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Question Number 206157 by cortano21 last updated on 08/Apr/24

Commented by cortano21 last updated on 08/Apr/24

$\underset{―}{\underset{⏟}{}}$
Commented by A5T last updated on 09/Apr/24

$T h i s i s g e n e r a l l y n o t u n i q u e l y d e f i n e d . O n e c a n$ $c o m e u p w i t h m a t h e m a t i c a l l y c o r r e c t s o l u t i o n s$ $w i t h d i f f e r e n t a n s w e r s . P o i n t s A s u c h t h a t$ $\frac{A B}{A P} = \frac{5}{3} l i e o n t h e A p o l l o n i u s c i r c l e .$
Commented by A5T last updated on 09/Apr/24

	Answered by dimentri last updated on 08/Apr/24

	$\frac{AP}{AB} = \frac{9}{15} = \frac{3}{5}$ $△ APB ⊥ in P$ $AP = 18, AB = 30$ ${BP}^{2} = AP . PC$ $PC = \frac{576}{18} = 32$ $area of △ ABC = \frac{50 \times 24}{2} = 600$
	Commented by mr W last updated on 08/Apr/24

	$i t h i n k i t i s n o t s a i d t h a t A P ⊥ B C .$
	Answered by mr W last updated on 08/Apr/24

	Commented by mr W last updated on 09/Apr/24

	$s a y A B = x$ $\frac{A P}{9} = \frac{x}{15} \Rightarrow A P = \frac{3 x}{5}$ ${(15 + 9)}^{2} = x^{2} + {(\frac{3 x}{5})}^{2} - 2 x \times \frac{3 x}{5} \cos α$ $\Rightarrow x^{2} = \frac{7200}{17 - 15 \cos α}$ $a r e a o f Δ A B C :$ $Δ = \frac{x^{2} \tan α}{2} = \frac{3600 \tan α}{17 - 15 \cos α}$ $w e s e e t h e t r i a n g l e A B C i s n o t$ $u n i q u e a n d i t s a r e a i s a l s o n o t$ $u n i q u e .$ $f o r l o c u s o f p o i n t A s e e a l s o$ $Q 181759$
	Commented by mr W last updated on 08/Apr/24

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