Question

two circles with radii a and b respectively touch each other externally and a smaller circle with radius c touch both the circles and also the common tanjent to the two circles,then prove that:
{(1/a)^0.5} +{(1/b)^0.5} ={(1/c)^0.5}

Answer 1

two circles with radii a and b respectively touch each other externally and a smaller circle with radius c touch both the circles and also the common tangent to the two circles, then prove that:

$$\{(1 / a) ^ {\wedge} 0. 5 \} + \{(1 / b) ^ {\wedge} 0. 5 \} = \{(1 / c) ^ {\wedge} 0. 5 \}$$

Solution

Circles are touch each other externally, thus:

$$A B = a + b$$ $$A C = a + c$$ $$B C = c + b$$

From right triangle $\triangle CDB$ , where $\angle D = 90{}^{\circ}$ , using Pythagorean theorem:

$$C D = \sqrt {(B C) ^ {2} - (B D) ^ {2}}$$

As:

$$B D = B L - D L = b - c$$

So:

$$C D = \sqrt {(c + b) ^ {2} - (c - b) ^ {2}} = \sqrt {c ^ {2} + 2 c b + b ^ {2} - c ^ {2} + 2 c b - b ^ {2}} = \sqrt {4 c b}$$

From rectangle CDLP:

$$C D = L P$$

Thus:

$$L P = 2 \sqrt {c b}$$

Similarly :

$$P K = 2 \sqrt {c a}$$

So:

$$L K = L P + P K = 2 \sqrt {c} (\sqrt {b} + \sqrt {a})$$

Similarly, from right triangle $\triangle AMB$ , where $\angle M = 90{}^{\circ}$ , using Pythagorean theorem:

$$B M = \sqrt {(b + a) ^ {2} - (b - a) ^ {2}} = 2 \sqrt {a b}$$

From rectangle KMBL:

$$B M = L K$$

Thus:

$$2 \sqrt {a b} = 2 \sqrt {c} (\sqrt {b} + \sqrt {a})$$

Dividing by $\frac{\sqrt{c}}{2\sqrt{ab}}$ gives:

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

Or :

$$\left(\frac {1}{a}\right) ^ {0. 5} + \left(\frac {1}{b}\right) ^ {0. 5} = \left(\frac {1}{c}\right) ^ {0. 5}$$