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Question Number 74713 by ajfour last updated on 29/Nov/19

Commented by ajfour last updated on 29/Nov/19

$I f p e r i m e t e r o f △ A B C = 18,$ $a n d a r e a i s m a x i m u m, f i n d$ $c o o r d i n a t e s o f p o i n t s A, B, C .$
	Answered by mr W last updated on 29/Nov/19

	$A B = B C = C A = 6$ $y_{C} = 3 \sqrt{3}$ $x_{C} = \frac{9 \sqrt{3}}{4}$ $x_{A} = \frac{9 \sqrt{3}}{4} - 3$ $x_{B} = \frac{9 \sqrt{3}}{4} + 3$
	Commented by mr W last updated on 29/Nov/19

	$t h i s i s o n l y b e c a u s e α ⩽ 60 °$ $f o r α > 60 ° {i t}^{'} s a d i f f e r e n t t r i a n g l e s h a p e .$
	Commented by ajfour last updated on 29/Nov/19

	$I t s a l r i g h t S i r, I d i n t t h i n k m u c h .$
	Commented by mr W last updated on 29/Nov/19

	Commented by mr W last updated on 30/Nov/19

	$f o r α > 60 °$ $c a s e 1 :$ $A C = B C = \frac{A B}{2 \cos α}$ $A B + \frac{A B}{\cos α} = p = 18$ $A B = \frac{p \cos α}{1 + \cos α}$ ${A r e a}_{1} = \frac{{A B}^{2} \tan α}{2} = \frac{p^{2} \sin α \cos α}{2 {(1 + \cos α)}^{2}}$ $c a s e 2 :$ $A B = A C$ $B C = A B \sin \frac{α}{2}$ $2 A B + A B \sin \frac{α}{2} = p = 18$ $A B = \frac{p}{2 + \sin \frac{α}{2}}$ ${A r e a}_{2} = \frac{{A B}^{2} \sin α}{2} = \frac{p^{2} \sin α}{2 {(2 + \sin \frac{α}{2})}^{2}}$ $\frac{{A r e a}_{1}}{{A r e a}_{2}} = \cos α {(\frac{2 + \sin \frac{α}{2}}{1 + \cos α})}^{2}$ $i f α ⩽ 77.68 ° c a s e 1 g i v e s m a x i m u m a r e a .$ $i f α > 77.68 ° c a s e 2 g i v e s m a x i m u m a r e a .$
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