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In the given figure, $X$ and $Y$ are two centres of two circles. They touch each other externally at a point $S$. $AB$ be the common tangent of both circles. $O$ be the centre of the third circle which touches the two circles at points $Q$ and $R$. If $AX=h$, $OM=f$ and $BY=g$, then prove that $$\frac {1}{\sqrt { f }}=\frac {1}{\sqrt { h }} +\frac {1}{\sqrt { g }}$$

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My attempt

If we join $OA$ we have, $$f^2={OA}^2-{AM}^2=$$; using pythagoras theorem

Again if we join $OB$ then $$f^2={OB}^2-{MB}^2$$ Now how should I move further?

pi-π
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2 Answers2

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Hint: $BM^2=(g+f)^2-(g-f)^2=4gf$ $AM^2=(h+f)^2-(h-f)^2=4hf$ $(AM+BM)^2=(g+h)^2-(g-h)^2=4gh$

Square root both sides of all three expressions, then plug the first two expressions into the third one, and you get the final equality you need to prove.

Huang
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How long is AB?

$XY = g + h$ you have a already constructed parallel to $AB$ from $Y$ to an unidentified point that I will call $Y'$

$YY' = h-g\\ (h-g)^2 + AB^2 = (h+g)^2\\ AB^2 = 4gh$

We can do a similar construction to get $BM$ and $AM$

$BM^2 = 4fg\\ AM^2 = 4fh\\ \sqrt {4fg} + \sqrt{4fh} = \sqrt{4gh}$

divide trough by $2\sqrt{fgh}$ and we get our desired result.

Doug M
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  • I couldn't understand your parallel concept, please clarify. – pi-π Jul 16 '16 at 09:21
  • You have constructed a line in your diagram that is parallel to AB and intersects Y, forming a narrow little right triangle. One leg is has the same length as AB, one leg has a length equal to h-g, and the hypotenuse = XY has a length equal to g+h – Doug M Jul 17 '16 at 21:46