Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 14940 by mrW1 last updated on 05/Jun/17

For those who are interested in   Geometry:   A triangle has an area of 1 unit. Each  of its sides is divided into 4 equal parts  through 3 points. The first and the last  point of each side will be connected  with each other to form 2 inscribed  triangles and these 2 triangles form  a hexagon. Find the area of the hexagon.    What is the result, if each side is  equally divided into 5 parts, or  generally into n parts?

$${For}\:{those}\:{who}\:{are}\:{interested}\:{in}\: \\ $$$${Geometry}:\: \\ $$$${A}\:{triangle}\:{has}\:{an}\:{area}\:{of}\:\mathrm{1}\:{unit}.\:{Each} \\ $$$${of}\:{its}\:{sides}\:{is}\:{divided}\:{into}\:\mathrm{4}\:{equal}\:{parts} \\ $$$${through}\:\mathrm{3}\:{points}.\:{The}\:{first}\:{and}\:{the}\:{last} \\ $$$${point}\:{of}\:{each}\:{side}\:{will}\:{be}\:{connected} \\ $$$${with}\:{each}\:{other}\:{to}\:{form}\:\mathrm{2}\:{inscribed} \\ $$$${triangles}\:{and}\:{these}\:\mathrm{2}\:{triangles}\:{form} \\ $$$${a}\:{hexagon}.\:{Find}\:{the}\:{area}\:{of}\:{the}\:{hexagon}. \\ $$$$ \\ $$$${What}\:{is}\:{the}\:{result},\:{if}\:{each}\:{side}\:{is} \\ $$$${equally}\:{divided}\:{into}\:\mathrm{5}\:{parts},\:{or} \\ $$$${generally}\:{into}\:{n}\:{parts}? \\ $$

Commented by mrW1 last updated on 05/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/Jun/17

Commented by mrW1 last updated on 06/Jun/17

you got X=((n(n−1)−3)/(n(n−1)))  and X=(1/2) by n=3  but the right answer is X=2/9 by n=3

$${you}\:{got}\:{X}=\frac{{n}\left({n}−\mathrm{1}\right)−\mathrm{3}}{{n}\left({n}−\mathrm{1}\right)} \\ $$$${and}\:{X}=\frac{\mathrm{1}}{\mathrm{2}}\:{by}\:{n}=\mathrm{3} \\ $$$${but}\:{the}\:{right}\:{answer}\:{is}\:{X}=\mathrm{2}/\mathrm{9}\:{by}\:{n}=\mathrm{3} \\ $$

Commented by mrW1 last updated on 06/Jun/17

error in your working (I think):  2(S−X)≠((6S)/(n(n−1)))⇒X≠S−((3S)/(n(n−1)))

$${error}\:{in}\:{your}\:{working}\:\left({I}\:{think}\right): \\ $$$$\mathrm{2}\left({S}−{X}\right)\neq\frac{\mathrm{6}{S}}{{n}\left({n}−\mathrm{1}\right)}\Rightarrow{X}\neq{S}−\frac{\mathrm{3}{S}}{{n}\left({n}−\mathrm{1}\right)} \\ $$

Commented by ajfour last updated on 06/Jun/17

could it be =((49)/(160)) ?

$${could}\:{it}\:{be}\:=\frac{\mathrm{49}}{\mathrm{160}}\:? \\ $$

Commented by mrW1 last updated on 06/Jun/17

((49)/(160)) for n=4 is correct, congratulation!

$$\frac{\mathrm{49}}{\mathrm{160}}\:{for}\:{n}=\mathrm{4}\:{is}\:{correct},\:{congratulation}! \\ $$

Commented by ajfour last updated on 06/Jun/17

thank you Sir, this question of  yours let me learn triangular  coordinates all by myself..

$${thank}\:{you}\:{Sir},\:{this}\:{question}\:{of} \\ $$$${yours}\:{let}\:{me}\:{learn}\:{triangular} \\ $$$${coordinates}\:{all}\:{by}\:{myself}.. \\ $$

Commented by mrW1 last updated on 06/Jun/17

I devised this question. That you got  the right answer in such a short time  shows that you have learnt quite a lot.

$${I}\:{devised}\:{this}\:{question}.\:{That}\:{you}\:{got} \\ $$$${the}\:{right}\:{answer}\:{in}\:{such}\:{a}\:{short}\:{time} \\ $$$${shows}\:{that}\:{you}\:{have}\:{learnt}\:{quite}\:{a}\:{lot}. \\ $$

Commented by mrW1 last updated on 06/Jun/17

This question can be solved with basic  geometry knowledge. It is not necessary  to use trigonometry or analytic geometry.

$$\mathrm{This}\:\mathrm{question}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{with}\:\mathrm{basic} \\ $$$$\mathrm{geometry}\:\mathrm{knowledge}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{not}\:\mathrm{necessary} \\ $$$$\mathrm{to}\:\mathrm{use}\:\mathrm{trigonometry}\:\mathrm{or}\:\mathrm{analytic}\:\mathrm{geometry}. \\ $$

Commented by mrW1 last updated on 06/Jun/17

To Behi′s father: you are very close at  the final solution. I′m sure you′ll be able  to get the general solution for case n.

$$\mathrm{To}\:\mathrm{Behi}'\mathrm{s}\:\mathrm{father}:\:\mathrm{you}\:\mathrm{are}\:\mathrm{very}\:\mathrm{close}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{final}\:\mathrm{solution}.\:\mathrm{I}'\mathrm{m}\:\mathrm{sure}\:\mathrm{you}'\mathrm{ll}\:\mathrm{be}\:\mathrm{able} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{case}\:\mathrm{n}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Jun/17

S_1 =(1/2).(b/n).((n−1)/n).c.sinA=((n−1)/n^2 ).S  S_1 ^′ =(1/2).((b′)/3).((c′)/3).sinA′=((S′)/9)  S′=S−3×((n−1)/n^2 )S=((n^2 −3n+3)/n^2 )S  S′_1 =((S′)/9)=((n^2 −3n+3)/(9n^2 ))  S_(hexagon) =S′−3S′_1 =(((n^2 −3n+3)/n^2 )−3×((n^2 −3n+3)/(9n^2 )))S=  =((2(n^2 −3n+3))/(3n^2 )).S  for:n=3⇒S_h =((2×3)/(3×9)).S=(2/9).S

$${S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}.\frac{{b}}{{n}}.\frac{{n}−\mathrm{1}}{{n}}.{c}.{sinA}=\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} }.{S} \\ $$$${S}_{\mathrm{1}} ^{'} =\frac{\mathrm{1}}{\mathrm{2}}.\frac{{b}'}{\mathrm{3}}.\frac{{c}'}{\mathrm{3}}.{sinA}'=\frac{{S}'}{\mathrm{9}} \\ $$$${S}'={S}−\mathrm{3}×\frac{{n}−\mathrm{1}}{{n}^{\mathrm{2}} }{S}=\frac{{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} }{S} \\ $$$${S}'_{\mathrm{1}} =\frac{{S}'}{\mathrm{9}}=\frac{{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{3}}{\mathrm{9}{n}^{\mathrm{2}} } \\ $$$${S}_{{hexagon}} ={S}'−\mathrm{3}{S}'_{\mathrm{1}} =\left(\frac{{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} }−\mathrm{3}×\frac{{n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{3}}{\mathrm{9}{n}^{\mathrm{2}} }\right){S}= \\ $$$$=\frac{\mathrm{2}\left({n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{3}\right)}{\mathrm{3}{n}^{\mathrm{2}} }.{S} \\ $$$${for}:{n}=\mathrm{3}\Rightarrow{S}_{{h}} =\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}×\mathrm{9}}.{S}=\frac{\mathrm{2}}{\mathrm{9}}.{S} \\ $$

Commented by mrW1 last updated on 07/Jun/17

please check again:  for n=4, the right answer should be  S_h =((49)/(160))S (as Mr. ajfour evaluated)    I think there is an error here:  S′_1 ≠((S′)/9) (this is only true for the case n=3)

$$\mathrm{please}\:\mathrm{check}\:\mathrm{again}: \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{4},\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{S}_{\mathrm{h}} =\frac{\mathrm{49}}{\mathrm{160}}\mathrm{S}\:\left(\mathrm{as}\:\mathrm{Mr}.\:\mathrm{ajfour}\:\mathrm{evaluated}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{an}\:\mathrm{error}\:\mathrm{here}: \\ $$$${S}'_{\mathrm{1}} \neq\frac{{S}'}{\mathrm{9}}\:\left(\mathrm{this}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for}\:\mathrm{the}\:\mathrm{case}\:\mathrm{n}=\mathrm{3}\right) \\ $$

Answered by ajfour last updated on 07/Jun/17

 ((Area_(hexagon) )/(Area_△ )) =((2(n^2 −3n+3)^2 )/(n^3 (2n−3))) .

$$\:\frac{{Area}_{{hexagon}} }{{Area}_{\bigtriangleup} }\:=\frac{\mathrm{2}\left(\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{n}}+\mathrm{3}\right)^{\mathrm{2}} }{\boldsymbol{{n}}^{\mathrm{3}} \left(\mathrm{2}\boldsymbol{{n}}−\mathrm{3}\right)}\:. \\ $$

Commented by mrW1 last updated on 07/Jun/17

your answer is correct and you have  used a different way as I originally  expected. you applied analytic geometry.  this is nice and easy, because you  don′t need any tricks.  but as I said one can also get the result  without analytic geometry or trigonometry.

$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{and}\:\mathrm{you}\:\mathrm{have} \\ $$$$\mathrm{used}\:\mathrm{a}\:\mathrm{different}\:\mathrm{way}\:\mathrm{as}\:\mathrm{I}\:\mathrm{originally} \\ $$$$\mathrm{expected}.\:\mathrm{you}\:\mathrm{applied}\:\mathrm{analytic}\:\mathrm{geometry}. \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{nice}\:\mathrm{and}\:\mathrm{easy},\:\mathrm{because}\:\mathrm{you} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{any}\:\mathrm{tricks}. \\ $$$$\mathrm{but}\:\mathrm{as}\:\mathrm{I}\:\mathrm{said}\:\mathrm{one}\:\mathrm{can}\:\mathrm{also}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result} \\ $$$$\mathrm{without}\:\mathrm{analytic}\:\mathrm{geometry}\:\mathrm{or}\:\mathrm{trigonometry}. \\ $$

Commented by mrW1 last updated on 07/Jun/17

x=0.222 for n=3  x=0.306 for n=4  x→1 when n→∞  the hexagon gets bigger and bigger  with increasing n

$$\mathrm{x}=\mathrm{0}.\mathrm{222}\:\mathrm{for}\:\mathrm{n}=\mathrm{3} \\ $$$$\mathrm{x}=\mathrm{0}.\mathrm{306}\:\mathrm{for}\:\mathrm{n}=\mathrm{4} \\ $$$$\mathrm{x}\rightarrow\mathrm{1}\:\mathrm{when}\:\mathrm{n}\rightarrow\infty \\ $$$$\mathrm{the}\:\mathrm{hexagon}\:\mathrm{gets}\:\mathrm{bigger}\:\mathrm{and}\:\mathrm{bigger} \\ $$$$\mathrm{with}\:\mathrm{increasing}\:\mathrm{n} \\ $$

Answered by mrW1 last updated on 08/Jun/17

Here my solution without usage of  trigonometry or analytic geometry.    The most important rule which I use  is shown in the next diagram.     Area of ΔABC=(1/2)ah_A =(1/2) a b sin C  Area of Δ123=(1/2)αah_1 =(1/2) αa βb sin C  ⇒(A_(Δ123) /A_(ΔABC) )=αβ

$$\mathrm{Here}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{without}\:\mathrm{usage}\:\mathrm{of} \\ $$$$\mathrm{trigonometry}\:\mathrm{or}\:\mathrm{analytic}\:\mathrm{geometry}. \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{most}\:\mathrm{important}\:\mathrm{rule}\:\mathrm{which}\:\mathrm{I}\:\mathrm{use} \\ $$$$\mathrm{is}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{next}\:\mathrm{diagram}.\: \\ $$$$ \\ $$$$\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{ABC}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ah}_{\mathrm{A}} =\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{a}\:\mathrm{b}\:\mathrm{sin}\:\mathrm{C} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{123}=\frac{\mathrm{1}}{\mathrm{2}}\alpha\mathrm{ah}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:\alpha\mathrm{a}\:\beta\mathrm{b}\:\mathrm{sin}\:\mathrm{C} \\ $$$$\Rightarrow\frac{\mathrm{A}_{\Delta\mathrm{123}} }{\mathrm{A}_{\Delta\mathrm{ABC}} }=\alpha\beta \\ $$

Commented by mrW1 last updated on 08/Jun/17

Commented by mrW1 last updated on 08/Jun/17

Commented by mrW1 last updated on 08/Jun/17

Each side of triangle ΔILG is equally  divided into n parts. n≥3.  Let a=((LG)/n)=length of each part of LG  Let us say the area of triangle ΔILG  is S=1.    According to the rule above we have  A_(ΔACG) =((n−1)/n)×(1/n)×S=((n−1)/n^2 )  =A_(ΔCHI) =A_(ΔNAL)   =A_(ΔBML) =A_(ΔMHI) =A_(ΔHBG)     ⇒A_(ΔMHB) =A_(ΔAHG) =S−3×A_(ΔACG)   =1−3×((n−1)/n^2 )=((n^2 −3n+3)/n^2 )  To get the area of the hexagon we now  only need to get the area of the “red”  as well as the “black” small triangles.

$$\mathrm{Each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle}\:\Delta\mathrm{ILG}\:\mathrm{is}\:\mathrm{equally} \\ $$$$\mathrm{divided}\:\mathrm{into}\:\mathrm{n}\:\mathrm{parts}.\:\mathrm{n}\geqslant\mathrm{3}. \\ $$$$\mathrm{Let}\:\mathrm{a}=\frac{\mathrm{LG}}{\mathrm{n}}=\mathrm{length}\:\mathrm{of}\:\mathrm{each}\:\mathrm{part}\:\mathrm{of}\:\mathrm{LG} \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:\Delta\mathrm{ILG} \\ $$$$\mathrm{is}\:\mathrm{S}=\mathrm{1}. \\ $$$$ \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rule}\:\mathrm{above}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{A}_{\Delta\mathrm{ACG}} =\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}×\frac{\mathrm{1}}{\mathrm{n}}×\mathrm{S}=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$=\mathrm{A}_{\Delta\mathrm{CHI}} =\mathrm{A}_{\Delta\mathrm{NAL}} \\ $$$$=\mathrm{A}_{\Delta\mathrm{BML}} =\mathrm{A}_{\Delta\mathrm{MHI}} =\mathrm{A}_{\Delta\mathrm{HBG}} \\ $$$$ \\ $$$$\Rightarrow\mathrm{A}_{\Delta\mathrm{MHB}} =\mathrm{A}_{\Delta\mathrm{AHG}} =\mathrm{S}−\mathrm{3}×\mathrm{A}_{\Delta\mathrm{ACG}} \\ $$$$=\mathrm{1}−\mathrm{3}×\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{we}\:\mathrm{now} \\ $$$$\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:``\mathrm{red}'' \\ $$$$\mathrm{as}\:\mathrm{well}\:\mathrm{as}\:\mathrm{the}\:``\mathrm{black}''\:\mathrm{small}\:\mathrm{triangles}. \\ $$

Commented by mrW1 last updated on 08/Jun/17

Using the rule above we have  A_(ΔACB) =((n−2)/n)×(1/n)×S=((n−2)/n^2 )  To get the area of A_(ΔDEB)  we need to  know how the line segment AC is  divided by points D and E.

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{rule}\:\mathrm{above}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{A}_{\Delta\mathrm{ACB}} =\frac{\mathrm{n}−\mathrm{2}}{\mathrm{n}}×\frac{\mathrm{1}}{\mathrm{n}}×\mathrm{S}=\frac{\mathrm{n}−\mathrm{2}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{To}\:\mathrm{get}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{A}_{\Delta\mathrm{DEB}} \:\mathrm{we}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{know}\:\mathrm{how}\:\mathrm{the}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{AC}\:\mathrm{is} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{points}\:\mathrm{D}\:\mathrm{and}\:\mathrm{E}. \\ $$

Commented by mrW1 last updated on 08/Jun/17

Commented by mrW1 last updated on 08/Jun/17

We introduce some auxiliary lines:  CK parallel to IF  CJ parallel to HB    ΔCJG is similar to ΔHBG  ⇒((JG)/(BG))=((CG)/(HG))  ⇒JG=((CG)/(HG))×BG=(1/(n−1))×BG=(a/(n−1))    ΔKCG is similar to ΔFIG  ⇒((KG)/(FG))=((CG)/(IG))  ⇒KG=((CG)/(IG))×FG=(1/n)×((LG)/2)=(1/n)×((na)/2)=(a/2)    ΔADF is similar to ΔACK  ⇒((AD)/(AF))=((AC)/(AK))  ⇒AD=((AF)/(AK))×AC=((LF−LA)/(AG−KG))×AC  =((((na)/2)−a)/(na−a−(a/2)))×AC=((n−2)/(2n−3))×AC    ΔAEB is similar to ΔACJ  ⇒((AE)/(AC))=((AB)/(AJ))  ⇒AE=((AB)/(AJ))×AC=((AB)/(AG−JG))×AC  =((na−2a)/(na−a−(a/(n−1))))×AC=((n−1)/n)×AC    DE=AE−AD=((n−1)/n)×AC−((n−2)/(2n−3))×AC  =(((n−1)/n)−((n−2)/(2n−3)))×AC  =(((n−1)(2n−3)−n(n−2))/(n(2n−3)))×AC  =((n^2 −3n+3)/(n(2n−3)))×AC    Using our rule abov we get  A_(ΔDEB) =((DE)/(AC))×A_(ΔACB)   =((n^2 −3n+3)/(n(2n−3)))×((n−2)/n^2 )=(((n^2 −3n+3)(n−2))/(n^3 (2n−3)))    Area of hexagon:  A_H =A_(ΔMHB) −3×A_(ΔDEB)   A_H =((n^2 −3n+3)/n^2 )−3×(((n^2 −3n+3)(n−2))/(n^3 (2n−3)))  =((n^2 −3n+3)/n^2 )[1−((3(n−2))/(n(2n−3)))]  =((n^2 −3n+3)/n^2 )[((n(2n−3)−3(n−2))/(n(2n−3)))]  =((n^2 −3n+3)/n^2 )[((2n^2 −6n+6)/(n(2n−3)))]  =((2(n^2 −3n+3)^2 )/(n^3 (2n−3)))

$$\mathrm{We}\:\mathrm{introduce}\:\mathrm{some}\:\mathrm{auxiliary}\:\mathrm{lines}: \\ $$$$\mathrm{CK}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{IF} \\ $$$$\mathrm{CJ}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{HB} \\ $$$$ \\ $$$$\Delta\mathrm{CJG}\:\mathrm{is}\:\mathrm{similar}\:\mathrm{to}\:\Delta\mathrm{HBG} \\ $$$$\Rightarrow\frac{\mathrm{JG}}{\mathrm{BG}}=\frac{\mathrm{CG}}{\mathrm{HG}} \\ $$$$\Rightarrow\mathrm{JG}=\frac{\mathrm{CG}}{\mathrm{HG}}×\mathrm{BG}=\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}×\mathrm{BG}=\frac{\mathrm{a}}{\mathrm{n}−\mathrm{1}} \\ $$$$ \\ $$$$\Delta\mathrm{KCG}\:\mathrm{is}\:\mathrm{similar}\:\mathrm{to}\:\Delta\mathrm{FIG} \\ $$$$\Rightarrow\frac{\mathrm{KG}}{\mathrm{FG}}=\frac{\mathrm{CG}}{\mathrm{IG}} \\ $$$$\Rightarrow\mathrm{KG}=\frac{\mathrm{CG}}{\mathrm{IG}}×\mathrm{FG}=\frac{\mathrm{1}}{\mathrm{n}}×\frac{\mathrm{LG}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{n}}×\frac{\mathrm{na}}{\mathrm{2}}=\frac{\mathrm{a}}{\mathrm{2}} \\ $$$$ \\ $$$$\Delta\mathrm{ADF}\:\mathrm{is}\:\mathrm{similar}\:\mathrm{to}\:\Delta\mathrm{ACK} \\ $$$$\Rightarrow\frac{\mathrm{AD}}{\mathrm{AF}}=\frac{\mathrm{AC}}{\mathrm{AK}} \\ $$$$\Rightarrow\mathrm{AD}=\frac{\mathrm{AF}}{\mathrm{AK}}×\mathrm{AC}=\frac{\mathrm{LF}−\mathrm{LA}}{\mathrm{AG}−\mathrm{KG}}×\mathrm{AC} \\ $$$$=\frac{\frac{\mathrm{na}}{\mathrm{2}}−\mathrm{a}}{\mathrm{na}−\mathrm{a}−\frac{\mathrm{a}}{\mathrm{2}}}×\mathrm{AC}=\frac{\mathrm{n}−\mathrm{2}}{\mathrm{2n}−\mathrm{3}}×\mathrm{AC} \\ $$$$ \\ $$$$\Delta\mathrm{AEB}\:\mathrm{is}\:\mathrm{similar}\:\mathrm{to}\:\Delta\mathrm{ACJ} \\ $$$$\Rightarrow\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{AB}}{\mathrm{AJ}} \\ $$$$\Rightarrow\mathrm{AE}=\frac{\mathrm{AB}}{\mathrm{AJ}}×\mathrm{AC}=\frac{\mathrm{AB}}{\mathrm{AG}−\mathrm{JG}}×\mathrm{AC} \\ $$$$=\frac{\mathrm{na}−\mathrm{2a}}{\mathrm{na}−\mathrm{a}−\frac{\mathrm{a}}{\mathrm{n}−\mathrm{1}}}×\mathrm{AC}=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}×\mathrm{AC} \\ $$$$ \\ $$$$\mathrm{DE}=\mathrm{AE}−\mathrm{AD}=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}×\mathrm{AC}−\frac{\mathrm{n}−\mathrm{2}}{\mathrm{2n}−\mathrm{3}}×\mathrm{AC} \\ $$$$=\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}−\frac{\mathrm{n}−\mathrm{2}}{\mathrm{2n}−\mathrm{3}}\right)×\mathrm{AC} \\ $$$$=\frac{\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{2n}−\mathrm{3}\right)−\mathrm{n}\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}×\mathrm{AC} \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}×\mathrm{AC} \\ $$$$ \\ $$$$\mathrm{Using}\:\mathrm{our}\:\mathrm{rule}\:\mathrm{abov}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}_{\Delta\mathrm{DEB}} =\frac{\mathrm{DE}}{\mathrm{AC}}×\mathrm{A}_{\Delta\mathrm{ACB}} \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}×\frac{\mathrm{n}−\mathrm{2}}{\mathrm{n}^{\mathrm{2}} }=\frac{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2n}−\mathrm{3}\right)} \\ $$$$ \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{hexagon}: \\ $$$$\mathrm{A}_{\mathrm{H}} =\mathrm{A}_{\Delta\mathrm{MHB}} −\mathrm{3}×\mathrm{A}_{\Delta\mathrm{DEB}} \\ $$$$\mathrm{A}_{\mathrm{H}} =\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }−\mathrm{3}×\frac{\left(\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2n}−\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }\left[\mathrm{1}−\frac{\mathrm{3}\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}\right] \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }\left[\frac{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)−\mathrm{3}\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}\right] \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}}{\mathrm{n}^{\mathrm{2}} }\left[\frac{\mathrm{2n}^{\mathrm{2}} −\mathrm{6n}+\mathrm{6}}{\mathrm{n}\left(\mathrm{2n}−\mathrm{3}\right)}\right] \\ $$$$=\frac{\mathrm{2}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{3n}+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} \left(\mathrm{2n}−\mathrm{3}\right)} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

Great job done mrW1.fantastic.  I love this proof very much.

$${Great}\:{job}\:{done}\:{mrW}\mathrm{1}.{fantastic}. \\ $$$${I}\:{love}\:{this}\:{proof}\:{very}\:{much}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

after this ,i am waiting for mr  Ajfour′s proof too.

$${after}\:{this}\:,{i}\:{am}\:{waiting}\:{for}\:{mr} \\ $$$${Ajfour}'{s}\:{proof}\:{too}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com