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Question Number 183951 by Acem last updated on 01/Jan/23

Commented by Acem last updated on 01/Jan/23

$∙ c A r e t h e p o i n t s A, F, N, M \in a c i r c l e ?$ $i f t h e y a r e t h e n f i n d i t c e n t e r p o s i t i o n a n d i t r a d i u s$ $* A N ⊥ B C, B A ⊥ C A$
	Answered by HeferH last updated on 01/Jan/23

	$a . S i n c e ∠ B A C = 90 ° a n d A M i s m e d i a n \Rightarrow$ $∠ N A M = 30 °$ $b . Y e s, t h e r a t i o o f t h e i r s i d e s i s \frac{1}{2},$ $t h e r e i s a s p e c i f i c t h e o r e m, i f y o u j o i n$ $t h e m i d p o i n t s o f a n y t w o s i d e s o f a t r i a n g l e$ $t h e t r i a n g l e t h a t y o u g e t i s g o i n g t o b e s i m i l a r$ $t o t h e f i r s t o n e .$ $c . Y e s, t h e q u a d r i l a t e r a l A F N M w o u l d n e e d t o$ $b e c y c l i c, f o r t h a t ∠ F + ∠ M = 180 ° \lor$ $∠ A + ∠ N = 180 ° . I n t h i s c a s e$ $60 ° (∠ M A F) + 120 ° (M N F) = 180 °$ $I w i l l s e n d a n i m a g e l a t e r :) a n d t h e r a d i u s$ $s t u f f$
	Commented by HeferH last updated on 01/Jan/23

	Commented by HeferH last updated on 01/Jan/23

	$A M N F e n d u p b e i n g a n i s o s c e l e s t r a p e z o i d .$ $\Rightarrow r = 1$ $A l s o :$ $(x, y) = (\frac{5}{2}, \frac{\sqrt{3}}{2}) (A c c o r d i n g t o t h e f i r s t d i a g r a m)$ $b u t I m a y h a v e a m i s t a k e s o m e w h e r e$
	Commented by Acem last updated on 01/Jan/23

	$B i g T h a n k s f o r y o u r e f f o r t s d e a r f r i e n d!$ $I o n l y {d i d n}^{'} t g e t t h e p o i n t (x, y), i f t h e o r i g i n w a s$ $M, t h e c e n t e r s h o u l d b e (\frac{1}{2}, \frac{\sqrt{3}}{2})$ $P . s : Y o u c a n c h e c k m y m e t h o d$
	Commented by HeferH last updated on 01/Jan/23

	$I t o u g h t B w a s t h e o r i g i n, s o t h a t s w h y$ $i g o t s o m e t h i n g e l s e t h a n k y o u f o r t h e$ $i n t e r e s t i n g p r o b l e m f r i e n d!$
	Commented by Acem last updated on 01/Jan/23

	$T h e r e i s n o t a n y p r o b l e m a t y o u r s o l u t i o n$ $I f y o u d e e m B a s t h e o r i g i n s o t h e p o i n t (\frac{5}{2}, \frac{\sqrt{3}}{2})$ $i s 100 % c o r r e c t .$ $Y o u r s o l u t i o n i s c o m p l e t l y c o r r e c t$ $T h a n k y o u v e r y m u c h!$
	Answered by Acem last updated on 01/Jan/23

	Commented by Acem last updated on 01/Jan/23

	Commented by Acem last updated on 01/Jan/23

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