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Question Number 23036 by ajfour last updated on 25/Oct/17

Commented by mrW1 last updated on 25/Oct/17

$\frac{[P Q R S T U]}{[A B C D E F]} = \frac{1}{7}$
Commented by ajfour last updated on 25/Oct/17

$T h e b l u e l i n e s e g m e n t f r o m A$ $s t a r t s a t A a n d e n d s i n m i d p o i n t$ $o f O E, b l u e l i n e s e g m e n t f r o m B$ $e n d s i n m i d p o i n t o f O F, . . a n d$ $s o o n . F i n d r a t i o a r e a o f r e g u l a r$ $h e x a g o n P Q R S T U s o f o r m e d,$ $t o t h e a r e a o f g i v e n r e g u l a r$ $h e x a g o n A B C D E F o f s i d e a .$
Commented by ajfour last updated on 25/Oct/17

$P l e a z e e x p l a i n S i r .$ $I c r e a t e d t h i s q u e s t i o n m y s e l f,$ ${h a v e n}^{'} t e v e n a t t e m p t e d, n i c e$ $a n s w e r s i r . I t w i l l t a k e a l o n g$ $m e t h o d i f i s o l v e w i t h v e c t o r d$ $o r c o o r d i n a t e g e o m e t r y .$ $K i n d l y e x p l a i n y o u r w a y s i r .$
Commented by mrW1 last updated on 25/Oct/17

Commented by mrW1 last updated on 25/Oct/17
$let a=1 to make the writing easier. OE=OD=...=a=1 OL=ON=...=(a/2)=(1/2) EN^2 =1^2 +((1/2))^2 −2×1×(1/2)×cos 120°=1+(1/4)+(1/2)=(7/4) ⇒EN=((√7)/2) OK is the angle bisector of ∠EON, ⇒((EK)/(KN))=((OE)/(ON))=2 ⇒EK=(2/3)×EN=(2/3)×((√7)/2)=((√7)/3) ⇒KN=(1/3)×EN=(1/3)×((√7)/2)=((√7)/6) { EK^2 =OE^2 +OK^2 −2×OE×OK×cos 60° (7/9)=1+OK^2 −2×1×OK×(1/2) OK^2 −OK+(2/9)=0 ⇒OK=(1/2)(1±(√(1−(8/9))))=(1/2)(1±(1/3))=((3±1)/6)= { ((1/3)),((2/3)) :} only (1/3) is suitable for us. ⇒OK=(1/3) } (one can also directly use formula for angle bisector to get OK instead of {...}) KD=OD−OK=1−(1/3)=(2/3) ΔDKR∼ΔEKO ((KR)/(KD))=((OK)/(EK)) ⇒KR=(1/3)×(3/(√7))×(2/3)=((2(√7))/(21)) ((RD)/(KD))=((OK)/(EO)) ⇒RD=((1×3)/(√7))×(2/3)=((2(√7))/7) RM=DM−RD=EK−RD=((√7)/3)−((2(√7))/7)=((√7)/(21)) SR=SK+KR=RM+KR=((√7)/(21))+((2(√7))/(21))=((√7)/7) =side length of the small hexagon (A_(small) /A_(big) )=(((side length small)/(side length big)))^2 =(((((√7)/7)a)/a))^2 =(1/7)$
$let a = 1 to make the writing easier .$ $OE = OD = . . . = a = 1$ $OL = ON = . . . = \frac{a}{2} = \frac{1}{2}$ ${EN}^{2} = 1^{2} + {(\frac{1}{2})}^{2} - 2 \times 1 \times \frac{1}{2} \times \cos 120 ° = 1 + \frac{1}{4} + \frac{1}{2} = \frac{7}{4}$ $\Rightarrow EN = \frac{\sqrt{7}}{2}$ $OK is the angle bisector of ∠ EON,$ $\Rightarrow \frac{EK}{KN} = \frac{OE}{ON} = 2$ $\Rightarrow EK = \frac{2}{3} \times EN = \frac{2}{3} \times \frac{\sqrt{7}}{2} = \frac{\sqrt{7}}{3}$ $\Rightarrow KN = \frac{1}{3} \times EN = \frac{1}{3} \times \frac{\sqrt{7}}{2} = \frac{\sqrt{7}}{6}$ ${$ ${EK}^{2} = {OE}^{2} + {OK}^{2} - 2 \times OE \times OK \times \cos 60 °$ $\frac{7}{9} = 1 + {OK}^{2} - 2 \times 1 \times OK \times \frac{1}{2}$ ${OK}^{2} - OK + \frac{2}{9} = 0$ $\Rightarrow OK = \frac{1}{2} (1 \pm \sqrt{1 - \frac{8}{9}}) = \frac{1}{2} (1 \pm \frac{1}{3}) = \frac{3 \pm 1}{6} = {\begin{cases} \frac{1}{3} \\ \frac{2}{3} \end{cases}$ $only \frac{1}{3} is suitable for us .$ $\Rightarrow OK = \frac{1}{3}$ $}$ $(one can also directly use formula for$ $angle bisector to get OK instead of {. . .})$ $KD = OD - OK = 1 - \frac{1}{3} = \frac{2}{3}$ $Δ DKR \sim Δ EKO$ $\frac{KR}{KD} = \frac{OK}{EK}$ $\Rightarrow KR = \frac{1}{3} \times \frac{3}{\sqrt{7}} \times \frac{2}{3} = \frac{2 \sqrt{7}}{21}$ $\frac{RD}{KD} = \frac{OK}{EO}$ $\Rightarrow RD = \frac{1 \times 3}{\sqrt{7}} \times \frac{2}{3} = \frac{2 \sqrt{7}}{7}$ $RM = DM - RD = EK - RD = \frac{\sqrt{7}}{3} - \frac{2 \sqrt{7}}{7} = \frac{\sqrt{7}}{21}$ $SR = SK + KR = RM + KR = \frac{\sqrt{7}}{21} + \frac{2 \sqrt{7}}{21} = \frac{\sqrt{7}}{7}$ $= side length of the small hexagon$ $\frac{A_{small}}{A_{big}} = {(\frac{side length small}{side length big})}^{2} = {(\frac{\frac{\sqrt{7}}{7} a}{a})}^{2} = \frac{1}{7}$
Commented by ajfour last updated on 25/Oct/17

$t h a n k y o u s i r, b u t i s h a l l a l s o$ $a t t e m p t n o w . V e c t o r w a y . .$ $p e r h a p s!$
Commented by mrW1 last updated on 25/Oct/17

$Here an other way to calculate OK$ $without solving a quadratic equation$ $which delievers 2 solutions to choose .$ $\frac{OK}{\sin ∠ OEN} = \frac{EK}{\sin α} (with α = 60 °)$ $\frac{ON}{\sin ∠ OEN} = \frac{EN}{\sin 2 α} = \frac{EN}{2 \sin α \cos α}$ $\Rightarrow \frac{ON}{EN} \times 2 \cos α = \frac{OK}{EK}$ $OK = EK \times \frac{ON}{EN} \times 2 \cos α$ $\cos 2 α = \frac{{OE}^{2} + {ON}^{2} - {EN}^{2}}{2 \times OE \times ON} = {2 \cos}^{2} α - 1$ $\Rightarrow OK = ON \times \frac{EK}{EN} \times \sqrt{2 (1 + \frac{{OE}^{2} + {ON}^{2} - {EN}^{2}}{2 \times OE \times ON})}$ $= \frac{1}{2} \times \frac{2}{3} \times \sqrt{2 (1 + \frac{1 + \frac{1}{4} - \frac{7}{4}}{2 \times 1 \times \frac{1}{2}})}$ $= \frac{1}{3} \times \sqrt{2 (1 - \frac{1}{2})}$ $= \frac{1}{3}$
Commented by ajfour last updated on 25/Oct/17

$L e t a = 1$ $e q . o f F L :$ $y = \frac{(\sqrt{3} / 4)}{(1 + 1 / 4)} (x + 1)$ $o r y = \frac{\sqrt{3}}{5} (x + 1) . . . . (i)$ $e q . o f E N :$ $y = \frac{- \sqrt{3} / 2}{1} (x - 1 / 2)$ $o r y = - \frac{\sqrt{3}}{4} (2 x - 1) . . . (i i)$ $l e t S (α, β); S l i e s o n F L a n d ? E N$ $\Rightarrow β = \frac{\sqrt{3}}{5} (α + 1) = - \frac{\sqrt{3}}{4} (2 α - 1)$ $\Rightarrow 4 α + 4 + 5 (2 α - 1) = 0$ $o r α = \frac{1}{14} a n d β = \frac{3 \sqrt{3}}{14}$ $O S = \sqrt{α^{2} + β^{2}} = \sqrt{\frac{28}{14 \times 14}} = \frac{1}{\sqrt{7}}$ $\frac{A r e a P Q R S T U}{A r e a A B C D E} = \frac{{O S}^{2}}{{O A}^{2}} = \frac{1}{7} .$
Commented by mrW1 last updated on 25/Oct/17

$very nice way!$
Commented by math solver last updated on 25/Oct/17

$s i r, i n e q . o f F L, w h i l e w r i t i n g s l o p e$ $i {d i d n}^{'} t g e t t h e n u m e r a t o r ? ?$ $i t h i n k n u m e r a t o r s h o u l d b e$ $t a n 30 \times \frac{5}{4} = 5 / 4 \times \sqrt{3}$
Commented by ajfour last updated on 25/Oct/17

$y e s .$
Commented by ajfour last updated on 25/Oct/17

$F (- a, 0) L (\frac{a}{2} \cos 60 °, \frac{a}{2} \sin 60 °)$ $\Rightarrow L (\frac{a}{4}, \frac{a \sqrt{3}}{4})$ $s l o p e o f F L = \frac{\frac{a \sqrt{3}}{4} - 0}{\frac{a}{4} - (- a)} = \frac{\sqrt{3}}{5} .$
Commented by math solver last updated on 25/Oct/17

$i s L m i d p o i n t o f O D ?$
Commented by math solver last updated on 25/Oct/17

${i s n}^{'} t E (- 0.5, 1) a n d N (0.5, 0) w h e n$ $a = 1 ? ? ?$
Commented by ajfour last updated on 26/Oct/17

$E (- 0.5, \frac{\sqrt{3}}{2})$
Commented by math solver last updated on 26/Oct/17

$ok, i got it now .$ $thanks!$
	Answered by Tinkutara last updated on 26/Oct/17

	Commented by Tinkutara last updated on 26/Oct/17

	$Is the method same or somewhat$ $different ?$
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