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Question Number 62542 by mr W last updated on 22/Jun/19

Commented by mr W last updated on 22/Jun/19

$t h e d i s t a n c e s o f a p o i n t M t o t h e$ $v e r t e x e s o f a t r i a n g l e a r e p, q, r .$ $f i n d t h e s i d e l e n g t h e s a n d h e n c e t h e$ $p e r i m e t e r o f t h e t r i a n g l e w i t h t h e$ $l a r g e s t p e r i m e t e r .$
	Answered by mr W last updated on 23/Jun/19
	$i think M should be center of incircle such that perimeter is maximum. let k=radius of incircle ∠A=2 sin^(−1) (k/p) ∠B=2 sin^(−1) (k/q) ∠C=2 sin^(−1) (k/r) ∠A+∠B+∠C=π ⇒sin^(−1) (k/p)+sin^(−1) (k/q)+sin^(−1) (k/r)=(π/2) ⇒cos (sin^(−1) (k/p)+sin^(−1) (k/q))=cos ((π/2)−sin^(−1) (k/r)) (√((1−(k^2 /p^2 ))(1−(k^2 /q^2 ))))−(k^2 /(pq))=(k/r) (√((1−(k^2 /p^2 ))(1−(k^2 /q^2 ))))=(k/r)+(k^2 /(pq)) (1−(k^2 /p^2 ))(1−(k^2 /q^2 ))=(k^2 /r^2 )+(k^4 /(p^2 q^2 ))+((2k^3 )/(pqr)) ((2k^3 )/(pqr))+((1/p^2 )+(1/q^2 )+(1/r^2 ))k^2 −1=0 k^3 +3((p^2 q^2 +q^2 r^2 +r^2 p^2 )/(6pqr))k^2 −2((pqr)/4)=0 ⇒k^3 +3uk^2 −2v=0 k=s−u ⇒s^3 −3u^2 s+2(u^3 −v)=0 s=2u sin {(1/3) sin^(−1) (1−(v/u^3 ))+((2nπ)/3)} k=u[2 sin {(1/3) sin^(−1) (1−(v/u^3 ))+((2π)/3)}−1]>0 ⇒k=((p^2 q^2 +q^2 r^2 +r^2 p^2 )/(6pqr))[2 sin {(1/3) sin^(−1) (1−((54p^4 q^4 r^4 )/((p^2 q^2 +q^2 r^2 +r^2 p^2 )^3 )))+((2π)/3)}−1] ⇒a=(√(q^2 −k^2 ))+(√(r^2 −k^2 )) ⇒b=(√(r^2 −k^2 ))+(√(p^2 −k^2 )) ⇒c=(√(p^2 −k^2 ))+(√(q^2 −k^2 ))$
	$i t h i n k M s h o u l d b e c e n t e r o f i n c i r c l e$ $s u c h t h a t p e r i m e t e r i s m a x i m u m .$ $l e t k = r a d i u s o f i n c i r c l e$ $∠ A = 2 \sin^{- 1} \frac{k}{p}$ $∠ B = 2 \sin^{- 1} \frac{k}{q}$ $∠ C = 2 \sin^{- 1} \frac{k}{r}$ $∠ A + ∠ B + ∠ C = π$ $\Rightarrow \sin^{- 1} \frac{k}{p} + \sin^{- 1} \frac{k}{q} + \sin^{- 1} \frac{k}{r} = \frac{π}{2}$ $\Rightarrow \cos (\sin^{- 1} \frac{k}{p} + \sin^{- 1} \frac{k}{q}) = \cos (\frac{π}{2} - \sin^{- 1} \frac{k}{r})$ $\sqrt{(1 - \frac{k^{2}}{p^{2}}) (1 - \frac{k^{2}}{q^{2}})} - \frac{k^{2}}{p q} = \frac{k}{r}$ $\sqrt{(1 - \frac{k^{2}}{p^{2}}) (1 - \frac{k^{2}}{q^{2}})} = \frac{k}{r} + \frac{k^{2}}{p q}$ $(1 - \frac{k^{2}}{p^{2}}) (1 - \frac{k^{2}}{q^{2}}) = \frac{k^{2}}{r^{2}} + \frac{k^{4}}{p^{2} q^{2}} + \frac{2 k^{3}}{p q r}$ $\frac{2 k^{3}}{p q r} + (\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}}) k^{2} - 1 = 0$ $k^{3} + 3 \frac{p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2}}{6 p q r} k^{2} - 2 \frac{p q r}{4} = 0$ $\Rightarrow k^{3} + 3 {u k}^{2} - 2 v = 0$ $k = s - u$ $\Rightarrow s^{3} - 3 u^{2} s + 2 (u^{3} - v) = 0$ $s = 2 u \sin {\frac{1}{3} \sin^{- 1} (1 - \frac{v}{u^{3}}) + \frac{2 n π}{3}}$ $k = u [2 \sin {\frac{1}{3} \sin^{- 1} (1 - \frac{v}{u^{3}}) + \frac{2 π}{3}} - 1] > 0$ $\Rightarrow k = \frac{p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2}}{6 p q r} [2 \sin {\frac{1}{3} \sin^{- 1} (1 - \frac{54 p^{4} q^{4} r^{4}}{{(p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2})}^{3}}) + \frac{2 π}{3}} - 1]$ $\Rightarrow a = \sqrt{q^{2} - k^{2}} + \sqrt{r^{2} - k^{2}}$ $\Rightarrow b = \sqrt{r^{2} - k^{2}} + \sqrt{p^{2} - k^{2}}$ $\Rightarrow c = \sqrt{p^{2} - k^{2}} + \sqrt{q^{2} - k^{2}}$
	Commented by ajfour last updated on 22/Jun/19

	$G r e a t, S i r, b u t a n y s u p p o r t i n g$ $i d e a t o w h y M s h o u l d b e i n c e n t r e$ $i f t h e p e r i n e t e r i s t o b e a m a x i m u m . .$
	Commented by mr W last updated on 23/Jun/19

	Commented by mr W last updated on 23/Jun/19

	$(s e e d i a g r a m a b o v e)$ ${l e t}^{'} s l o o k a t t h e c a s e w h e n B, C a r e$ $g i v e n a n d A i s t o d e t e r m i n e, w h i c h$ $m u s t l i e o n a c i r c l e w i t h c e n t e r M .$ $t h e l a w o f n a t u r e t e l l s u s : w h e n t h e$ $p a t h f r o m B t o C v i a A s h o u l d b e$ $o p t i m u m (m i n i m u m o r m a x i m u m),$ $i t m u s t f o l l o w t h e p a t h o f a l i g h t r a y,$ $i . e . a s i f t h e c i r c l e w e r e a m i r r o r .$ $p o i n t A s h o u l d b e t h e r e f l e c t i o n p o i n t$ $o f t h e l i g h t r a y a n d t h e r e f o r e A M$ $i s t h e b i s e c t o r o f a n g l e ∠ B A C .$ $w h e n w e c o n s i d e r n o t o n l y p o i n t A,$ $b u t a l s o o t h e r v e r t e x e s, w e w i l l$ $c o n c l u d e t h a t t h e t r i a n g l e h a s o p t i m u m$ $p e r i m e t e r w h e n M i s o n t h e b i s e c t o r s$ $o f a l l a n g l e s o f t h e t r i a n g l e, t h a t$ $m e a n s M i s t h e c e n t e r o f i t s i n c i r c l e .$
	Commented by mr W last updated on 23/Jun/19

	Commented by mr W last updated on 23/Jun/19

	$(s e e d i a g r a m a b o v e)$ $w e c a n a l s o c o n s i d e r l i k e t h i s :$ $p o i n t A l i e s o n c i r c l e w i t h c e n t e r M .$ $w e k n o w t h e l o c u s o f a p o i n t, w h o s e$ $s u m o f d i s t a n c e s t o B a n d C i s$ $c o n s t a n t, i s a n e l l i p s e w i t h f o c u s e s$ $a t p o i n t B a n d C . i n f i n i t e s u c h$ $e l l i p s e s a r e p o s s i b l e . t h e s u m o f t h e$ $d i s t a n c e s A B + A C i s m a x i m u m,$ $w h e n A a l s o l i e s o n s u c h a n e l l i p s e$ $w h i c h t o u c h e s t h e c i r c l e w i t h c e n t e r M .$ $i n t h i s c a s e A M i s a l s o a n o r m a l$ $o f t h e e l l i p s e . w e k n o w a l i g h t r a y$ $f r o m a f o c u s o f a n e l l i p s e p a s s e s a l s o$ $t h r o u g h t h e o t h e r f o c u s a f t e r r e f l e c t i o n$ $o n t h e e l l i p s e, i . e . t h e n o r m a l a t$ $p o i n t A i s t h e a n g l e b i s e c t o r o f t h e$ $l i g h t r a y B A a n d A C, s e e a l s o$ $f o l l o w i n g d i a g r a m .$
	Commented by mr W last updated on 23/Jun/19

	Commented by mr W last updated on 23/Jun/19

	$d u e t o t h i s c o n s i d e r a t i o n i t h i n k t h e$ $t r i a n g l e w e s e a r c h i s t h a t o n e w h o s e$ $i n c i r c l e c e n t e r i s t h e p o i n t M .$ $w e {d o n}^{'} t n e e d t o a p p l y c a l c u l u s m e t h o d$ $l i k e \frac{\partial P}{\partial θ} = 0 a n d \frac{\partial P}{\partial φ} = 0 .$
	Commented by ajfour last updated on 23/Jun/19

	$P l e n t i f u l e v i d e n c e, S i r . T h a n k s$ $a l o t . B u t y e t a n a n a l y t i c a l w a y$ $t o p r o v e t h i s {i s n}^{'} t e a s y (i t r i e d)!$
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