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 136415 by ajfour last updated on 21/Mar/21

Commented by ajfour last updated on 21/Mar/21

$$Findmaximumareaof$$$$△APQintermsofa,b,c,h.$$

Answered by mr W last updated on 24/Mar/21

Commented by mr W last updated on 27/Mar/21

$$\mathit{c}\mathit{l}\mathit{a}\mathit{s}\mathit{s}\mathit{i}\mathit{c}\mathit{w}\mathit{a}\mathit{y}$$$$z_C=p$$$$y_C=0$$$$x_C=\sqrt{b^2-p^2}$$$$z_B=q$$$$y_B=r$$$$x_B^2=c^2-q^2-r^2$$$${BC}^2={(q-p)}^2+r^2+{(x_B-\sqrt{b^2-p^2})}^2=a^2$$$$c^2+b^2-2pq-2x_B\sqrt{b^2-p^2}=a^2$$$$pq+x_B\sqrt{b^2-p^2}=\frac{b^2+c^2-a^2}{2}=k$$$$x_B=\frac{k-pq}{\sqrt{b^2-p^2}}$$$$\Rightarrow r^2=c^2-q^2-\frac{{(k-pq)}^2}{b^2-p^2}$$$$\frac{x_Q}{x_C}=\frac{h}{h-p}$$$$x_Q=\frac{h\sqrt{b^2-p^2}}{h-p}$$$$\frac{y_P}{y_B}=\frac{h}{h-q}$$$$y_P=\frac{hr}{h-q}$$$$shadowareaAPQA=\frac{x_Q{y}_P}{2}$$$$=\frac{h^{2}r\sqrt{b^2-p^2}}{2(h-p)(h-q)}$$$$A=\frac{h^2\sqrt{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}}{2(h-p)(h-q)}$$$$let{A}^2=\frac{h^4\mathrm{\Phi }}{4}$$$$\mathrm{\Phi }=\frac{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}{{(h-p)}^2{(h-q)}^2}$$$$\frac{\partial \mathrm{\Phi }}{\partial p}=\frac{-2p(c^2-q^2)+2q(k-pq)}{{(h-p)}^2{(h-q)}^2}+2\times \frac{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}{{(h-p)}^3{(h-q)}^2}=0$$$$(hp-b^2)(c^2-q^2)=(hq-k)(k-pq)$$$${hc}^{2}p+b^2{q}^2=k(h+p)q+b^2{c}^2-k^2...\left(i\right)$$$$\frac{\partial \mathrm{\Phi }}{\partial q}=\frac{-2q(b^2-p^2)+2p(k-pq)}{{(h-p)}^2{(h-q)}^2}+2\times \frac{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}{{(h-p)}^2{(h-q)}^3}=0$$$$(hq-c^2)(b^2-p^2)=(hp-k)(k-pq)$$$${hb}^{2}q+c^2{p}^2=k(h+q)p+b^2{c}^2-k^2...\left(ii\right)$$$$$$$$\left(ii\right)-\left(i\right):$$$$h(b^{2}q-c^{2}p)+c^2{p}^2-b^2{q}^2=kh(p-q)$$$$c^2{p}^2-h(c^2+k)p-b^2{q}^2+h(b^2+k)q=0\left(I\right)$$$$\left(ii\right)+\left(i\right):$$$$c^{2}p(h+p)+b^{2}q(h+q)=k[h(p+q)+2pq]+2b^2{c}^2-2k^2$$$$c^2{p}^2+h(c^2-k)p+b^2{q}^2+h(b^2-k)q=2kpq+2b^2{c}^2-2k^2\left(II\right)$$$$$$$$\left(I\right)+\left(II\right):$$$$c^2{p}^2-hkp+{hb}^{2}q=kpq+b^2{c}^2-k^2$$$$({hb}^2-kp)q=b^2{c}^2-k^2+hkp-c^2{p}^2$$$$\Rightarrow q=\frac{b^2{c}^2-k^2+hkp-c^2{p}^2}{{hb}^2-kp}\left(III\right)$$$$putthisinto\left(I\right)tosolveforp.$$$$$$$$\underset{―}{example1}:a=5,b=4,c=7,h=10$$$${\mathrm{\Delta }}_{ABC}=9.7980$$$$\Rightarrow p=2.5897,q=5.2984,r=4.0847$$$$\Rightarrow A_{max}=17.8706$$$$\underset{―}{example2}:a=5,b=7,c=7,h=10$$$${\mathrm{\Delta }}_{ABC}=16.3459$$$$\Rightarrow p=q=5.5119,r=4.0753$$$$\Rightarrow A_{max}=43.6497$$$$====================$$$$A=\frac{h^2\sqrt{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}}{2(h-p)(h-q)}$$$$with$$$$q=\frac{b^2{c}^2-k^2+hkp-c^2{p}^2}{{hb}^2-kp}$$$$k=\frac{b^2+c^2-a^2}{2}$$

Commented by ajfour last updated on 22/Mar/21

$$itrieditsir,wasntmuchsuccess$$$$sodintpost.Howdoyou$$$$guesstheanswer,Sir?$$

Commented by mr W last updated on 28/Mar/21

$$thanksforcomfirmingsir!$$

Commented by mr W last updated on 24/Mar/21

$$exactsolutionseemsimpossibleto$$$$me.icouldnotfindageometricway$$$$either.ihope{you}^{\prime }llgetmore$$$$successthroughvectorway.$$

Commented by ajfour last updated on 24/Mar/21

$$Thanksforpursuingittillhere$$$$Sir..$$

Commented by ajfour last updated on 24/Mar/21

Commented by ajfour last updated on 24/Mar/21

$$\sin \theta =\frac{b}{h},\cos \theta =\frac{b}{s}$$$$s=\frac{b}{\sqrt{1-{\left(\frac{b}{h}\right)}^2}}$$$$t=\frac{c}{\sqrt{1-{\left(\frac{c}{h}\right)}^2}}$$$$nextweneedtofind\mathrm{\angle }PAQ$$$$....$$

Commented by ajfour last updated on 24/Mar/21

Commented by mr W last updated on 25/Mar/21

$$iunderstandyourthought.ididthe$$$$same,butitisnotcorrect.forareaof$$$$shadowitisnotobviourswhatmakes$$$$itsareamaximum.{let}^{\prime }slookthe$$$$symmetricalcasethatAB=AC.$$$$wemaythinktheareaoftheshadow$$$$ismaximum,when\mathrm{\Delta }ABCstands$$$$perpendicularlytothelineSR,since$$$$thelengthofAR,aswellasAPand$$$$AQaremaximum.butthisisnot$$$$correct,sincethemaximallength$$$$ofAPandAQ{doesn}^{\prime }tleadtothe$$$$maximumareaofAPQ.$$

Commented by mr W last updated on 24/Mar/21

Commented by mr W last updated on 24/Mar/21

Commented by mr W last updated on 25/Mar/21

$$AB=AC=b$$$$BC=a$$$$AD=d=\frac{2\mathrm{\Delta }}{a}with\mathrm{\Delta }=areaofABC$$$$ED=d\cos \theta$$$$AE=d\sin \theta$$$$ES=h-d\sin \theta$$$$\frac{AR}{ED}=\frac{AS}{ES}=\frac{h}{h-d\sin \theta }$$$$\Rightarrow AR=\frac{hd\cos \theta }{h-d\sin \theta }$$$$\frac{PQ}{AB}=\frac{h}{h-d\sin \theta }$$$$\Rightarrow PQ=\frac{ha}{h-d\sin \theta }$$$$areaofshadowAPQ=A$$$$A=\frac{AR\times PQ}{2}=\frac{h^{2}ad\cos \theta }{2{(h-d\sin \theta )}^2}=\frac{\mathrm{\Delta }\cos \theta }{{(1-\frac{d}{h}\sin \theta )}^2}$$$$let\lambda =\frac{d}{h}=\frac{2\mathrm{\Delta }}{ah}$$$$\frac{A}{\mathrm{\Delta }}=\mathrm{\Phi }=\frac{\cos \theta }{{(1-\lambda \sin \theta )}^2}$$$$\frac{d\mathrm{\Phi }}{d\theta }=-\frac{\sin \theta }{{(1-\lambda \sin \theta )}^2}+\frac{2\lambda {\cos }^2\theta }{{(1-\lambda \sin \theta )}^3}=0$$$$\sin \theta =\frac{2\lambda {\cos }^2\theta }{1-\lambda \sin \theta }$$$$\sin \theta -\lambda {\sin }^2\theta =2\lambda -2\lambda {\sin }^2\theta$$$${\sin }^2\theta +\frac{1}{\lambda }\sin \theta -2=0$$$$\Rightarrow \sin \theta =\frac{\sqrt{1+8{\lambda }^2}-1}{2\lambda }$$$$\Rightarrow {\theta }_m={\sin }^{-1}\frac{\sqrt{1+8{\lambda }^2}-1}{2\lambda }$$$$at{\theta }_{m}theareaofshadowismaximum.$$$${\mathrm{\Phi }}_{max}=\frac{\sqrt{2(\sqrt{1+8{\lambda }^2}-2{\lambda }^2-1)}}{\lambda (4{\lambda }^2+5-3\sqrt{1+8{\lambda }^2})}$$$$example:a=5,b=c=7,h=10$$$$\Rightarrow \mathrm{\Delta }=16.3459$$$$\lambda =\frac{2\times 16.3459}{5\times 10}=0.653836$$$${\mathrm{\Phi }}_{max}=2.670394$$$$\Rightarrow A_{max}=2.670394\times 16.3459=43.6499$$$$thesameresultasbefore.$$$$$$$$whenSDtangentsAD,seedotlines,$$$$\sin \theta =\frac{d}{h}=\lambda \Rightarrow \sin {\theta }_m,thatmeansthe$$$$areaofshadowisherenotmaximum.$$

Commented by mr W last updated on 25/Mar/21

$$if\sin {\theta }_m=\lambda =\frac{d}{h}=\frac{2\mathrm{\Delta }}{ah}$$$$A_{max}=\mathrm{\Delta }\times \frac{\cos \theta }{{(1-\lambda \sin \theta )}^2}=\mathrm{\Delta }\times \frac{\sqrt{1-{\lambda }^2}}{{(1-{\lambda }^2)}^2}$$$$=\frac{\mathrm{\Delta }}{{(1-{\lambda }^2)}^{\frac{3}{2}}}=\frac{\mathrm{\Delta }}{{(1-\frac{4{\mathrm{\Delta }}^2}{a^2{h}^2})}^{\frac{3}{2}}}$$$$thisistheformulawhichigave$$$$atfirst.butnowweknowitisnot$$$$correct.$$

Commented by mr W last updated on 25/Mar/21

Commented by mr W last updated on 26/Mar/21

$$\underset{―}{\mathit{S}\mathit{u}\mathit{m}\mathit{m}\mathit{a}\mathit{r}\mathit{y}}$$$$wecanexpresstheareaofshadowas$$$$afunctionofasinglevariablep:$$$$A=\frac{h^2\sqrt{(b^2-p^2)(c^2-q^2)-{(k-pq)}^2}}{2(h-p)(h-q)}$$$$with$$$$q=\frac{b^2{c}^2-k^2+hkp-c^2{p}^2}{{hb}^2-kp}$$$$k=\frac{b^2+c^2-a^2}{2}$$$$themaximumareaoftheshadow$$$$canbedeterminedapproximately.$$

Commented by mr W last updated on 26/Mar/21

Commented by mr W last updated on 26/Mar/21

Commented by ajfour last updated on 28/Mar/21

$$YesSir,itsallcorrect$$$$Ijustcheckedforb=c$$$$A=\frac{{ah}^2\sqrt{4b^2\sin {}^2\theta -a^2}}{4{(h-b\cos \theta )}^2}$$$$forb=c=7,a=5,h=10$$$$iget{A}_{max}=43.6497$$$$at\cos \theta =0.7874$$

Answered by mr W last updated on 25/Mar/21

Commented by mr W last updated on 25/Mar/21

$$\mathit{v}\mathit{e}\mathit{c}\mathit{t}\mathit{o}\mathit{r}\mathit{w}\mathit{a}\mathit{y}$$$$$$$$\mathrm{\Delta }=areaofABC$$$$$$$$\mathit{S}\mathit{B}=\mathit{b}-\mathit{h}$$$$\mathit{S}\mathit{P}=\xi (\mathit{b}-\mathit{h})$$$$\mathit{A}\mathit{P}=\mathit{h}+\mathit{S}\mathit{P}=\mathit{h}+\xi (\mathit{b}-\mathit{h})=(1-\xi )\mathit{h}+\xi \mathit{b}$$$$\mathit{A}\mathit{P}\cdot \mathit{h}=0$$$$(1-\xi )h^2+\xi \mathit{b}\cdot \mathit{h}=0$$$$\xi =\frac{1}{1-\frac{\mathit{b}\cdot \mathit{h}}{h^2}}$$$$\mathit{p}=\mathit{A}\mathit{P}=\frac{1}{1-\frac{\mathit{b}\cdot \mathit{h}}{h^2}}(\frac{\mathit{b}\cdot \mathit{h}}{h^2}\mathit{h}+\mathit{b})$$$$similarly$$$$\mathit{q}=\mathit{A}\mathit{Q}=\frac{1}{1-\frac{\mathit{c}\cdot \mathit{h}}{h^2}}(\frac{\mathit{c}\cdot \mathit{h}}{h^2}\mathit{h}+\mathit{c})$$$$areaofshadowA=\frac{1}{2}\mid \mathit{p}\times \mathit{q}\mid$$$$\mathit{b}=(b,0,0)$$$$\mathit{c}=(c\cos \mathrm{\angle }A,c\sin \mathrm{\angle }A,0)$$$$\mathit{h}=(\alpha h,\beta h,\gamma h)with\gamma =\sqrt{1-{\alpha }^2-{\beta }^2}$$$$\mathit{b}\cdot \mathit{h}=\alpha hb$$$$\mathit{c}\cdot \mathit{h}=\alpha hc\cos \mathrm{\angle }A+\beta hc\sin \mathrm{\angle }A$$$$\Rightarrow \mathit{p}=\frac{bh}{h-\alpha b}(1+{\alpha }^2,\alpha \beta ,\alpha \gamma )$$$$\Rightarrow \mathit{q}=\frac{ch}{h-c(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)}((1+{\alpha }^2)\cos \mathrm{\angle }A+\alpha \beta \sin \mathrm{\angle }A,(1+{\beta }^2)\sin \mathrm{\angle }A+\alpha \beta \cos \mathrm{\angle }A,(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)\gamma )$$$$u=\alpha \beta \times (\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)\gamma -\alpha \gamma \times [(1+{\beta }^2)\sin \mathrm{\angle }A+\alpha \beta \cos \mathrm{\angle }A]$$$$u=-\alpha \gamma \sin \mathrm{\angle }A$$$$v=(1+{\alpha }^2)\times (\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)\gamma -\alpha \gamma \times [(1+{\alpha }^2)\cos \mathrm{\angle }A+\alpha \beta \sin \mathrm{\angle }A]$$$$v=-\beta \gamma \sin \mathrm{\angle }A$$$$w=(1+{\alpha }^2)\times [(1+{\beta }^2)\sin \mathrm{\angle }A+\alpha \beta \cos \mathrm{\angle }A]-\alpha \beta \times [(1+{\alpha }^2)\cos \mathrm{\angle }A+\alpha \beta \sin A]$$$$w=(1+{\alpha }^2+{\beta }^2)\sin \mathrm{\angle }A$$$$\frac{1}{2}\mid \mathit{p}\times \mathit{q}\mid =\frac{1}{2}\times \frac{bh}{h-\alpha b}\times \frac{ch}{h-c(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)}\times \sqrt{u^2+v^2+z^2}$$$$\mid \mathit{p}\times \mathit{q}\mid =\frac{{bch}^2\sin \mathrm{\angle }A\sqrt{({\alpha }^2+{\beta }^2){\gamma }^2+{(1+{\alpha }^2+{\beta }^2)}^2}}{2(h-\alpha b)[h-c(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)]}$$$$\mid \mathit{p}\times \mathit{q}\mid =\frac{h^2\mathrm{\Delta }\sqrt{(1-{\gamma }^2){\gamma }^2+{(2-{\gamma }^2)}^2}}{(h-\alpha b)[h-c(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)]}$$$$\mid \mathit{p}\times \mathit{q}\mid =\frac{\mathrm{\Delta }\sqrt{1+3{\alpha }^2+3{\beta }^2}}{(1-\frac{b}{h}\alpha )[1-\frac{c}{h}(\alpha \cos \mathrm{\angle }A+\beta \sin \mathrm{\angle }A)]}$$$$let\frac{b}{h}=l,\frac{c\cos \mathrm{\angle }A}{h}=m,\frac{c\sin \mathrm{\angle }A}{h}=n$$$$areaofshadowA=\frac{\mathrm{\Delta }\sqrt{1+3{\alpha }^2+3{\beta }^2}}{(1-l\alpha )(1-m\alpha -n\beta )}$$$$=\sqrt{\mathrm{\Phi }}\mathrm{\Delta }$$$$\mathrm{\Phi }=\frac{1+3{\alpha }^2+3{\beta }^2}{{(1-l\alpha )}^2{(1-m\alpha -n\beta )}^2}$$$$.....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com