Question-23036

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

Commented by mrW1 last updated on 25/Oct/17

$$\frac{\left[\boldsymbol{{PQRSTU}}\right]}{\left[\boldsymbol{{ABCDEF}}\right]}=\frac{\mathrm{1}}{\mathrm{7}} \\ $$

Commented by ajfour last updated on 25/Oct/17

The blue line segment from A starts at A and ends in midpoint of OE, blue line segment from B ends in midpoint of OF,.. and so on. Find ratio area of regular hexagon PQRSTU so formed, to the area of given regular hexagon ABCDEF of side a.

$${The}\:{blue}\:{line}\:{segment}\:{from}\:{A} \\ $$$${starts}\:{at}\:{A}\:{and}\:{ends}\:{in}\:{midpoint} \\ $$$${of}\:{OE},\:{blue}\:{line}\:{segment}\:{from}\:{B} \\ $$$${ends}\:{in}\:{midpoint}\:{of}\:{OF},..\:{and} \\ $$$${so}\:{on}.\:{Find}\:{ratio}\:\boldsymbol{{area}}\:\boldsymbol{{of}}\:{regular} \\ $$$${hexagon}\:\boldsymbol{{PQRSTU}}\:{so}\:{formed}, \\ $$$${to}\:{the}\:\boldsymbol{{area}}\:\boldsymbol{{of}}\:{given}\:{regular} \\ $$$${hexagon}\:\boldsymbol{{ABCDEF}}\:{of}\:{side}\:\boldsymbol{{a}}. \\ $$

Commented by ajfour last updated on 25/Oct/17

$${Pleaze}\:{explain}\:{Sir}. \\ $$$${I}\:{created}\:{this}\:{question}\:{myself}, \\ $$$${haven}'{t}\:{even}\:{attempted},\:{nice} \\ $$$${answer}\:{sir}.\:{It}\:{will}\:{take}\:{a}\:{long} \\ $$$${method}\:{if}\:{i}\:{solve}\:{with}\:{vectord} \\ $$$${or}\:{coordinate}\:{geometry}. \\ $$$${Kindly}\:{explain}\:{your}\:{way}\:{sir}. \\ $$

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)

$$\mathrm{let}\:\mathrm{a}=\mathrm{1}\:\mathrm{to}\:\mathrm{make}\:\mathrm{the}\:\mathrm{writing}\:\mathrm{easier}. \\ $$$$ \\ $$$$\mathrm{OE}=\mathrm{OD}=…=\mathrm{a}=\mathrm{1} \\ $$$$\mathrm{OL}=\mathrm{ON}=…=\frac{\mathrm{a}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{EN}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{cos}\:\mathrm{120}°=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{EN}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{OK}\:\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{of}\:\angle\mathrm{EON}, \\ $$$$\Rightarrow\frac{\mathrm{EK}}{\mathrm{KN}}=\frac{\mathrm{OE}}{\mathrm{ON}}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{EK}=\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{EN}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{KN}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{EN}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{6}} \\ $$$$\left\{\right. \\ $$$$\mathrm{EK}^{\mathrm{2}} =\mathrm{OE}^{\mathrm{2}} +\mathrm{OK}^{\mathrm{2}} −\mathrm{2}×\mathrm{OE}×\mathrm{OK}×\mathrm{cos}\:\mathrm{60}° \\ $$$$\frac{\mathrm{7}}{\mathrm{9}}=\mathrm{1}+\mathrm{OK}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{OK}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{OK}^{\mathrm{2}} −\mathrm{OK}+\frac{\mathrm{2}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{OK}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\sqrt{\mathrm{1}−\frac{\mathrm{8}}{\mathrm{9}}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{3}\pm\mathrm{1}}{\mathrm{6}}=\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}}}\\{\frac{\mathrm{2}}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{only}\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{is}\:\mathrm{suitable}\:\mathrm{for}\:\mathrm{us}. \\ $$$$\Rightarrow\mathrm{OK}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left.\right\} \\ $$$$\left(\mathrm{one}\:\mathrm{can}\:\mathrm{also}\:\mathrm{directly}\:\mathrm{use}\:\mathrm{formula}\:\mathrm{for}\right. \\ $$$$\left.\mathrm{angle}\:\mathrm{bisector}\:\mathrm{to}\:\mathrm{get}\:\mathrm{OK}\:\mathrm{instead}\:\mathrm{of}\:\left\{…\right\}\right) \\ $$$$\mathrm{KD}=\mathrm{OD}−\mathrm{OK}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$$$\Delta\mathrm{DKR}\sim\Delta\mathrm{EKO} \\ $$$$\frac{\mathrm{KR}}{\mathrm{KD}}=\frac{\mathrm{OK}}{\mathrm{EK}} \\ $$$$\Rightarrow\mathrm{KR}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{7}}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{21}} \\ $$$$\frac{\mathrm{RD}}{\mathrm{KD}}=\frac{\mathrm{OK}}{\mathrm{EO}} \\ $$$$\Rightarrow\mathrm{RD}=\frac{\mathrm{1}×\mathrm{3}}{\:\sqrt{\mathrm{7}}}×\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}} \\ $$$$\mathrm{RM}=\mathrm{DM}−\mathrm{RD}=\mathrm{EK}−\mathrm{RD}=\frac{\sqrt{\mathrm{7}}}{\mathrm{3}}−\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{21}} \\ $$$$ \\ $$$$\mathrm{SR}=\mathrm{SK}+\mathrm{KR}=\mathrm{RM}+\mathrm{KR}=\frac{\sqrt{\mathrm{7}}}{\mathrm{21}}+\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{21}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{7}} \\ $$$$=\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{hexagon} \\ $$$$ \\ $$$$\frac{\mathrm{A}_{\mathrm{small}} }{\mathrm{A}_{\mathrm{big}} }=\left(\frac{\mathrm{side}\:\mathrm{length}\:\mathrm{small}}{\mathrm{side}\:\mathrm{length}\:\mathrm{big}}\right)^{\mathrm{2}} =\left(\frac{\frac{\sqrt{\mathrm{7}}}{\mathrm{7}}\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{7}} \\ $$

Commented by ajfour last updated on 25/Oct/17

$${thank}\:{you}\:{sir},\:{but}\:{i}\:{shall}\:{also} \\ $$$${attempt}\:{now}.\:{Vector}\:{way}.. \\ $$$${perhaps}! \\ $$

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. ((OK)/(sin ∠OEN))=((EK)/(sin α)) (with α=60°) ((ON)/(sin ∠OEN))=((EN)/(sin 2α))=((EN)/(2sin α cos α)) ⇒((ON)/(EN))×2cos α=((OK)/(EK)) OK=EK×((ON)/(EN))×2cos α cos 2α=((OE^2 +ON^2 −EN^2 )/(2×OE×ON))=2cos^2 α−1 ⇒OK=ON×((EK)/(EN))×(√(2(1+((OE^2 +ON^2 −EN^2 )/(2×OE×ON))))) =(1/2)×(2/3)×(√(2(1+((1+(1/4)−(7/4))/(2×1×(1/2)))))) =(1/3)×(√(2(1−(1/2)))) =(1/3)

$$\mathrm{Here}\:\mathrm{an}\:\mathrm{other}\:\mathrm{way}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{OK} \\ $$$$\mathrm{without}\:\mathrm{solving}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{equation} \\ $$$$\mathrm{which}\:\mathrm{delievers}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{choose}. \\ $$$$ \\ $$$$\frac{\mathrm{OK}}{\mathrm{sin}\:\angle\mathrm{OEN}}=\frac{\mathrm{EK}}{\mathrm{sin}\:\alpha}\:\:\:\left(\mathrm{with}\:\alpha=\mathrm{60}°\right) \\ $$$$\frac{\mathrm{ON}}{\mathrm{sin}\:\angle\mathrm{OEN}}=\frac{\mathrm{EN}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{\mathrm{EN}}{\mathrm{2sin}\:\alpha\:\mathrm{cos}\:\alpha} \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{ON}}{\mathrm{EN}}×\mathrm{2cos}\:\alpha=\frac{\mathrm{OK}}{\mathrm{EK}} \\ $$$$\mathrm{OK}=\mathrm{EK}×\frac{\mathrm{ON}}{\mathrm{EN}}×\mathrm{2cos}\:\alpha \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{\mathrm{OE}^{\mathrm{2}} +\mathrm{ON}^{\mathrm{2}} −\mathrm{EN}^{\mathrm{2}} }{\mathrm{2}×\mathrm{OE}×\mathrm{ON}}=\mathrm{2cos}^{\mathrm{2}} \:\alpha−\mathrm{1} \\ $$$$\Rightarrow\mathrm{OK}=\mathrm{ON}×\frac{\mathrm{EK}}{\mathrm{EN}}×\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{OE}^{\mathrm{2}} +\mathrm{ON}^{\mathrm{2}} −\mathrm{EN}^{\mathrm{2}} }{\mathrm{2}×\mathrm{OE}×\mathrm{ON}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}×\sqrt{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{7}}{\mathrm{4}}}{\mathrm{2}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\sqrt{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by ajfour last updated on 25/Oct/17

Let a=1 eq. of FL: y=((((√3)/4))/((1+1/4)))(x+1) or y=((√3)/5)(x+1) ....(i) eq. of EN: y=((−(√3)/2)/1)(x−1/2) or y=−((√3)/4)(2x−1) ...(ii) let S(α,β) ; S lies on FL and?EN ⇒ 𝛃=((√3)/5)(𝛂+1)=−((√3)/4)(2𝛂−1) ⇒ 4𝛂+4+5(2𝛂−1)=0 or 𝛂=(1/(14)) and 𝛃=((3(√3))/(14)) OS =(√(𝛂^2 +𝛃^2 )) =(√(((28)/(14×14)) )) = (1/( (√7))) ((Area PQRSTU)/(Area ABCDE))=((OS^( 2) )/(OA^2 )) = (1/7) .

$$\:{Let}\:{a}=\mathrm{1} \\ $$$${eq}.\:{of}\:{FL}: \\ $$$$\:\:\:{y}=\frac{\left(\sqrt{\mathrm{3}}/\mathrm{4}\right)}{\left(\mathrm{1}+\mathrm{1}/\mathrm{4}\right)}\left({x}+\mathrm{1}\right) \\ $$$${or}\:\:\:\:\:\boldsymbol{{y}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\:\:\:\:….\left({i}\right) \\ $$$${eq}.\:{of}\:{EN}: \\ $$$$\:\:\:{y}=\frac{−\sqrt{\mathrm{3}}/\mathrm{2}}{\mathrm{1}}\left({x}−\mathrm{1}/\mathrm{2}\right) \\ $$$${or}\:\:\:\boldsymbol{{y}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\:\:\:\:…\left({ii}\right) \\ $$$${let}\:{S}\left(\alpha,\beta\right)\:\:;\:{S}\:{lies}\:{on}\:{FL}\:{and}?{EN} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{\beta}=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}}\left(\boldsymbol{\alpha}+\mathrm{1}\right)=−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{2}\boldsymbol{\alpha}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{4}\boldsymbol{\alpha}+\mathrm{4}+\mathrm{5}\left(\mathrm{2}\boldsymbol{\alpha}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:{or}\:\:\:\boldsymbol{\alpha}=\frac{\mathrm{1}}{\mathrm{14}}\:\:{and}\:\:\boldsymbol{\beta}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{14}} \\ $$$$\:\:{OS}\:=\sqrt{\boldsymbol{\alpha}^{\mathrm{2}} +\boldsymbol{\beta}^{\mathrm{2}} }\:=\sqrt{\frac{\mathrm{28}}{\mathrm{14}×\mathrm{14}}\:}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}} \\ $$$$\frac{\boldsymbol{{A}}{rea}\:{PQRSTU}}{\boldsymbol{{A}}{rea}\:{ABCDE}}=\frac{{OS}^{\:\mathrm{2}} }{{OA}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{7}}\:. \\ $$

Commented by mrW1 last updated on 25/Oct/17

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{way}! \\ $$

Commented by math solver last updated on 25/Oct/17

sir , in eq. of FL , while writing slope i didn′t get the numerator?? i think numerator should be tan30×(5/4)=5/4 ×(√( 3))

$${sir}\:,\:{in}\:{eq}.\:{of}\:{FL}\:,\:{while}\:{writing}\:{slope} \\ $$$${i}\:{didn}'{t}\:{get}\:{the}\:{numerator}?? \\ $$$${i}\:{think}\:{numerator}\:{should}\:{be}\: \\ $$$${tan}\mathrm{30}×\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{5}/\mathrm{4}\:×\sqrt{\:\mathrm{3}} \\ $$

Commented by ajfour last updated on 25/Oct/17

$${yes}. \\ $$

Commented by ajfour last updated on 25/Oct/17

F(−a,0) L((a/2)cos 60° , (a/2)sin 60° ) ⇒ L((a/4), ((a(√3))/4)) slope of FL = ((((a(√3))/4)−0)/((a/4)−(−a))) = ((√3)/5) .

$${F}\left(−{a},\mathrm{0}\right)\:\:\:\:{L}\left(\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\mathrm{60}°\:,\:\frac{{a}}{\mathrm{2}}\mathrm{sin}\:\mathrm{60}°\:\right) \\ $$$$\Rightarrow\:\:{L}\left(\frac{{a}}{\mathrm{4}},\:\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$$${slope}\:{of}\:{FL}\:=\:\frac{\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{4}}−\mathrm{0}}{\frac{{a}}{\mathrm{4}}−\left(−{a}\right)}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{5}}\:. \\ $$

Commented by math solver last updated on 25/Oct/17

$${is}\:{L}\:{mid}\:{point}\:{of}\:{OD}? \\ $$

Commented by math solver last updated on 25/Oct/17

isn′t E(−0.5,1) and N(0.5,0) when a=1???

$${isn}'{t}\:{E}\left(−\mathrm{0}.\mathrm{5},\mathrm{1}\right)\:{and}\:{N}\left(\mathrm{0}.\mathrm{5},\mathrm{0}\right)\:{when}\: \\ $$$${a}=\mathrm{1}??? \\ $$

Commented by ajfour last updated on 26/Oct/17