Circle: Beautiful problem

Three circles touch the same straight line and touch each other, as shown . Prove that the radii  a , b and c , where c is smallest , satisfy the equation $ \displaystyle \frac{1}{{\sqrt{a}}}+\frac{1}{{\sqrt{b}}}=\frac{1}{{\sqrt{c}}}$ .

Solution

$ \displaystyle \text{Let}\ PC=x\ \ ,\ DQ=y\ \ \ ,\ \ QE=z$ 

$ \displaystyle \text{In}\ \text{rt}\ \Delta OCP\ ,\ \ O{{P}^{2}}=O{{C}^{2}}+P{{C}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{(a+b)}^{2}}={{(a-b)}^{2}}+{{x}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}=4ab$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=2\sqrt{{ab}}$

$ \displaystyle \text{In}\ \text{rt}\ \Delta ODQ\ ,\ \ {{y}^{2}}=O{{Q}^{2}}-O{{D}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{y}^{2}}={{(a+c)}^{2}}-{{(a-c)}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{y}^{2}}=4ac$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=2\sqrt{{ac}}$

$ \displaystyle \text{In}\ \text{rt}\ \Delta QEP\ ,\ \ \ {{z}^{2}}=Q{{P}^{2}}-P{{E}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{z}^{2}}={{(b+c)}^{2}}-{{(b-c)}^{2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{z}^{2}}=4bc$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z=2\sqrt{{bc}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ x=y+z$

$ \displaystyle \ \ \ \ 2\sqrt{{ab}}=2\sqrt{{ac}}+2\sqrt{{bc}}$

$ \displaystyle \ \ \ \ \ \sqrt{{ab}}=\sqrt{{ac}}+\sqrt{{bc}}$

$ \displaystyle \ \ \ \ \ \frac{{\sqrt{{ab}}}}{{\sqrt{{abc}}}}=\frac{{\sqrt{{ac}}}}{{\sqrt{{abc}}}}+\frac{{\sqrt{{bc}}}}{{\sqrt{{abc}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \frac{1}{{\sqrt{c}}}=\frac{1}{{\sqrt{b}}}+\frac{1}{{\sqrt{a}}}$

$ \displaystyle \ \therefore \ \ \ \frac{1}{{\sqrt{a}}}+\frac{1}{{\sqrt{b}}}=\frac{1}{{\sqrt{c}}}$

*Note
      If two circles are tangent internally or externally , the line containing their centres also   contains the point of tangency .