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Question Number 75296 by ajfour last updated on 09/Dec/19

Commented by ajfour last updated on 09/Dec/19

$I f p e r i m e t e r o f △ P Q R i s p, f i n d$ $m a x i m u m a r e a o f △ P Q R i n$ $t e r m s o f a, b, c, p . (p < a + b + c)$
Commented by mr W last updated on 09/Dec/19

$p_{m i n} = \frac{a^{2} (b^{2} + c^{2} - a^{2}) + b^{2} (c^{2} + a^{2} - b^{2}) + c^{2} (a^{2} + b^{2} - c^{2})}{2 a b c}$ $p_{m a c} = a + b + c$ $p_{m i n} < p < p_{m a x}$
	Answered by mr W last updated on 10/Dec/19

	Commented by mr W last updated on 10/Dec/19

	$p = \sqrt{{(1 - β)}^{2} b^{2} + γ^{2} c^{2} - 2 (1 - β) γ b c \cos A}$ $+ \sqrt{{(1 - γ)}^{2} c^{2} + α^{2} a^{2} - 2 (1 - γ) α c a \cos B}$ $+ \sqrt{{(1 - α)}^{2} a^{2} + β^{2} b^{2} - 2 (1 - α) β a b \cos C}$ $A_{Δ P Q R} = A_{Δ A B C} - \frac{1}{2} (1 - β) γ b c \sin A$ $- \frac{1}{2} (1 - γ) α c a \sin B$ $- \frac{1}{2} (1 - α) β a b \sin C$ $A_{Δ P Q R} = A_{Δ A B C} - \frac{1}{2} A$ $A = (1 - β) γ b c \sin A + (1 - γ) α c a \sin B + (1 - α) β a b \sin C$ $. . .$
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