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-connec

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

$$Forthosewhoareinterestedin$$$$Geometry:$$$$Atrianglehasanareaof1unit.Each$$$$ofitssidesisdividedinto4equalparts$$$$through3points.Thefirstandthelast$$$$pointofeachsidewillbeconnected$$$$witheachothertoform2inscribed$$$$trianglesandthese2trianglesform$$$$ahexagon.Findtheareaofthehexagon.$$$$$$$$Whatistheresult,ifeachsideis$$$$equallydividedinto5parts,or$$$$generallyintonparts?$$

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

$$yougotX=\frac{n(n-1)-3}{n(n-1)}$$$$andX=\frac{1}{2}byn=3$$$$buttherightanswerisX=2/9byn=3$$

Commented by mrW1 last updated on 06/Jun/17

$$errorinyourworking\left(Ithink\right):$$$$2(S-X)\ne \frac{6S}{n(n-1)}\Rightarrow X\ne S-\frac{3S}{n(n-1)}$$

Commented by ajfour last updated on 06/Jun/17

$$coulditbe=\frac{49}{160}?$$

Commented by mrW1 last updated on 06/Jun/17

$$\frac{49}{160}forn=4iscorrect,congratulation!$$

Commented by ajfour last updated on 06/Jun/17

$$thankyouSir,thisquestionof$$$$yoursletmelearntriangular$$$$coordinatesallbymyself..$$

Commented by mrW1 last updated on 06/Jun/17

$$Idevisedthisquestion.Thatyougot$$$$therightanswerinsuchashorttime$$$$showsthatyouhavelearntquitealot.$$

Commented by mrW1 last updated on 06/Jun/17

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

Commented by mrW1 last updated on 06/Jun/17

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

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

$$S_1=\frac{1}{2}.\frac{b}{n}.\frac{n-1}{n}.c.sinA=\frac{n-1}{n^2}.S$$$$S_1^{{}^{\prime }}=\frac{1}{2}.\frac{b^{\prime }}{3}.\frac{c^{\prime }}{3}.{sinA}^{\prime }=\frac{S^{\prime }}{9}$$$$S^{\prime }=S-3\times \frac{n-1}{n^2}S=\frac{n^2-3n+3}{n^2}S$$$$S_1^{\prime }=\frac{S^{\prime }}{9}=\frac{n^2-3n+3}{9n^2}$$$$S_{hexagon}=S^{\prime }-3S_1^{\prime }=(\frac{n^2-3n+3}{n^2}-3\times \frac{n^2-3n+3}{9n^2})S=$$$$=\frac{2(n^2-3n+3)}{3n^2}.S$$$$for:n=3\Rightarrow S_h=\frac{2\times 3}{3\times 9}.S=\frac{2}{9}.S$$

Commented by mrW1 last updated on 07/Jun/17

$$\mathrm{please}\mathrm{check}\mathrm{again}:$$$$\mathrm{for}\mathrm{n}=4,\mathrm{the}\mathrm{right}\mathrm{answer}\mathrm{should}\mathrm{be}$$$${\mathrm{S}}_{\mathrm{h}}=\frac{49}{160}\mathrm{S}(\mathrm{as}\mathrm{Mr}.\mathrm{ajfour}\mathrm{evaluated})$$$$$$$$\mathrm{I}\mathrm{think}\mathrm{there}\mathrm{is}\mathrm{an}\mathrm{error}\mathrm{here}:$$$$S_1^{\prime }\ne \frac{S^{\prime }}{9}(\mathrm{this}\mathrm{is}\mathrm{only}\mathrm{true}\mathrm{for}\mathrm{the}\mathrm{case}\mathrm{n}=3)$$

Answered by ajfour last updated on 07/Jun/17

$$\frac{{Area}_{hexagon}}{{Area}_{△}}=\frac{2{({\mathit{n}}^2-3\mathit{n}+3)}^2}{{\mathit{n}}^3(2\mathit{n}-3)}.$$

Commented by mrW1 last updated on 07/Jun/17

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

Commented by mrW1 last updated on 07/Jun/17

$$\mathrm{x}=0.222\mathrm{for}\mathrm{n}=3$$$$\mathrm{x}=0.306\mathrm{for}\mathrm{n}=4$$$$\mathrm{x}\to 1\mathrm{when}\mathrm{n}\to \mathrm{\infty }$$$$\mathrm{the}\mathrm{hexagon}\mathrm{gets}\mathrm{bigger}\mathrm{and}\mathrm{bigger}$$$$\mathrm{with}\mathrm{increasing}\mathrm{n}$$

Answered by mrW1 last updated on 08/Jun/17

$$\mathrm{Here}\mathrm{my}\mathrm{solution}\mathrm{without}\mathrm{usage}\mathrm{of}$$$$\mathrm{trigonometry}\mathrm{or}\mathrm{analytic}\mathrm{geometry}.$$$$$$$$\mathrm{The}\mathrm{most}\mathrm{important}\mathrm{rule}\mathrm{which}\mathrm{I}\mathrm{use}$$$$\mathrm{is}\mathrm{shown}\mathrm{in}\mathrm{the}\mathrm{next}\mathrm{diagram}.$$$$$$$$\mathrm{Area}\mathrm{of}\mathrm{\Delta }\mathrm{ABC}=\frac{1}{2}{\mathrm{ah}}_{\mathrm{A}}=\frac{1}{2}\mathrm{a}\mathrm{b}\sin \mathrm{C}$$$$\mathrm{Area}\mathrm{of}\mathrm{\Delta }123=\frac{1}{2}\alpha {\mathrm{ah}}_1=\frac{1}{2}\alpha \mathrm{a}\beta \mathrm{b}\sin \mathrm{C}$$$$\Rightarrow \frac{{\mathrm{A}}_{\mathrm{\Delta }123}}{{\mathrm{A}}_{\mathrm{\Delta }\mathrm{ABC}}}=\alpha \beta$$

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

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

Commented by mrW1 last updated on 08/Jun/17

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

Commented by mrW1 last updated on 08/Jun/17

Commented by mrW1 last updated on 08/Jun/17

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

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

$$GreatjobdonemrW1.fantastic.$$$$Ilovethisproofverymuch.$$

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

$$afterthis,iamwaitingformr$$$${Ajfour}^{\prime }sprooftoo.$$

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