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Question Number 80015 by mr W last updated on 30/Jan/20

Commented by mr W last updated on 30/Jan/20

$1 . P r o v e t h a t b o t h e q u i l a t e r i a l t r i a n g l e s$ $t o u c h e a c h o t h e r i n t h e s q u a r e a s s h o w n .$ $2 . F i n d r_{1} / r_{2}$
Commented by jagoll last updated on 30/Jan/20

$c i r c l e a g a i n t$
	Answered by mind is power last updated on 30/Jan/20
	$let ABCD square A(0,0);B(1,0) C(1,1) D(0,1) triangle ABE de Bas E is image of B by rootation center A angle 60 ⇒Z_E −Z_A =e^(i(π/3)) (Z_B −Z_A ) ⇒Z_E =((1/2)+((i(√3))/2))1 E((1/2),((√3)/2)) we get head of triangle the last one has C(1,1) and F=(CE)∩(AD) (CE):y=((((√3)/2)−1)/(−(1/2)))x+b y=(2−(√3))x+(√3)−1 F= { ((y=(2−(√3))x+(√3)−1)),((x=0)) :} ⇒y=(√3)−1⇒F(0,(√3)−1) the last Head G is obtined as Rootation of F by (π/3) center C Z_G =Z_c +((1/2)+((i(√3))/2))(Z_F −Z_c ) =(1+i)+(1/2)(1+i(√3))(i((√3)−1)−1−i) =(1+i)+(1/2)(1+i(√3))(i((√3)−2)−1) =1+i+(1/2)(−4+2(√3)−2i) =(−2+(√3)−i)+1+i =(√3)−1 G((√3)−1,0) ∈(AB) as show in diagramme This proof existence of such Triangle r_1 ,r_2 juste circle inside Two triangle witche we can find coordinate of head too be continued$
	$l e t$ $A B C D s q u a r e$ $A (0, 0); B (1, 0) C (1, 1) D (0, 1)$ $t r i a n g l e$ $A B E d e B a s$ $E i s i m a g e o f B b y r o o t a t i o n c e n t e r A a n g l e 60$ $\Rightarrow Z_{E} - Z_{A} = e^{i \frac{π}{3}} (Z_{B} - Z_{A})$ $\Rightarrow Z_{E} = (\frac{1}{2} + \frac{i \sqrt{3}}{2}) 1$ $E (\frac{1}{2}, \frac{\sqrt{3}}{2})$ $w e g e t h e a d o f t r i a n g l e$ $t h e l a s t o n e h a s C (1, 1) a n d F = (C E) \cap (A D)$ $(C E) : y = \frac{\frac{\sqrt{3}}{2} - 1}{- \frac{1}{2}} x + b$ $y = (2 - \sqrt{3}) x + \sqrt{3} - 1$ $F = {\begin{cases} y = (2 - \sqrt{3}) x + \sqrt{3} - 1 \\ x = 0 \end{cases}$ $\Rightarrow y = \sqrt{3} - 1 \Rightarrow F (0, \sqrt{3} - 1)$ $t h e l a s t H e a d G i s o b t i n e d a s R o o t a t i o n o f F b y \frac{π}{3} c e n t e r C$ $Z_{G} = Z_{c} + (\frac{1}{2} + \frac{i \sqrt{3}}{2}) (Z_{F} - Z_{c})$ $= (1 + i) + \frac{1}{2} (1 + i \sqrt{3}) (i (\sqrt{3} - 1) - 1 - i)$ $= (1 + i) + \frac{1}{2} (1 + i \sqrt{3}) (i (\sqrt{3} - 2) - 1)$ $= 1 + i + \frac{1}{2} (- 4 + 2 \sqrt{3} - 2 i)$ $= (- 2 + \sqrt{3} - i) + 1 + i$ $= \sqrt{3} - 1$ $G (\sqrt{3} - 1, 0) \in (A B) a s s h o w i n d i a g r a m m e$ $T h i s p r o o f e x i s t e n c e o f s u c h T r i a n g l e$ $r_{1}, r_{2} j u s t e c i r c l e i n s i d e T w o t r i a n g l e w i t c h e w e c a n f i n d$ $c o o r d i n a t e o f h e a d t o o b e c o n t i n u e d$
	Answered by ajfour last updated on 30/Jan/20

	Commented by ajfour last updated on 30/Jan/20

	$l e t s q u a r e s i d e b e 2 .$ $C (1, \sqrt{3})$ $l e t D E = 2 - x$ $\Rightarrow 2 x^{2} = 4 + {(2 - x)}^{2}$ $\Rightarrow x^{2} + 4 x - 8 = 0$ $x = - 2 + \sqrt{4 + 8} = 2 (\sqrt{3} - 1)$ $\Rightarrow O F = O D = x = 2 \sqrt{3} - 2$ $H F = 2 - x = 4 - 2 \sqrt{3}$ $\underline{e q . o f F G}$ $y = (2 - \sqrt{3}) x + 2 \sqrt{3} - 2$ $C (1, \sqrt{3}) l e t s c h e c k i f i t l i e s$ $o n t h i s l i n e : x = 1$ $y = 2 - \sqrt{3} + 2 \sqrt{3} - 2 = \sqrt{3}$ $\Rightarrow t r i a n g l e s t o u c h i n t h e$ $m a n n e r s h o w n .$ $r_{1} (\cot \frac{π}{6} + \cot \frac{π}{8}) = O D = x$ $= 2 \sqrt{3} - 2$ $r_{2} (\cot \frac{π}{12} + \cot \frac{π}{24}) = 2$ $\frac{r_{1}}{r_{2}} = \frac{(2 \sqrt{3} - 2)}{2} \frac{(\cot \frac{π}{12} + \cot \frac{π}{24})}{(\cot \frac{π}{6} + \cot \frac{π}{8})}$ $= (\sqrt{3} - 1) (\sqrt{3} + 1) = 2 .$
	Commented by mr W last updated on 30/Jan/20

	$t h a n k y o u s i r! p e r f e c t l y s o l v e d!$
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