Given a convex hexagon ABCDEG satisfy: AB=BC, CD=DE, EF=FA. Suppose △ACE is a right triangle. (((BC)/(BE))+((DE)/(DA))+((FA)/(FC)))


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Question Number 139254 by mathdanisur last updated on 25/Apr/21

$G i v e n a c o n v e x h e x a g o n A B C D E G$ $s a t i s f y : A B = B C, C D = D E, E F = F A .$ $S u p p o s e △ A C E i s a r i g h t t r i a n g l e .$ ${(\frac{B C}{B E} + \frac{D E}{D A} + \frac{F A}{F C})}_{m i n} = ?$
	Answered by mr W last updated on 27/Apr/21

	Commented by mr W last updated on 27/Apr/21

	$B C = \sqrt{a^{2} + p^{2}}$ $B E = \sqrt{a^{2} + {(2 b + p)}^{2}}$ $P = \frac{B E}{B C} = \frac{\sqrt{a^{2} + {(2 b + p)}^{2}}}{\sqrt{a^{2} + p^{2}}} = \frac{\sqrt{1 + {(\frac{2 b}{a} + \frac{p}{a})}^{2}}}{\sqrt{1 + {(\frac{p}{a})}^{2}}}$ $l e t x = \frac{p}{a}, λ = \frac{b}{a}$ $P = \frac{\sqrt{1 + {(2 λ + x)}^{2}}}{\sqrt{1 + x^{2}}}$ $\frac{d P}{d x} = \frac{2 (2 λ + x)}{2 \sqrt{1 + {(2 λ + x)}^{2}} \sqrt{1 + x^{2}}} - \frac{2 x \sqrt{1 + {(2 λ + x)}^{2}}}{2 (1 + x^{2}) \sqrt{1 + x^{2}}} = 0$ $\frac{(2 λ + x)}{\sqrt{1 + {(2 λ + x)}^{2}}} = \frac{x \sqrt{1 + {(2 λ + x)}^{2}}}{(1 + x^{2})}$ $(1 + x^{2}) (2 λ + x) = x [1 + {(2 λ + x)}^{2}]$ $x^{2} + 2 λ x - 1 = 0$ $x = \sqrt{λ^{2} + 1} - λ = 1 f o r λ = \frac{b}{a} = 1$ $P_{m a x} = \frac{\sqrt{1 + {(2 + 1)}^{2}}}{\sqrt{1 + 1}} = \sqrt{5}$ ${(\frac{B C}{B E})}_{m i n} = \frac{1}{P_{m a x}} = \frac{1}{\sqrt{5}}$ $s i m i l a r l y$ ${(\frac{D E}{D A})}_{m i n} = \frac{1}{\sqrt{5}}$ $w i t h b = a, c = \sqrt{2} a$ $F A = \sqrt{c^{2} + r^{2}}$ $F C = c + r$ $S = \frac{F A}{F C} = \frac{\sqrt{c^{2} + r^{2}}}{c + r} = \frac{\sqrt{1 + {(\frac{r}{c})}^{2}}}{1 + \frac{r}{c}}$ $y = \frac{r}{c}$ $S = \frac{\sqrt{1 + y^{2}}}{1 + y}$ $\frac{d S}{d y} = \frac{2 y}{2 \sqrt{1 + y^{2}} (1 + y)} - \frac{\sqrt{1 + y^{2}}}{{(1 + y)}^{2}} = 0$ $\frac{y}{\sqrt{1 + y^{2}}} = \frac{\sqrt{1 + y^{2}}}{1 + y}$ $y^{2} + y = 1 + y^{2}$ $y = 1$ $S_{m i n} = \frac{\sqrt{1 + 1^{2}}}{1 + 1} = \frac{1}{\sqrt{2}}$ ${(\frac{B C}{B E} + \frac{D E}{D A} + \frac{F A}{F C})}_{m i n} = \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{2}} = \frac{2 \sqrt{2} + \sqrt{5}}{\sqrt{10}}$
	Commented by mathdanisur last updated on 30/Apr/21

	$t h a n k y o u s i r$
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