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Question-48338




Question Number 48338 by ajfour last updated on 22/Nov/18
Commented by mr W last updated on 23/Nov/18
nice working!
$${nice}\:{working}! \\ $$
Commented by ajfour last updated on 22/Nov/18
For example   p=2 , q=4, r=3.  Find  x= ?.
$${For}\:{example}\:\:\:{p}=\mathrm{2}\:,\:{q}=\mathrm{4},\:{r}=\mathrm{3}. \\ $$$${Find}\:\:{x}=\:?. \\ $$
Commented by ajfour last updated on 22/Nov/18
Will this do Sir ?
$${Will}\:{this}\:{do}\:{Sir}\:? \\ $$
Commented by ajfour last updated on 22/Nov/18
Commented by ajfour last updated on 22/Nov/18
R^2 cos^(−1) (h/R)−h(√(R^2 −h^2 )) = p+q  R^2 cos^(−1) (h/R)−k(√(R^2 −k^2 )) = r+p     x = πR^2 −p−q−r  And     ((πR^2 )/4) = ((q+r)/2)+hk     ⇒ ((πR^2 )/4) =((q+r)/2)+h′k′R^2        R^( 2)  = (((q+r)/2)/((π/4)−h′k′))  ________________________   let (h/R) = h′   & (p/R^2 ) = p′ and so on  cos^(−1) h′−h′(√(1−h′^( 2) )) = (((p+q)/(q+r)))((π/2)−2h′k′)  cos^(−1) k′−k′(√(1−k′^( 2) )) = (((r+p)/(q+r)))((π/2)−2h′k′)   x = ((𝛑(q+r))/(((𝛑/2)−2h′k′)))−(p+q+r) .
$${R}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \frac{{h}}{{R}}−{h}\sqrt{{R}^{\mathrm{2}} −{h}^{\mathrm{2}} }\:=\:{p}+{q} \\ $$$${R}^{\mathrm{2}} \mathrm{cos}^{−\mathrm{1}} \frac{{h}}{{R}}−{k}\sqrt{{R}^{\mathrm{2}} −{k}^{\mathrm{2}} }\:=\:{r}+{p} \\ $$$$\:\:\:{x}\:=\:\pi{R}^{\mathrm{2}} −{p}−{q}−{r} \\ $$$${And}\:\:\:\:\:\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{{q}+{r}}{\mathrm{2}}+{hk} \\ $$$$\:\:\:\Rightarrow\:\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}\:=\frac{{q}+{r}}{\mathrm{2}}+{h}'{k}'{R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{R}^{\:\mathrm{2}} \:=\:\frac{\left({q}+{r}\right)/\mathrm{2}}{\frac{\pi}{\mathrm{4}}−{h}'{k}'} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:{let}\:\frac{{h}}{{R}}\:=\:{h}'\:\:\:\&\:\frac{{p}}{{R}^{\mathrm{2}} }\:=\:{p}'\:{and}\:{so}\:{on} \\ $$$$\mathrm{cos}^{−\mathrm{1}} {h}'−{h}'\sqrt{\mathrm{1}−{h}'^{\:\mathrm{2}} }\:=\:\left(\frac{{p}+{q}}{{q}+{r}}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{h}'{k}'\right) \\ $$$$\mathrm{cos}^{−\mathrm{1}} {k}'−{k}'\sqrt{\mathrm{1}−{k}'^{\:\mathrm{2}} }\:=\:\left(\frac{{r}+{p}}{{q}+{r}}\right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}{h}'{k}'\right) \\ $$$$\:\boldsymbol{{x}}\:=\:\frac{\boldsymbol{\pi}\left(\boldsymbol{{q}}+\boldsymbol{{r}}\right)}{\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}−\mathrm{2}\boldsymbol{{h}}'\boldsymbol{{k}}'\right)}−\left(\boldsymbol{{p}}+\boldsymbol{{q}}+\boldsymbol{{r}}\right)\:. \\ $$
Commented by MJS last updated on 22/Nov/18
tried with integrals  for your example I get  x=5.68110∧R=2.16174 ∨ x=144.410∧R=6.98799;  in the first case the center is shifted by   (((.143930)),((.253156)) ), in the second case by  (((.835304)),((.854448)) )
$$\mathrm{tried}\:\mathrm{with}\:\mathrm{integrals} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{example}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=\mathrm{5}.\mathrm{68110}\wedge{R}=\mathrm{2}.\mathrm{16174}\:\vee\:{x}=\mathrm{144}.\mathrm{410}\wedge{R}=\mathrm{6}.\mathrm{98799}; \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{case}\:\mathrm{the}\:\mathrm{center}\:\mathrm{is}\:\mathrm{shifted}\:\mathrm{by} \\ $$$$\begin{pmatrix}{.\mathrm{143930}}\\{.\mathrm{253156}}\end{pmatrix},\:\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{case}\:\mathrm{by}\:\begin{pmatrix}{.\mathrm{835304}}\\{.\mathrm{854448}}\end{pmatrix} \\ $$

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