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Question Number 104101 by ajfour last updated on 19/Jul/20

Commented by ajfour last updated on 19/Jul/20

$I f t h e o u t e r t r i a n g l e i s e q u i l a t e r a l,$ $a n d t h e t w o e l l i p s e s o f s a m e s i z e$ $a n d s h a p e, f i n d r a t i o o f a r e a o f$ $o u t e r t r i a n g l e t o i n n e r (b l u e)$ $t r i a n g l e .$
Commented by mr W last updated on 19/Jul/20

$o n l y i f b l u e e l l i p s e = y e l l o w$ $e l l i p s e = i n c i r c l e o f t r i a n g l e$ $\Rightarrow \frac{s m a l l t r i a n g l e}{b i g t r i a n g l e} = \frac{1}{4}$
	Answered by ajfour last updated on 19/Jul/20

	$l e t e q . o f b l u e e l l i p s e b e$ $\frac{x^{2}}{a^{2}} + \frac{{(y - b)}^{2}}{b^{2}} = 1$ $L e t r i g h t s i d e o f o u t e r t r i a n g l e h a s$ $e q : y = x \sqrt{3} + c a n d a s t h i s i s$ $t a n g e n t t o b o t h e l l i p s e s, f i r s t$ $b^{2} x^{2} + a^{2} {(x \sqrt{3} + c - b)}^{2} - a^{2} b^{2} = 0$ $o r$ $(3 a^{2} + b^{2}) x^{2} + 2 \sqrt{3} (a^{2}) (c - b) x$ $+ a^{2} {(c - b)}^{2} - a^{2} b^{2} = 0$ $h a s d o u b l e r o o t; \Rightarrow$ $12 a^{4} {(c - b)}^{2} = 4 a^{2} (3 a^{2} + b^{2}) [{(c - b)}^{2} - b^{2}]$ $\Rightarrow 3 a^{2} {(c - b)}^{2} = (3 a^{2} + b^{2}) [{(c - b)}^{2} - b^{2}]$ $\Rightarrow {(c - b)}^{2} = 3 a^{2} + b^{2} . . . . . (I)$ $N o w e q . o f t h e y e l l o w e l l i p s e i s$ $\frac{x^{2}}{b^{2}} + \frac{{(y - a)}^{2}}{a^{2}} = 1$ $y = x \sqrt{3} + c i s t a n g e n t t o t h i s e l l i p s e$ $t o o, h e n c e$ $a^{2} x^{2} + b^{2} {(x \sqrt{3} + c - a)}^{2} - a^{2} b^{2} = 0$ $(a^{2} + 3 b^{2}) x^{2} + 2 \sqrt{3} (b^{2}) (c - a) x$ $+ b^{2} [{(c - a)}^{2} - a^{2}] = 0$ $h a s a l s o a d o u b l e r o o t; \Rightarrow$ $12 b^{4} {(c - a)}^{2} = 4 b^{2} (a^{2} + 3 b^{2}) [{(c - a)}^{2} - a^{2}]$ $\Rightarrow 3 b^{2} {(c - a)}^{2} = (a^{2} + 3 b^{2}) [{(c - a)}^{2} - a^{2}]$ $\Rightarrow {(c - a)}^{2} = a^{2} + 3 b^{2} . . . . . (I I)$ $N o w e l i m i n a t i n g c a m o n g (I) & (I I)$ $b + \sqrt{3 a^{2} + b^{2}} = a + \sqrt{a^{2} + 3 b^{2}}$ $l e t \frac{b}{a} = μ$ $\Rightarrow μ + \sqrt{3 + μ^{2}} = 1 + \sqrt{1 + 3 μ^{2}} . . . . (i)$ $\Rightarrow μ = 1 ★$ $\Rightarrow c - a = c - b = 2 a = 2 b$ $\Rightarrow a = b = \frac{c}{3}$ $\Rightarrow \frac{△_{s m a l l}}{△_{l a r g e}} = \frac{1}{4} ◼$ $m r W S i r y o u a r e p e r f e c t l y r i g h t!$
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