Question-204970

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 = \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}} \dots (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}} \dots (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}} \dots (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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