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Question Number 166617 by mr W last updated on 23/Feb/22

Commented by mr W last updated on 23/Feb/22

$t h r e e c i r c l e s w i t h r a d i i a, b, c$ $t o u c h e a c h o t h e r a s s h o w n .$ $f i n d t h e m a x i m u m s i d e l e n g t h o f$ $t h e e q u i l a t e r a l t r i a n g l e w h o s e$ $v e r t i c e s l i e o n t h e t h r e e c i r c l e s .$
	Answered by ajfour last updated on 25/Feb/22

	Answered by mr W last updated on 20/Mar/22

	Commented by mr W last updated on 22/Mar/22
	$area of ABC Δ=(√(abc(a+b+c))) incircle radius r=(Δ/(a+b+c))=(√((abc)/(a+b+c))) α′=2 tan^(−1) (r/a) β′=2 tan^(−1) (r/b) γ′=2 tan^(−1) (r/c) PQ=s: s^2 =[a+b+a cos (α+α′)+b cos β]^2 +[a sin (α+α′)+b sin β]^2 s^2 =(a+b)^2 +a^2 +b^2 +2a(a+b)cos (α+α′)+2b(a+b) cos β+2ab[cos (α+α′)cos β+ sin (α+α′) sin β] (s^2 /2)−(a^2 +b^2 +ab)=(a+b)[a cos (α+α′)+b cos β]+ab[cos (α+α′)cos β+ sin (α+α′) sin β] (s^2 /2)−(a^2 +b^2 +ab)−a(a+b)cos (α+α′)=b{[a+b+a cos (α+α′)]cos β+a sin (α+α′) sin β} (s^2 /2)−(a^2 +b^2 +ab)−a(a+b)cos (α+α′)=b(√((a+b)^2 +a^2 +2a(a+b)cos (α+α′))) sin [β+tan^(−1) ((a+b+a cos (α+α′))/(a sin (α+α′)))] determinant (((β=−tan^(−1) ((a+b+a cos (α+α′))/(a sin (α+α′)))+sin^(−1) (((s^2 /2)−(a^2 +b^2 +ab)−a(a+b)cos (α+α′))/(b(√((a+b)^2 +a^2 +2a(a+b)cos (α+α′))))) ))) similarly RP=s: (s^2 /2)−(c^2 +a^2 +ca)=(c+a)[c cos (γ+γ′)+a cos α]+ca[cos (γ+γ′)cos α+ sin (γ+γ′) sin α] (s^2 /2)−(c^2 +a^2 +ca)−a(c+a)cos α=c{(c+a+a cos α) cos (γ+γ′)+a sin (γ+γ′) sin α} (s^2 /2)−(c^2 +a^2 +ca)−a(c+a)cos α=c(√((c+a)^2 +a^2 +2a(c+a)cos α)) sin (γ+γ′+tan^(−1) ((c+a+a cos α)/(a sin α))) determinant (((γ=−γ′−tan^(−1) ((c+a+a cos α)/(a sin α))+sin^(−1) (((s^2 /2)−(c^2 +a^2 +ca)−a(c+a)cos α)/(c(√((c+a)^2 +a^2 +2a(c+a)cos α))))))) similarly QR=s: (s^2 /2)−(b^2 +c^2 +bc)=(b+c)[b cos (β+β′)+c cos γ]+bc cos (β+β′−γ) determinant ((((s^2 /(2bc))−((b/c)+(c/b)+1)=(b+c)[((cos (β+β′))/c)+((cos γ)/b)]+cos (β+β′−γ))))$
	$a r e a o f A B C Δ = \sqrt{a b c (a + b + c)}$ $i n c i r c l e r a d i u s r = \frac{Δ}{a + b + c} = \sqrt{\frac{a b c}{a + b + c}}$ $α^{'} = 2 \tan^{- 1} \frac{r}{a}$ $β^{'} = 2 \tan^{- 1} \frac{r}{b}$ $γ^{'} = 2 \tan^{- 1} \frac{r}{c}$ $P Q = s :$ $s^{2} = {[a + b + a \cos (α + α^{'}) + b \cos β]}^{2} + {[a \sin (α + α^{'}) + b \sin β]}^{2}$ $s^{2} = {(a + b)}^{2} + a^{2} + b^{2} + 2 a (a + b) \cos (α + α^{'}) + 2 b (a + b) \cos β + 2 a b [\cos (α + α^{'}) \cos β + \sin (α + α^{'}) \sin β]$ $\frac{s^{2}}{2} - (a^{2} + b^{2} + a b) = (a + b) [a \cos (α + α^{'}) + b \cos β] + a b [\cos (α + α^{'}) \cos β + \sin (α + α^{'}) \sin β]$ $\frac{s^{2}}{2} - (a^{2} + b^{2} + a b) - a (a + b) \cos (α + α^{'}) = b {[a + b + a \cos (α + α^{'})] \cos β + a \sin (α + α^{'}) \sin β}$ $\frac{s^{2}}{2} - (a^{2} + b^{2} + a b) - a (a + b) \cos (α + α^{'}) = b \sqrt{{(a + b)}^{2} + a^{2} + 2 a (a + b) \cos (α + α^{'})} \sin [β + \tan^{- 1} \frac{a + b + a \cos (α + α^{'})}{a \sin (α + α^{'})}]$ $\begin{matrix} β = - \tan^{- 1} \frac{a + b + a \cos (α + α^{'})}{a \sin (α + α^{'})} + \sin^{- 1} \frac{\frac{s^{2}}{2} - (a^{2} + b^{2} + a b) - a (a + b) \cos (α + α^{'})}{b \sqrt{{(a + b)}^{2} + a^{2} + 2 a (a + b) \cos (α + α^{'})}} \end{matrix}$ $s i m i l a r l y$ $R P = s :$ $\frac{s^{2}}{2} - (c^{2} + a^{2} + c a) = (c + a) [c \cos (γ + γ^{'}) + a \cos α] + c a [\cos (γ + γ^{'}) \cos α + \sin (γ + γ^{'}) \sin α]$ $\frac{s^{2}}{2} - (c^{2} + a^{2} + c a) - a (c + a) \cos α = c {(c + a + a \cos α) \cos (γ + γ^{'}) + a \sin (γ + γ^{'}) \sin α}$ $\frac{s^{2}}{2} - (c^{2} + a^{2} + c a) - a (c + a) \cos α = c \sqrt{{(c + a)}^{2} + a^{2} + 2 a (c + a) \cos α} \sin (γ + γ^{'} + \tan^{- 1} \frac{c + a + a \cos α}{a \sin α})$ $\begin{matrix} γ = - γ^{'} - \tan^{- 1} \frac{c + a + a \cos α}{a \sin α} + \sin^{- 1} \frac{\frac{s^{2}}{2} - (c^{2} + a^{2} + c a) - a (c + a) \cos α}{c \sqrt{{(c + a)}^{2} + a^{2} + 2 a (c + a) \cos α}} \end{matrix}$ $s i m i l a r l y$ $Q R = s :$ $\frac{s^{2}}{2} - (b^{2} + c^{2} + b c) = (b + c) [b \cos (β + β^{'}) + c \cos γ] + b c \cos (β + β^{'} - γ)$ $\begin{matrix} \frac{s^{2}}{2 b c} - (\frac{b}{c} + \frac{c}{b} + 1) = (b + c) [\frac{\cos (β + β^{'})}{c} + \frac{\cos γ}{b}] + \cos (β + β^{'} - γ) \end{matrix}$
	Commented by mr W last updated on 20/Mar/22

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