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Question Number 5354 by Rasheed Soomro last updated on 11/May/16

Commented by Rasheed Soomro last updated on 11/May/16

$ABCD and EFGH are squares in$ $above figure . The centres of the arcs$ $are vertices of the external square .$ $If EF = t then AB = ?$
Commented by Yozzii last updated on 11/May/16

$L e t s i d e s D A a n d A B l i e o n t h e y a n d x$ $a x e s r e s p e c t i v e l y, w i t h A h a v i n g$ $c o o r d i n a t e s (0, 0) . S o l e t B h a v e c o o r d i n a t e s B (r, 0)$ $r > 0 . S i n c e F G = t > 0, f o r E G F H b e i n g$ $a s q u a r e o f s i d e t, i t s s i d e s a r e p a r a l l e l$ $t o t h o s e o f ◻ A B C D a n d i t s c e n t r e$ $c o i n c i d e s w i t h t h a t o f ◻ A B C D, t h e n$ $t h e l i n e x = r / 2 r e v e a l s a s y m m e t r y$ $o f t h e f i g u r e . H e n c e, t h e x - c o o r d i n a t e$ $o f p o i n t F i s g i v e n b y x = \frac{r - t}{2} . S i n c e F$ $l i e s o n t h e d i a g o n a l A C w h i c h c o i n c i d e s$ $w i t h t h e l i n e y = x, t h e n a t F, y = \frac{r - t}{2} .$ $S o, p o i n t F h a s c o o r d i n a t e s (\frac{r - t}{2}, \frac{r - t}{2}) .$ $N o w, t h e p o i n t H i s t h e p o i n t F t r a n s l a t e d$ $t u n i t s h o r i z o n t a l l y a n d v e r t i c a l l y t o w a r d s$ $t h e s i d e B C o f ◻ A B C D . S o, r e l a t i v e t o A,$ $H h a s c o o r d i n a t e s (\frac{r - t}{2} + t, \frac{r - t}{2} + t) = (\frac{r + t}{2}, \frac{r + t}{2}) .$ $B u t, A H = A B = r s i n c e A B H f o r m s a s e c t o r o f a c i r c l e w i t h A a s t h e c e n t r e . T h e r e f o r e,$ ${(\frac{r + t}{2} - 0)}^{2} + {(\frac{r + t}{2} - 0)}^{2} = r^{2}$ $2 {(\frac{r + t}{2})}^{2} = r^{2} \Rightarrow \frac{r + t}{2} = \frac{r}{\sqrt{2}} \Rightarrow r + t = r \sqrt{2}$ $r (\sqrt{2} - 1) = t \Rightarrow r = \frac{t}{\sqrt{2} - 1} = t (1 + \sqrt{2})$ $S o, i f E F = t \Rightarrow A B = t (1 + \sqrt{2}) .$
Commented by Rasheed Soomro last updated on 12/May/16

$Good approach!$
	Answered by Rasheed Soomro last updated on 12/May/16
	$Non-analytical approach. { ((AF+FH=AB)),((CH+FH=CD=AB)) :} { (),() :}AF=CH Let AF=x and Let AB=T ∴CH=x , BC=CD=DA=T since, EF=t So, FG=GH=HE=t FH=(√(t^2 +t^2 ))=(√2) t AF+FH=AB⇒x+(√2) t=T..........................I AC=AF+FH+CH=2x+(√2) t=(√(T^2 +T^2 )) =(√2) T....II 2x+2(√2) t=2T I×2.........III 2x+(√2) t=(√2) T II×1.........IV (√2) t=(2−(√2))T III −IV T=(((√2) t)/(2−(√2)))×((2+(√2))/(2+(√2)))=t((√2)+1)$
	$Non - analytical approach .$ ${\begin{cases} AF + FH = AB \\ CH + FH = CD = AB \end{cases} {\begin{cases} \end{cases} AF = CH$ $Let AF = x and Let AB = T$ $∴ CH = x, BC = CD = DA = T$ $since,$ $EF = t$ $So, FG = GH = HE = t$ $FH = \sqrt{t^{2} + t^{2}} = \sqrt{2} t$ $AF + FH = AB \Rightarrow x + \sqrt{2} t = T . . . . . . . . . . . . . . . . . . . . . . . . . . I$ $AC = AF + FH + CH = 2 x + \sqrt{2} t = \sqrt{T^{2} + T^{2}} = \sqrt{2} T . . . . II$ $2 x + 2 \sqrt{2} t = 2 T I \times 2 . . . . . . . . . III$ $2 x + \sqrt{2} t = \sqrt{2} T II \times 1 . . . . . . . . . IV$ $\sqrt{2} t = (2 - \sqrt{2}) T III - IV$ $T = \frac{\sqrt{2} t}{2 - \sqrt{2}} \times \frac{2 + \sqrt{2}}{2 + \sqrt{2}} = t (\sqrt{2} + 1)$
	Commented by Yozzii last updated on 12/May/16

	$N i c e! M y u n d e r s t a n d i n g o f a n d c o n f i d e n c e i n E u c l i d e a n g e o m e t r y s t i l l$ $n e e d s s o m e w o r k . T h e e d u c a t i o n s y s t e m I^{'} v e g o n e t h r o u g h$ $p l a c e d e m p h a s i s o n a n a l y t i c a l m e t h o d s .$
	Commented by Rasheed Soomro last updated on 12/May/16

	Commented by Rasheed Soomro last updated on 12/May/16

	$I have been observing you since your joining$ $to this forum . Your logic is very powerful, so$ $Euclidean geometry is simple for you!$ $. . . . .$ $Have you qualified the cambridge exam / test ?$
	Commented by Yozzii last updated on 13/May/16

	$I^{'} v e o n l y s a t t h e C I E F u r t h e r M a t h$ $A L e v e l e x a m i n 2015, a n d t h a t {d o e s n}^{'} t$ $h a v e a n e m p h a s i s o n E u c l i d e a n g e o m e t r y .$ $I t a p p e a r s a b i t i n s t u d y i n g m e c h a n i c s,$ $b u t a t t h a t l e v e l r i g o r o u s p r o o f w r i t i n g$ ${i s n}^{'} t r e q u i r e d . {T h a t}^{'} s t h e p a s t .$ $N o w I^{'} m j u s t h e l p i n g m y s e l f g r o w$ $i n m a t h e m a t i c a l s k i l l, r i g o u r a n d p r e s e n t a t i o n$ $b e f o r e I b e g i n u n d e r g r a d u a t e u n i v e r s i t y s t u d i e s$ $f r o m S e p t e m b e r 2016 i n m a t h e m a t i c s .$
	Commented by Rasheed Soomro last updated on 13/May/16

	$T h α n k S! Actually, I {didn}^{'} t consider any connection$ $between your exam and Eucledean geometry .$ $I only wanted to be informed about your$ $progress .$
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