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Question Number 204970 by mr W last updated on 04/Mar/24

Commented by mr W last updated on 04/Mar/24

$t h e v e r t i c e s o f t r i a n g l e P Q R l i e o n$ $e a c h o f t h e t h r e e t o u c h i n g c i r c l e s$ $w i t h r a d i i a, b, c r e s p e c t i v e l y .$ $f i n d t h e m a x i m u m a r e a o f t h e$ $t r i a n g l e .$
	Answered by mr W last updated on 02/May/24

	Commented by mr W last updated on 02/May/24
	$Δ=(√(abc(a+b+c))) R=((abc)/(4Δ))=(1/4)(√((abc)/(a+b+c))) sin ∠A=((2R)/(b+c))=(1/(2(b+c)))(√((abc)/(a+b+c))) sin ∠B=((2R)/(c+a))=(1/(2(c+a)))(√((abc)/(a+b+c))) sin ∠C=((2R)/(a+b))=(1/(2(a+b)))(√((abc)/(a+b+c))) cos ∠A=(((c+a)^2 +(a+b)^2 −(b+c)^2 )/(2(c+a)(a+b)))=((a(a+b+c)−bc)/((c+a)(a+b))) cos ∠B=(((a+b)^2 +(b+c)^2 −(c+a)^2 )/(2(a+b)(b+c)))=((b(a+b+c)−ca)/((a+b)(b+c))) cos ∠C=(((b+c)^2 +(c+a)^2 −(a+b)^2 )/(2(b+c)(c+a)))=((c(a+b+c)−ab)/((b+c)(c+a))) α+β+γ=2π according to Q206922 sin (α−∠A)=sin α cos ∠A−cos α sin ∠A sin (α−∠A)=((a(a+b+c)−bc)/((c+a)(a+b)))×sin α−(1/(2(b+c)))(√((abc)/(a+b+c)))×cos α (c+a)(a+b) sin (α−∠A)=[a(a+b+c)−bc] sin α−(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×cos α cos (α−∠A)=cos α cos ∠A+sin α sin ∠A cos (α−∠A)=((a(a+b+c)−bc)/((c+a)(a+b)))×cos α+(1/(2(b+c)))(√((abc)/(a+b+c)))×sin α (c+a)(a+b)cos (α−∠A)=[a(a+b+c)−bc] cos α+(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×sin α x=(((c+a)(a+b) sin (α−∠A))/( (√((c+a)^2 sin^2 γ+(a+b)^2 sin^2 β+2(c+a)(a+b) sin β sin γ cos (α−∠A))))) ⇒x=(([a(a+b+c)−bc] sin α−(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×cos α)/( (√((c+a)^2 sin^2 γ+(a+b)^2 sin^2 β+2 sin β sin γ {[a(a+b+c)−bc] cos α+(((c+a)(a+b))/(2(b+c)))(√((abc)/(a+b+c)))×sin α})))) y=(((a+b)(b+c) sin (β−∠B))/( (√((a+b)^2 sin^2 α+(b+c)^2 sin^2 γ+2(a+b)(b+c) sin γ sin a cos (β−∠B))))) z=(((b+c)(c+a) sin (γ−∠C))/( (√((b+c)^2 sin^2 β+(c+a)^2 sin^2 α+2(b+c)(c+a) sin α sin β cos (γ−∠C))))) p=(√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)) q=(√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)) r=(√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)) ((sin ∠Q_2 )/(z+c))=((sin ∠R_1 )/(y+b))=((sin α)/p) ⇒sin ∠Q_2 =(((z+c) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ⇒sin ∠R_1 =(((y+b) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) similarly sin ∠R_2 =(((x+a) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) sin ∠P_1 =(((z+c) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) sin ∠P_2 =(((y+b) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) sin ∠Q_1 =(((x+a) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ∠P_1 =∠Q_2 ⇒(((z+c) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β))))=(((z+c) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ⇒((sin α)/(sin β))=((√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α))/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) ...(i) ∠Q_1 =∠R_2 ⇒(((x+a) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ))))=(((x+a) sin β)/( (√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β)))) ⇒((sin β)/(sin γ))=((√((z+c)^2 +(x+a)^2 −2(z+c)(x+a) cos β))/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ...(ii) ∠R_1 =∠P_2 ⇒(((y+b) sin α)/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α))))=(((y+b) sin γ)/( (√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ)))) ⇒((sin γ)/(sin α))=((√((x+a)^2 +(y+b)^2 −2(x+a)(y+b) cos γ))/( (√((y+b)^2 +(z+c)^2 −2(y+b)(z+c) cos α)))) ...(iii) two of these three equations can be solved to get α, β and γ.$
	$Δ = \sqrt{a b c (a + b + c)}$ $R = \frac{a b c}{4 Δ} = \frac{1}{4} \sqrt{\frac{a b c}{a + b + c}}$ $\sin ∠ A = \frac{2 R}{b + c} = \frac{1}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}}$ $\sin ∠ B = \frac{2 R}{c + a} = \frac{1}{2 (c + a)} \sqrt{\frac{a b c}{a + b + c}}$ $\sin ∠ C = \frac{2 R}{a + b} = \frac{1}{2 (a + b)} \sqrt{\frac{a b c}{a + b + c}}$ $\cos ∠ A = \frac{{(c + a)}^{2} + {(a + b)}^{2} - {(b + c)}^{2}}{2 (c + a) (a + b)} = \frac{a (a + b + c) - b c}{(c + a) (a + b)}$ $\cos ∠ B = \frac{{(a + b)}^{2} + {(b + c)}^{2} - {(c + a)}^{2}}{2 (a + b) (b + c)} = \frac{b (a + b + c) - c a}{(a + b) (b + c)}$ $\cos ∠ C = \frac{{(b + c)}^{2} + {(c + a)}^{2} - {(a + b)}^{2}}{2 (b + c) (c + a)} = \frac{c (a + b + c) - a b}{(b + c) (c + a)}$ $α + β + γ = 2 π$ $a c c o r d i n g t o Q 206922$ $\sin (α - ∠ A) = \sin α \cos ∠ A - \cos α \sin ∠ A$ $\sin (α - ∠ A) = \frac{a (a + b + c) - b c}{(c + a) (a + b)} \times \sin α - \frac{1}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \cos α$ $(c + a) (a + b) \sin (α - ∠ A) = [a (a + b + c) - b c] \sin α - \frac{(c + a) (a + b)}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \cos α$ $\cos (α - ∠ A) = \cos α \cos ∠ A + \sin α \sin ∠ A$ $\cos (α - ∠ A) = \frac{a (a + b + c) - b c}{(c + a) (a + b)} \times \cos α + \frac{1}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \sin α$ $(c + a) (a + b) \cos (α - ∠ A) = [a (a + b + c) - b c] \cos α + \frac{(c + a) (a + b)}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \sin α$ $x = \frac{(c + a) (a + b) \sin (α - ∠ A)}{\sqrt{{(c + a)}^{2} \sin^{2} γ + {(a + b)}^{2} \sin^{2} β + 2 (c + a) (a + b) \sin β \sin γ \cos (α - ∠ A)}}$ $\Rightarrow x = \frac{[a (a + b + c) - b c] \sin α - \frac{(c + a) (a + b)}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \cos α}{\sqrt{{(c + a)}^{2} \sin^{2} γ + {(a + b)}^{2} \sin^{2} β + 2 \sin β \sin γ {[a (a + b + c) - b c] \cos α + \frac{(c + a) (a + b)}{2 (b + c)} \sqrt{\frac{a b c}{a + b + c}} \times \sin α}}}$ $y = \frac{(a + b) (b + c) \sin (β - ∠ B)}{\sqrt{{(a + b)}^{2} \sin^{2} α + {(b + c)}^{2} \sin^{2} γ + 2 (a + b) (b + c) \sin γ \sin a \cos (β - ∠ B)}}$ $z = \frac{(b + c) (c + a) \sin (γ - ∠ C)}{\sqrt{{(b + c)}^{2} \sin^{2} β + {(c + a)}^{2} \sin^{2} α + 2 (b + c) (c + a) \sin α \sin β \cos (γ - ∠ C)}}$ $p = \sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}$ $q = \sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}$ $r = \sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}$ $\frac{\sin ∠ Q_{2}}{z + c} = \frac{\sin ∠ R_{1}}{y + b} = \frac{\sin α}{p}$ $\Rightarrow \sin ∠ Q_{2} = \frac{(z + c) \sin α}{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}}$ $\Rightarrow \sin ∠ R_{1} = \frac{(y + b) \sin α}{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}}$ $s i m i l a r l y$ $\sin ∠ R_{2} = \frac{(x + a) \sin β}{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}}$ $\sin ∠ P_{1} = \frac{(z + c) \sin β}{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}}$ $\sin ∠ P_{2} = \frac{(y + b) \sin γ}{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}}$ $\sin ∠ Q_{1} = \frac{(x + a) \sin γ}{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}}$ $∠ P_{1} = ∠ Q_{2}$ $\Rightarrow \frac{(z + c) \sin β}{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}} = \frac{(z + c) \sin α}{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}}$ $\Rightarrow \frac{\sin α}{\sin β} = \frac{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}}{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}} . . . (i)$ $∠ Q_{1} = ∠ R_{2}$ $\Rightarrow \frac{(x + a) \sin γ}{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}} = \frac{(x + a) \sin β}{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}}$ $\Rightarrow \frac{\sin β}{\sin γ} = \frac{\sqrt{{(z + c)}^{2} + {(x + a)}^{2} - 2 (z + c) (x + a) \cos β}}{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}} . . . (i i)$ $∠ R_{1} = ∠ P_{2}$ $\Rightarrow \frac{(y + b) \sin α}{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}} = \frac{(y + b) \sin γ}{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}}$ $\Rightarrow \frac{\sin γ}{\sin α} = \frac{\sqrt{{(x + a)}^{2} + {(y + b)}^{2} - 2 (x + a) (y + b) \cos γ}}{\sqrt{{(y + b)}^{2} + {(z + c)}^{2} - 2 (y + b) (z + c) \cos α}} . . . (i i i)$ $t w o o f t h e s e t h r e e e q u a t i o n s c a n b e$ $s o l v e d t o g e t α, β a n d γ .$
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