Question-155770

Question Number 155770 by amin96 last updated on 04/Oct/21

Commented by amin96 last updated on 04/Oct/21

$$\boldsymbol{\mathrm{P}}{rove}\:{that}\:\:\:\mathrm{2}\boldsymbol{\mathrm{mn}}=\boldsymbol{\mathrm{D}}×\boldsymbol{\mathrm{d}} \\ $$

Answered by mr W last updated on 05/Oct/21

Commented by mr W last updated on 04/Oct/21

say radius of big circle is R. OA^2 =R^2 −(D^2 /4) OB^2 =R^2 −(d^2 /4) CD=((m−n)/2) OD^2 =R^2 −(((m+n)/2))^2 OB^2 =(OD+CB)^2 +CD^2 OB^2 =OD^2 +CB^2 +2×OD×CB+CD^2 R^2 −(d^2 /4)=R^2 −(((m+n)/2))^2 +((d/2))^2 +d(√(R^2 −(((m+n)/2))^2 ))+(((m−n)/2))^2 ⇒2mn=d^2 +2d(√(R^2 −(((m+n)/2))^2 )) ...(i) OA^2 =(CA−OD)^2 +CD^2 OA^2 =CA^2 +OD^2 −2×OD×CA+CD^2 R^2 −(D^2 /4)=((D/2))^2 +R^2 −(((m+n)/2))^2 −D(√(R^2 −(((m+n)/2))^2 ))+(((m−n)/2))^2 ⇒2mn=D^2 −2D(√(R^2 −(((m+n)/2))^2 )) ...(ii) (i)×D+(ii)×d: 2(D+d)mn=d^2 D+D^2 d=(D+d)Dd ⇒2mn=Dd

$${say}\:{radius}\:{of}\:{big}\:{circle}\:{is}\:{R}. \\ $$$${OA}^{\mathrm{2}} ={R}^{\mathrm{2}} −\frac{{D}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${OB}^{\mathrm{2}} ={R}^{\mathrm{2}} −\frac{{d}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${CD}=\frac{{m}−{n}}{\mathrm{2}} \\ $$$${OD}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${OB}^{\mathrm{2}} =\left({OD}+{CB}\right)^{\mathrm{2}} +{CD}^{\mathrm{2}} \\ $$$${OB}^{\mathrm{2}} ={OD}^{\mathrm{2}} +{CB}^{\mathrm{2}} +\mathrm{2}×{OD}×{CB}+{CD}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\frac{{d}^{\mathrm{2}} }{\mathrm{4}}={R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} +{d}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} }+\left(\frac{{m}−{n}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{mn}={d}^{\mathrm{2}} +\mathrm{2}{d}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$${OA}^{\mathrm{2}} =\left({CA}−{OD}\right)^{\mathrm{2}} +{CD}^{\mathrm{2}} \\ $$$${OA}^{\mathrm{2}} ={CA}^{\mathrm{2}} +{OD}^{\mathrm{2}} −\mathrm{2}×{OD}×{CA}+{CD}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\frac{{D}^{\mathrm{2}} }{\mathrm{4}}=\left(\frac{{D}}{\mathrm{2}}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} −{D}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} }+\left(\frac{{m}−{n}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{mn}={D}^{\mathrm{2}} −\mathrm{2}{D}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{m}+{n}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×{D}+\left({ii}\right)×{d}: \\ $$$$\mathrm{2}\left({D}+{d}\right){mn}={d}^{\mathrm{2}} {D}+{D}^{\mathrm{2}} {d}=\left({D}+{d}\right){Dd} \\ $$$$\Rightarrow\mathrm{2}{mn}={Dd} \\ $$

Commented by mr W last updated on 04/Oct/21