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Question Number 3324 by prakash jain last updated on 10/Dec/15

$$\mathrm{Prove}\mathrm{that}\mathrm{among}\mathrm{all}\mathrm{triangle}\mathrm{of}\mathrm{equal}$$$$\mathrm{perimeter},\mathrm{equilateral}\mathrm{triangle}\mathrm{has}\mathrm{the}$$$$\mathrm{largest}\mathrm{area}.$$

Commented by 123456 last updated on 10/Dec/15

$$2s(x,y,z)=x+y+z$$$$\mathrm{A}(x,y,z)=\sqrt{s(s-a)(s-b)(s-c)}$$

Commented by Rasheed Soomro last updated on 11/Dec/15

$$Letx,x+landx+l+maremeasuresofthreesides$$$$wherel,m⩾0$$$$Perimeter=3x+2l+m$$$$S=\frac{3x+2l+m}{2}$$$$\mathcal{I}n\mathcal{E}quilateraltriangleS=\frac{3x}{2}forl,m=0$$$$A_{△}=\sqrt{\frac{3x}{2}{(\frac{3x}{2}-x)}^3}=\sqrt{\frac{3x}{2}\times \frac{x^3}{2^3}}$$$$=\sqrt{\frac{3x^4}{2^4}}=\frac{\sqrt{3}{x}^2}{4}$$$$AsSisconstant(becauseperimeterisconstant$$$$HenceforanytriangleS$$$$S=\frac{3x}{2}$$$$Areaofanytriangle$$$$A=\sqrt{\frac{3x}{2}(\frac{3x}{2}-x)(\frac{3x}{2}-x-l)(\frac{3x}{2}-x-l-m)}$$$$=\sqrt{\frac{3x}{2}\times \frac{x}{2}\times \frac{x-2l}{2}\times \frac{x-2l-2m}{2}}$$$$=\frac{\sqrt{3x^2(x-2l)(x-2l-2m)}}{4}$$$$Wehavetoprove$$$$A_{△}⩾A$$$$\frac{\sqrt{3}{x}^2}{4}⩾\frac{\sqrt{3x^2(x-2l)(x-2l-2m)}}{4}forl,m⩾0$$$$C^{\mathcal{ONTINUE}}$$

Commented by prakash jain last updated on 11/Dec/15

$$\mathrm{From}\mathrm{your}\mathrm{derivation}$$$$\mathrm{In}\mathrm{your}\mathrm{formula}\mathrm{for}\mathrm{area}$$$$a=x,b=x-l,c=x-l-m$$$$hows=\frac{3x}{2}$$

Commented by Rasheed Soomro last updated on 12/Dec/15

$$ForequilateraltriangleofsidexS=\frac{3x}{2}$$$$andIhaveusedthisalsoincaseofcommon$$$$triangle,becausetheperimeterisconstant.$$$$\mathcal{B}utnowIhavechangedmymind.Iamworking$$$$ondifferentway.Plseemyanswerbelow.$$$$Iamalways\mathbf{thankful}toyouforguidance.$$

Answered by Rasheed Soomro last updated on 13/Dec/15

$$\mathcal{A}reaof\mathcal{C}ommon\mathcal{T}riangle$$$$Letthemeasuresofsidesofacommontriangleare$$$$x,y,z$$$$\mathcal{A}rea\left(A\right)willbe$$$$A=\sqrt{S(S-x)(S-y)(S-x)};S=\frac{1}{2}(x+y+z)$$$$=\sqrt{\frac{x+y+z}{2}(\frac{x+y+z}{2}-x)(\frac{x+y+z}{2}-y)(\frac{x+y+z}{2}-z)}$$$$=\sqrt{\left(\frac{x+y+z}{2}\right)\left(\frac{x+y+z-2x}{2}\right)\left(\frac{x+y+z-2y}{2}\right)\left(\frac{x+y+z-2z}{2}\right)}$$$$=\sqrt{\left(\frac{x+y+z}{2}\right)\left(\frac{-x+y+z}{2}\right)\left(\frac{x-y+z}{2}\right)\left(\frac{x+y-z}{2}\right)}$$$$=\frac{1}{4}\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$$$$\mathcal{A}reaof\mathcal{E}quilateral\mathcal{T}riangle$$$$\therefore Everysideofofanequilateraltrianglehaving$$$$sameperimeter(x+y+z)is\frac{x+y+z}{3}$$$$\therefore A_{△}=\sqrt{S{(S-\frac{x+y+z}{3})}^3}$$$$=\sqrt{\frac{x+y+z}{2}\times {(\frac{x+y+z}{2}-\frac{x+y+z}{3})}^3}$$$$=\sqrt{\frac{x+y+z}{2}\times {\left(\frac{3x+3y+3z-2x-2y-2z}{6}\right)}^3}$$$$=\sqrt{\frac{x+y+z}{2}\times {\left(\frac{x+y+z}{6}\right)}^3}$$$$=\sqrt{\frac{{(x+y+z)}^4}{2^4\times 3^2\times 3}}=\frac{{(x+y+z)}^2}{12}\times \frac{1}{\sqrt{3}}$$$$A_{△}=\frac{{(x+y+z)}^2}{12\sqrt{3}}$$$$-----<><><><>-----$$$$Nowwehavetoprove:$$$$A_{△}⩾A$$$$\frac{{(x+y+z)}^2}{12\sqrt{3}}⩾\frac{1}{4}\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$$$$forx,y,z>0$$$$\Rightarrow \frac{{(x+y+z)}^2}{3\sqrt{3}}⩾\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$$$${\mathbf{Yozzi}}^{\prime }s\mathbf{contribution}:$$$$Yozzihasprovedthat$$$$\Rightarrow \frac{{(x+y+z)}^2}{4}⩾\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$$$$(SeeanswerbyYozzi.AnapproachI{can}^{\prime }teven$$$$thinkof!)$$$$Nowsince4<3\sqrt{3}$$$$\therefore \frac{{(x+y+z)}^2}{3\sqrt{3}}⩾\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$$$$\therefore A_{△}⩾A$$$$QED$$

Commented by Rasheed Soomro last updated on 14/Dec/15

$$\frac{{(x+y+z)}^2}{4}⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}$$$$itcannotsolelyleadonetosafelystating$$$$that\frac{{(x+y+z)}^2}{3\sqrt{3}}⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}.$$$$⋮$$$$Whytransitivepropertyofinequality{doesn}^{\prime }t$$$$help?$$$$IfwehavetoprovethatA>Bandwehave$$$$bysomeway\frac{A}{2}>B$$$$\frac{A}{2}>B\Rightarrow A>B?$$$$$$

Commented by Yozzii last updated on 13/Dec/15

$$If4<3\sqrt{3}\Rightarrow \frac{1}{4}>\frac{1}{3\sqrt{3}}\Rightarrow \frac{{(x+y+z)}^2}{4}>\frac{{(x+y+z)}^2}{3\sqrt{3}}$$$$Now,wereally{weren}^{\prime }tgiventhat$$$$A_{△}⩾A.Evenifweweretocheck$$$$thatthisisthecase,wewouldhave$$$$todecomposetheinequalitytoa$$$$formthattheinformationwehave$$$$aboutthesituationissatisfies.The$$$$AM-GMsadly{didn}^{\prime }thelpasIhad$$$$hopeditwouldbecausewhileit$$$$showsthat$$$$\frac{{(x+y+z)}^2}{4}⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}$$$$itcannotsolelyleadonetosafelystating$$$$that\frac{{(x+y+z)}^2}{3\sqrt{3}}⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}.$$$$$$$$Thetransitivepropertyofinequalities$$$${wouldn}^{\prime }thaveworkedforusinthis$$$$approach:(.$$$$$$$$$$$$$$

Commented by Yozzii last updated on 14/Dec/15

$$Saywehavethefollowingproblem.$$$$---------------------$$$$Wewanttoshowp>q.Wehavethat$$$$r>qandr=npwheren,p,q\in {\mathbb{R}}^{+}$$$$\therefore np>q\Rightarrow p>\frac{1}{n}q.q=\sqrt{\underset{r=1}{\prod ^4}{a}_r}.$$$$$$$$Letp=\frac{u}{3\sqrt{3}}andr=\frac{u}{4},u={(x+y+z)}^2.$$$$u=3p\sqrt{3}\Rightarrow r=\frac{3p\sqrt{3}}{4}\Rightarrow n=\frac{3\sqrt{3}}{4}.$$$$\therefore p>\frac{4}{3\sqrt{3}}q$$$$\frac{16}{27}={\left(\frac{4}{3\sqrt{3}}\right)}^2\Rightarrow 0<\frac{4}{3\sqrt{3}}<1.$$$$\therefore Theinformationonlyshowsthat$$$$p>\frac{4}{3\sqrt{3}}qandnotp>qsince\frac{4}{3\sqrt{3}}q<q.$$$$Onanumberlinewehavethreecases$$$$duetothisresult:$$$$\left(i\right)\overrightarrow{0}...............\frac{4}{3\sqrt{3}}q........p....q....$$$$\left(ii\right)\overrightarrow{0}...............\frac{4}{3\sqrt{3}}q........q....p.....$$$$\left(iii\right)\overrightarrow{0}..............\frac{4}{3\sqrt{3}}q.......p=q......$$$$Whichiscorrect?$$

Commented by Rasheed Soomro last updated on 15/Dec/15

$${\mathcal{T}}^{{\mathcal{H}}^{\mathcal{A}}\mathcal{N}}{\mathcal{K}}_{\mathcal{S}sss}!$$

Commented by Yozzii last updated on 14/Dec/15

$$Letr>7andr=1.2p.$$$$\therefore 1.2p>7\Rightarrow p>\frac{7}{1.2}butthis$$$${doesn}^{\prime }tshowthatp>7.$$$$$$$$Similarly,$$$$\frac{{(z+y+x)}^2}{4}⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}$$$$\Rightarrow {(x+y+z)}^2⩾4\sqrt{\underset{r=1}{\prod ^4}{a}_r}$$$$\Rightarrow \frac{{(x+y+z)}^2}{3\sqrt{3}}⩾\frac{4}{3\sqrt{3}}\sqrt{\underset{r=1}{\prod ^4}{a}_r}\ngeqq \sqrt{\underset{r=1}{\prod ^4}{a}_r}since\frac{4}{3\sqrt{3}}\ngtr 1$$$$$$

Commented by Rasheed Soomro last updated on 15/Dec/15

$$\mathcal{T}\mathcal{H}\alpha n\mathcal{N}\mathbb{k}{\mathcal{S}}^{\mathcal{SS}Sssss!}$$

Answered by Yozzi last updated on 12/Dec/15

$$Foratrianglewithsidelengths$$$$x,yandz,thelengthofoneside$$$$cannotexceedthesumofthelengths$$$$oftheothertwosides.Symbolically,$$$$x+y-z⩾0\wedge x+z-y⩾0\wedge z+y-x⩾0.$$$$$$$$\mathrm{PROOF}:Supposeforcontradiction$$$$thatonesideoflength\mathit{a}ofagiven$$$$triangleismorethanthesumofthe$$$$lengthsoftheothertwosides,denoted$$$$by\mathit{b}and\mathit{c}.$$$$\therefore \mathit{a}>\mathit{b}+\mathit{c}$$$$with\mathit{a},\mathit{b},\mathit{c}>0.$$$$\Rightarrow {\mathit{a}}^2>{\mathit{b}}^2+{\mathit{c}}^2+2\mathit{b}\mathit{c}$$$$\Rightarrow {\mathit{a}}^2-{\mathit{b}}^2-{\mathit{c}}^2>2\mathit{b}\mathit{c}(\ast )$$$$Bythecosinerule,$$$${\mathit{a}}^2={\mathit{b}}^2+{\mathit{c}}^2-2\mathit{b}\mathit{c}cos\theta$$$$where\theta istheanglesubtendedbythe$$$$meetingofthesides\mathit{b}and\mathit{c},opposite$$$$to\mathit{a}.$$$$\Rightarrow -2\mathit{b}\mathit{c}cos\theta ={\mathit{a}}^2-{\mathit{b}}^2-{\mathit{c}}^2$$$$\therefore -2\mathit{b}\mathit{c}cos\theta >2\mathit{b}\mathit{c}\Rightarrow cos\theta <-1$$$$\Rightarrow \mid cos\theta \mid >1.But,\mathrm{\forall }\theta \in \mathbb{R},\mid cos\theta \mid ⩽1.$$$$Thisisacontradiction,sothat$$$$indeed\mathit{a}⩽\mathit{b}+\mathit{c}ingeneralandwe$$$$canshowthat\mathit{b}⩽\mathit{a}+\mathit{c}and\mathit{c}⩽\mathit{a}+\mathit{b}$$$$likewise,for\mathit{a},\mathit{b},\mathit{c}>0.◻$$$$Hence,alltermsunderthesquare$$$$rootofMr.{Rasheed}^{\prime }srhsexpression$$$$arenon-negative.$$$$$$$$$$$$Sincealltermsarepositivereals,byAM-GM,$$$$\frac{1}{4}(\{x+y+z\}+\{x+y-z\}+\{x+z-y\}+\{z+y-x\})⩾{\left((x+y+z)(x+y-z)(z+y-x)(x+z-y)\right)}^{1/4}$$$$a_1=x+y+z,a_2=x+y-z$$$$a_3=x+z-y,a_4=z+y-x$$$$\frac{1}{4}\left(2(x+y+z)\right)⩾{\left(\underset{r=1}{\prod ^4}{a}_r\right)}^{1/4}$$$$\Rightarrow \frac{1}{4}{(x+y+z)}^2⩾\sqrt{\underset{r=1}{\prod ^4}{a}_r}$$$$$$$$$$

Commented by Rasheed Soomro last updated on 12/Dec/15

$$\mathcal{I}{didn}^{\prime }tunderstandlasttwosteps.$$$$\mathcal{W}herehas{a}_{r}comefrom?$$

Commented by Yozzi last updated on 12/Dec/15

$$I^{\prime }veaddeddefinitionsfor{a}_r,1⩽r⩽4.$$

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