Question-17963

Question Number 17963 by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

AB=2a , CD=2b Find r in terms of a,b,φ, and R. Hence find AP^2 +BP^2 +CP^2 +DP^2 .

$$\mathrm{AB}=\mathrm{2a}\:\:,\:\:\mathrm{CD}=\mathrm{2b} \\ $$$$\mathrm{Find}\:\boldsymbol{\mathrm{r}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:\:\mathrm{a},\mathrm{b},\phi,\:\mathrm{and}\:\mathrm{R}. \\ $$$$\mathrm{Hence}\:\mathrm{find}\:\mathrm{AP}^{\mathrm{2}} +\mathrm{BP}^{\mathrm{2}} +\mathrm{CP}^{\mathrm{2}} +\mathrm{DP}^{\mathrm{2}} . \\ $$

Commented by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

x=(√(R^2 −a^2 )) , y=(√(R^2 −b^2 )) sin 𝛗_1 =(x/r) , sin 𝛗_2 =(y/r) 𝛗_1 +𝛗_2 = 𝛗 , so sin^(−1) (((√(R^2 −a^2 ))/r) )+sin^(−1) (((√(R^2 −b^2 ))/r))=𝛗 ...(i) cannot r be obtained explicitly ? AP=a−PM , BP=a+PM CP=b−NP , DP=b+NP AP^2 +BP^2 +CP^2 +DP^2 = 2a^2 +2PM^2 +2b^2 +2NP^2 ...(ii) Now, PM^2 +NP^2 =r^2 −x^2 +r^2 −y^2 and with x^2 =R^2 −a^2 , y^2 =R^2 −b^2 we have PM^2 +NP^2 =2r^2 −(x^2 +y^2 ) =2r^2 −(2R^2 −a^2 −b^2 ) =2r^2 +a^2 +b^2 −2R^2 substituting in (ii), we get AP^2 +BP^2 +CP^2 +DP^2 =(2a)^2 +(2b)^2 −4(R^2 −r^2 ) ; r is obtained from (i) .

$$\mathrm{x}=\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:\:\:,\:\:\:\:\mathrm{y}=\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }\: \\ $$$$\mathrm{sin}\:\boldsymbol{\phi}_{\mathrm{1}} =\frac{\mathrm{x}}{\mathrm{r}}\:\:\:\:,\:\:\:\:\:\mathrm{sin}\:\boldsymbol{\phi}_{\mathrm{2}} =\frac{\mathrm{y}}{\mathrm{r}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} \:=\:\boldsymbol{\phi}\:\:\:,\:\mathrm{so} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}{\mathrm{r}}\:\right)+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{R}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }}{\mathrm{r}}\right)=\boldsymbol{\phi}\:\:…\left(\mathrm{i}\right) \\ $$$$\:\:\mathrm{cannot}\:\mathrm{r}\:\mathrm{be}\:\mathrm{obtained}\:\mathrm{explicitly}\:? \\ $$$$\mathrm{AP}=\mathrm{a}−\mathrm{PM}\:\:\:\:,\:\:\mathrm{BP}=\mathrm{a}+\mathrm{PM} \\ $$$$\mathrm{CP}=\mathrm{b}−\mathrm{NP}\:\:\:\:\:,\:\:\mathrm{DP}=\mathrm{b}+\mathrm{NP} \\ $$$$\mathrm{AP}^{\mathrm{2}} +\mathrm{BP}^{\mathrm{2}} +\mathrm{CP}^{\mathrm{2}} +\mathrm{DP}^{\mathrm{2}} =\:\: \\ $$$$\:\:\:\:\:\mathrm{2a}^{\mathrm{2}} +\mathrm{2PM}^{\mathrm{2}} +\mathrm{2b}^{\mathrm{2}} +\mathrm{2NP}^{\mathrm{2}} \:\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{Now},\:\mathrm{PM}^{\mathrm{2}} +\mathrm{NP}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \\ $$$$\:\mathrm{and}\:\mathrm{with}\:\:\mathrm{x}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \:,\:\:\mathrm{y}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\:\mathrm{PM}^{\mathrm{2}} +\mathrm{NP}^{\mathrm{2}} =\mathrm{2r}^{\mathrm{2}} −\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2r}^{\mathrm{2}} −\left(\mathrm{2R}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2r}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2R}^{\mathrm{2}} \\ $$$$\mathrm{substituting}\:\mathrm{in}\:\left(\mathrm{ii}\right),\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{AP}^{\mathrm{2}} +\mathrm{BP}^{\mathrm{2}} +\mathrm{CP}^{\mathrm{2}} +\mathrm{DP}^{\mathrm{2}} \\ $$$$\:\:\:\:=\left(\mathrm{2a}\right)^{\mathrm{2}} +\left(\mathrm{2b}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{R}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \right)\:; \\ $$$$\:\:\:\:\mathrm{r}\:\mathrm{is}\:\mathrm{obtained}\:\mathrm{from}\:\left(\mathrm{i}\right)\:. \\ $$

Commented by b.e.h.i.8.3.417@gmail.com last updated on 14/Jul/17

$${nice}\:{and}\:{beautiful}\:{mr}\:{Ajfour}. \\ $$

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