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Question Number 62486 by ajfour last updated on 21/Jun/19

Commented by ajfour last updated on 21/Jun/19

$$Findmaximumperimeterof$$$$quadrilateralABCD.$$

Answered by mr W last updated on 22/Jun/19

$$p=\sqrt{a^2+b^2-2ab\cos \theta }+\sqrt{b^2+c^2-2bc\cos \phi }+\sqrt{c^2+d^2-2cd\cos \varphi }+\sqrt{d^2+a^2-2da\cos (2\pi -\theta -\phi -\varphi )}$$$$\frac{\partial p}{\partial \theta }=\frac{ab\sin \theta }{\sqrt{a^2+b^2-2ab\cos \theta }}-\frac{da\sin (2\pi -\theta -\phi -\varphi )}{\sqrt{d^2+a^2-2da\cos (2\pi -\theta -\phi -\varphi )}}=0$$$$\frac{\partial p}{\partial \phi }=\frac{bc\sin \phi }{\sqrt{b^2+c^2-2bc\cos \phi }}-\frac{da\sin (2\pi -\theta -\phi -\varphi )}{\sqrt{d^2+a^2-2da\cos (2\pi -\theta -\phi -\varphi )}}=0$$$$\frac{\partial p}{\partial \varphi }=\frac{cd\sin \varphi }{\sqrt{c^2+d^2-2cd\cos \varphi }}-\frac{da\sin (2\pi -\theta -\phi -\varphi )}{\sqrt{d^2+a^2-2da\cos (2\pi -\theta -\phi -\varphi )}}=0$$$$\frac{ab\sin \theta }{\sqrt{a^2+b^2-2ab\cos \theta }}=\frac{da\sin (2\pi -\theta -\phi -\varphi )}{\sqrt{d^2+a^2-2da\cos (2\pi -\theta -\phi -\varphi )}}=\sqrt{k}$$$$a^2{b}^2{\sin }^2\theta =k(a^2+b^2-2ab\cos \theta )$$$$a^2{b}^2(1-{\cos }^2\theta )=k(a^2+b^2-2ab\cos \theta )$$$$a^2{b}^2{\cos }^2\theta -2abk\cos \theta +(a^2+b^2)k-a^2{b}^2=0$$$$\cos \theta =\frac{abk-\sqrt{a^2{b}^2{k}^2-a^2{b}^2[(a^2+b^2)k-a^2{b}^2]}}{a^2{b}^2}$$$$\Rightarrow \cos \theta =\frac{k-\sqrt{k^2-(a^2+b^2)k+a^2{b}^2}}{ab}$$$$\Rightarrow \theta ={\cos }^{-1}\frac{k-\sqrt{k^2-(a^2+b^2)k+a^2{b}^2}}{ab}$$$$\Rightarrow \phi ={\cos }^{-1}\frac{k-\sqrt{k^2-(b^2+c^2)k+b^2{c}^2}}{bc}$$$$\Rightarrow \varphi ={\cos }^{-1}\frac{k-\sqrt{k^2-(c^2+d^2)k+c^2{d}^2}}{cd}$$$$\Rightarrow 2\pi -(\theta +\phi +\varphi )={\cos }^{-1}\frac{k-\sqrt{k^2-(d^2+a^2)k+d^2{a}^2}}{da}$$$$wegetk...$$$$example:a=10,b=8,c=6,d=5$$$$\Rightarrow k=21.2423$$$$\Rightarrow \theta =117.3773°$$$$\Rightarrow \phi =94.6335°$$$$\Rightarrow \varphi =62.6226°$$

Commented by peter frank last updated on 22/Jun/19

$$congratulationsirforhardwork$$

Commented by ajfour last updated on 22/Jun/19

$$ThanksSir,butiwished,wecould$$$$crackitbysomegeometricalway?$$$$Thankyouanyway.$$

Answered by mr W last updated on 23/Jun/19

$$Oshouldbetheincirclecentersuch$$$$thatthequadrilateralhasmaximum$$$$perimeter\left(seeQ62542\right)$$$$letr=radiusoftheincircle$$$$\mathrm{\angle }A=2{\sin }^{-1}\frac{r}{a}$$$$\mathrm{\angle }B=2{\sin }^{-1}\frac{r}{b}$$$$\mathrm{\angle }C=2{\sin }^{-1}\frac{r}{c}$$$$\mathrm{\angle }D=2{\sin }^{-1}\frac{r}{d}$$$$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D=2\pi$$$${\sin }^{-1}\frac{r}{a}+{\sin }^{-1}\frac{r}{b}+{\sin }^{-1}\frac{r}{c}+{\sin }^{-1}\frac{r}{d}=\pi$$$$\cos ({\sin }^{-1}\frac{r}{a}+{\sin }^{-1}\frac{r}{b})=\cos (\pi -{\sin }^{-1}\frac{r}{c}-{\sin }^{-1}\frac{r}{d})$$$$\sqrt{(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})}-\frac{r^2}{ab}=-\sqrt{(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})}+\frac{r^2}{cd}$$$$\sqrt{(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})}-\sqrt{(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})}=\frac{r^2}{ab}+\frac{r^2}{cd}$$$$(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})+(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})-2\sqrt{(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})}=\frac{r^4}{a^2{b}^2}+\frac{r^4}{c^2{d}^2}+\frac{2r^4}{abcd}$$$$2-(\frac{r^2}{a^2}+\frac{r^2}{b^2}+\frac{r^2}{c^2}+\frac{r^2}{d^2})+\frac{r^4}{a^2{b}^2}+\frac{r^4}{c^2{d}^2}-2\sqrt{(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})}=\frac{r^4}{a^2{b}^2}+\frac{r^4}{c^2{d}^2}+\frac{2r^4}{abcd}$$$$\sqrt{(1-\frac{r^2}{a^2})(1-\frac{r^2}{b^2})(1-\frac{r^2}{c^2})(1-\frac{r^2}{d^2})}=1-\frac{1}{2}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^2-\frac{r^4}{abcd}$$$$[1-(\frac{r^2}{a^2}+\frac{r^2}{b^2})+\frac{r^4}{a^2{b}^2}][1-(\frac{r^2}{c^2}+\frac{r^2}{d^2})+\frac{r^4}{c^2{d}^2}]=1+\frac{1}{4}{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})}^2{r}^4+\frac{r^8}{a^2{b}^2{c}^2{d}^2}-\frac{2r^4}{abcd}-(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^2+\frac{1}{abcd}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^6$$$$1-(\frac{r^2}{a^2}+\frac{r^2}{b^2})+\frac{r^4}{a^2{b}^2}-(\frac{r^2}{c^2}+\frac{r^2}{d^2})+(\frac{r^2}{a^2}+\frac{r^2}{b^2})(\frac{r^2}{c^2}+\frac{r^2}{d^2})-\frac{r^4}{a^2{b}^2}(\frac{r^2}{c^2}+\frac{r^2}{d^2})+\frac{r^4}{c^2{d}^2}-\frac{r^4}{c^2{d}^2}(\frac{r^2}{a^2}+\frac{r^2}{b^2})+\frac{r^8}{a^2{b}^2{c}^2{d}^2}=1+\frac{1}{4}{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})}^2{r}^4+\frac{r^8}{a^2{b}^2{c}^2{d}^2}-\frac{2r^4}{abcd}-(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^2+\frac{1}{abcd}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^6$$$$(\frac{1}{a^2{b}^2}+\frac{1}{a^2{c}^2}+\frac{1}{a^2{d}^2}+\frac{1}{b^2{c}^2}+\frac{1}{b^2{d}^2}+\frac{1}{c^2{d}^2})-\left(\frac{a^2+b^2+c^2+d^2}{a^2{b}^2{c}^2{d}^2}\right)r^2=\frac{1}{4}{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})}^2-\frac{2}{abcd}+\frac{1}{abcd}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})r^2$$$$\Rightarrow [\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+\frac{a^2+b^2+c^2+d^2}{abcd}]r^2=2+(\frac{cd}{ab}+\frac{ab}{cd}+\frac{bd}{ac}+\frac{bc}{ad}+\frac{ad}{bc}+\frac{ac}{bd})-\frac{abcd}{4}{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})}^2$$$$\Rightarrow r=\sqrt{\frac{2+(\frac{cd}{ab}+\frac{ab}{cd}+\frac{bd}{ac}+\frac{bc}{ad}+\frac{ad}{bc}+\frac{ac}{bd})-\frac{abcd}{4}{(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})}^2}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}+\frac{a^2+b^2+c^2+d^2}{abcd}}}$$

Commented by ajfour last updated on 24/Jun/19

$$ImmenselyBeautifulSir!$$

Commented by mr W last updated on 23/Jun/19

$$thanksforcheckingsir!ihadamistake$$$$intheworking,nowfixed.$$$$witha=b=c=d{we}^{\prime }llgetr=\frac{a}{\sqrt{2}}.$$

Commented by mr W last updated on 23/Jun/19

$$example:a=10,b=8,c=6,d=5$$$$r=\sqrt{\frac{2+(\frac{30}{80}+\frac{80}{30}+\frac{40}{60}+\frac{48}{50}+\frac{50}{48}+\frac{60}{40})-\frac{2400}{4}{(\frac{1}{100}+\frac{1}{64}+\frac{1}{36}+\frac{1}{25})}^2}{\frac{1}{100}+\frac{1}{64}+\frac{1}{36}+\frac{1}{25}+\frac{100+64+36+25}{2400}}}$$$$r=\sqrt{\frac{2+\frac{721}{100}-600{\left(\frac{269}{2880}\right)}^2}{\frac{269}{2880}+\frac{225}{2400}}}=4.60894$$$$\theta ={180}^{°}-{\sin }^{-1}\frac{r}{a}-{\sin }^{-1}\frac{r}{b}=117.3773°$$$$\phi =180°-{\sin }^{-1}\frac{r}{b}-{\sin }^{-1}\frac{r}{c}=94.6335°$$$$\varphi =180°-{\sin }^{-1}\frac{r}{c}-{\sin }^{-1}\frac{r}{d}=62.6227°$$$$theresultsarethesameasusing$$$$calculusmethod.$$

Commented by ajfour last updated on 24/Jun/19

$$\mathcal{I}tsa\mathcal{D}iscoverythenSir,$$$$\mathcal{C}ongratulations,andthanks$$$$forsharingit.$$

Commented by mr W last updated on 24/Jun/19

$$thankssir!$$$$i{didn}^{\prime }texpectthatthesolutionfor$$$$quadrilateraliseveneasierthanfor$$$$triangle.imeanwithoutcubicequation.$$

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