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 204944 by mr W last updated on 02/Mar/24

Commented by mr W last updated on 03/Mar/24

$$thedistancesfromapointinside$$$$atriangletoitsvertexesarep,q,r$$$$respectively.findthemaximum$$$$areaofthetriangleanditssides.$$

Answered by mr W last updated on 03/Mar/24

Commented by mr W last updated on 03/Mar/24

$$\alpha +\beta +\gamma =2\pi$$$$\mathrm{\Delta }=areaoftriangle$$$$\mathrm{\Delta }=\frac{qr\sin \alpha }{2}+\frac{rp\sin \beta }{2}+\frac{pq\sin \gamma }{2}$$$$\mathrm{\Delta }=\frac{pqr}{2}(\frac{\sin \alpha }{p}+\frac{\sin \beta }{q}+\frac{\sin \gamma }{r})=\frac{pqr}{2}\mathrm{\Phi }$$$$with\mathrm{\Phi }=\frac{\sin \alpha }{p}+\frac{\sin \beta }{q}+\frac{\sin \gamma }{r}$$$${\mathrm{\Delta }}_{max}\iff {\mathrm{\Phi }}_{max}$$$$F=\frac{\sin \alpha }{p}+\frac{\sin \beta }{q}+\frac{\sin \gamma }{r}-\lambda (\alpha +\beta +\gamma -2\pi )$$$$\frac{\partial F}{\partial \alpha }=\frac{\cos \alpha }{p}-\lambda =0\Rightarrow \cos \alpha =p\lambda$$$$similarly$$$$\cos \beta =q\lambda$$$$\cos \gamma =r\lambda$$$$$$$$weseer\cos \beta =q\cos \gamma =qr\lambda ,that$$$$meansAS\mathrm{\perp }BC.$$$$similarlyBS\mathrm{\perp }AC,CS\mathrm{\perp }AB.$$$$thatmeansSistheorthocenterof$$$$thetriangle.$$$$$$$$\cos (\alpha +\beta )=\cos (2\pi -\gamma )=\cos \gamma$$$$pq{\lambda }^2-\sqrt{(1-p^2{\lambda }^2)(1-q^2{\lambda }^2)}=r\lambda$$$$pq{\lambda }^2-r\lambda =\sqrt{(1-p^2{\lambda }^2)(1-q^2{\lambda }^2)}$$$$p^2{q}^2{\lambda }^4-2pqr{\lambda }^3+r^2{\lambda }^2=(1-p^2{\lambda }^2)(1-q^2{\lambda }^2)$$$$p^2{q}^2{\lambda }^4-2pqr{\lambda }^3+r^2{\lambda }^2=1-(p^2+q^2){\lambda }^2+p^2{q}^2{\lambda }^4$$$$\Rightarrow \frac{1}{{\lambda }^3}-\frac{p^2+q^2+r^2}{\lambda }+2pqr=0$$$$\Rightarrow \frac{1}{\lambda }=-2\sqrt{\frac{p^2+q^2+r^2}{3}}\sin [\frac{\pi }{3}+\frac{1}{3}{\sin }^{-1}\frac{pqr}{{\left(\frac{p^2+q^2+r^2}{3}\right)}^{\frac{3}{2}}}]$$$$\Rightarrow \lambda =-\frac{1}{2\sqrt{\frac{p^2+q^2+r^2}{3}}\sin [\frac{\pi }{3}+\frac{1}{3}{\sin }^{-1}\frac{pqr}{{\left(\frac{p^2+q^2+r^2}{3}\right)}^{\frac{3}{2}}}]}$$$$\sin \alpha =\sqrt{1-p^2{\lambda }^2}$$$$\sin \beta =\sqrt{1-q^2{\lambda }^2}$$$$\sin \gamma =\sqrt{1-r^2{\lambda }^2}$$$$\Rightarrow {\mathrm{\Delta }}_{max}=\frac{pqr}{2}(\sqrt{\frac{1}{p^2}-{\lambda }^2}+\sqrt{\frac{1}{q^2}-{\lambda }^2}+\sqrt{\frac{1}{r^2}-{\lambda }^2})$$$$$$$$a=\sqrt{q^2+r^2-2qr\cos \alpha }=\sqrt{q^2+r^2-2pqr\lambda }$$$$similarly$$$$b=\sqrt{r^2+p^2-2pqr\lambda }$$$$c=\sqrt{p^2+q^2-2pqr\lambda }$$

Commented by mr W last updated on 03/Mar/24

$$example:p=2,q=3,r=5$$

Commented by mr W last updated on 03/Mar/24

Commented by mr W last updated on 04/Mar/24

$$with\lambda =\frac{1}{2\sqrt{\frac{p^2+q^2+r^2}{3}}\sin [\frac{\pi }{3}-\frac{1}{3}{\sin }^{-1}\frac{pqr}{{\left(\frac{p^2+q^2+r^2}{3}\right)}^{\frac{3}{2}}}]}$$$$wegettheminimumarea.inthis$$$$casetheorthocenterliesoutside$$$$thetriangle.$$

Commented by mr W last updated on 03/Mar/24

Terms of Service

Privacy Policy

Contact: info@tinkutara.com