{2} - k^{2}} + \sqrt{r^{2} p^{2} - k^{2}})$ $a^{2} = q^{2} + r^{2} - 2 q r \cos β = q^{2} + r^{2} - 2 k$ $a = \sqrt{q^{2} + r^{2} - 2 k}$ $b = \sqrt{r^{2} + p^{2} - 2 k}$ $c = \sqrt{p^{2} + q^{2} - 2 k}$ $\underset{―}{e x a m p l e : p = 10, q = 8, r = 4}$ $Δ_{m a x} \approx 67.303409$ $a \approx 11.084677$ $b \approx 12.604367$ $c \approx 14.382978$ $c o r e s p o n d i n g e q u i l a t e r a l t r i a n g l e :$ $s \approx 11.943609, Δ \approx 61.769169$
	Commented by mr W last updated on 27/Dec/21

	Commented by mr W last updated on 27/Dec/21

	Commented by mr W last updated on 27/Dec/21

	Commented by mr W last updated on 27/Dec/21

	Commented by mr W last updated on 27/Dec/21

	$w e s e e g e n e r a l l y t h e c o r e s p o n d i n g$ $e q u i l a t e r a l t r i a n g l e (b l u e m a r k e d i n$ $d i a g r a m s) (s e e a l s o Q 123040) i s n o t$ $t h e t r i a n g l e w i t h t h e m a x i m u m$ $a r e a .$
	Commented by mr W last updated on 27/Dec/21

	$f r o m p q \cos α = q r \cos β = r p \cos γ$ $w e g e t q \cos α = r \cos γ, t h a t m e a n s$ $A O ⊥ B C . s i m i l a r l y B O ⊥ C A, C O ⊥ A B .$ $t h a t m e a n s O i s t h e o r t h o c e n t e r o f$ $t r i a n g l e A B C w h e n t h e t r i a n g l e$ $h a s t h e m a x i m u m a r e a .$ $e x a m p l e : p = 7, q = 6, r = 5$ $a \approx 9.7924505$ $b \approx 10.4351372$ $c \approx 10.9495245$ $t h i s i s t h e s a m e r e s u l t a s i n Q 161015 .$
	Commented by Tawa11 last updated on 27/Dec/21

	$Great sir .$
	Commented by behi834171 last updated on 27/Dec/21

	$g r e a a a a a a a a a a t d e a r m a s t e r .$ $t h a n k s a l o t .$
	Commented by behi834171 last updated on 27/Dec/21

	Thank you so much,dear master,for your favor. I'm here,from now on.
	Commented by behi834171 last updated on 27/Dec/21

	$γ = 90^{∙} i s a n i n t e r e s t i n g c a s e . I t h i n k .$
	Commented by mr W last updated on 27/Dec/21

	$w h e n o n e o f p, q, r i s z e r o, t h e n t h e$ $c o r e s p o n d i n g v e r t e x i s a l s o t h e$ $o r t h o c e n t e r . i n t h i s c a s e t h e t r i a n g l e$ $i s r i g h t a n g l e d .$
	Commented by mr W last updated on 27/Dec/21

	Commented by behi834171 last updated on 27/Dec/21

	Commented by behi834171 last updated on 27/Dec/21

	$γ = 90^{∙}, p, q, r \neq 0$
	Commented by mr W last updated on 27/Dec/21

	$t h e t a s k h e r e i s t o f i n d t h e t r i a n g l e$ $w i t h m a x i m u m a r e a w h e n p, q, r a r e$ $g i v e n . i t i s n o t t o f i n d t h e t r i a n g l e$ $w i t h g i v e n α, β o r γ .$ $s o f a r a s p, q, r \neq 0, t h e t r i a n g l e w i t h$ $m a x i x u m a r e a c a n n e v e r b e a t r i a n g l e$ $w i t h γ = 90 °!$ $t h e t r i a n g l e y o u g a v e w i t h γ = 90 ° i s$ $n o t t h e t r i a n g l e w h i c h h a s t h e$ $m a x i m u m a r e a .$
	Commented by mr W last updated on 27/Dec/21

	Commented by behi834171 last updated on 27/Dec/21

	$t h a n k y o u v e r y m u c h f o r y o u r a t t e n t i o n .$ $c a n w e f i n d t h e s i d e s o f s u c h t r i a n g l e$ $w h i t : γ = 90^{∙} ?$ $i h a v e t h e s u c h u n s o l v e d p r o b l e m i n f o r u m$ $b u t i {d o n}^{'} t r e m e m b e r i t s q . I D . . .$
	Commented by mr W last updated on 27/Dec/21

	$y o u m u s t s p e c i f y e x a c t l y w h a t i s y o u r$ $t a s k . i n t h e c u r r e n t q u e s t i o n t h e$ $p o i n t O i s t h e o r t h o c e n t e r o f t h e$ $t r i a n g l e . w e c a n n o t f i n d a t r i a n g l e$ $w h o s e o r t h o c e n t e r l i e s i n s i d e t h e$ $t r i a n g l e a n d m a k e s a n a n g l e γ = 90 ° .$ $t h i s i s o n l y t h e c a s e w h e n t h e$ $o r t h o c e n t e r l i e s a t o n e v e r t e x o f a$ $r i g h t a n g l e d t r i a n g l e, b u t i n t h i s$ $c a s e o n e o f p, q, r i s z e r o .$
	Commented by behi834171 last updated on 27/Dec/21

	$t h e t a s k i s : t o f i n d s i d e s o f a t r i a n g l e$ $w i t h g i v e n : p, q, r a n d : γ = 90^{∙} .$ $i s i t p o s s i b l e ?$
	Commented by mr W last updated on 27/Dec/21

	$t h e n i n f i n i t e t r i a n g l e s a r e p o s s i b l e .$ $o n l y A C = \sqrt{r^{2} + p^{2}} i s f i x e d .$
	Commented by mr W last updated on 27/Dec/21

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