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3 circles of radius 3 cm, 4cm, 5 cm touch each other externally at $A$, $B$, $C$. Tangents drawn at $A$, $B$, $C$ intersect at $P$.

Find $ PA + PB + PC$ .

Thanks.

My thoughts and approach:

Well I made the figure but still could not proceed ahead.

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  • $\begingroup$ First hint: what can you say on the length of $PA$ compared to the length of $PB$? $\endgroup$ – TZakrevskiy Jul 5 '14 at 13:51
  • $\begingroup$ I dont know , And I cant get any idea :( $\endgroup$ – AJ_ Jul 5 '14 at 13:58
  • $\begingroup$ Ok, let's start from another angle: what level of geometry do you know? $\endgroup$ – TZakrevskiy Jul 5 '14 at 14:05
  • $\begingroup$ Thats a bit hard to explain but maybe relative to the standards of this site I dont know much. $\endgroup$ – AJ_ Jul 5 '14 at 15:26
  • $\begingroup$ This is not an answer. Do you know about equilateral triangles? Right triangles? Properties of tangent lines to circles? Anything? $\endgroup$ – TZakrevskiy Jul 5 '14 at 15:29
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$PA+PB+PC$ is just three times the inradius of a triangle having side lengths $(a,b,c)=(7,8,9)$. Since $$ r(a+b+c) = 2\Delta,$$ we only need to find the area of the triangle thorugh Heron's formula. We have: $$\Delta^2 = p(p-a)(p-b)(p-c) = 12\cdot 3\cdot 4\cdot 5,$$ hence $\Delta=12\sqrt{5}$ and: $$PA+PB+PC = 3r=\frac{6\Delta}{a+b+c} = 3\sqrt{5}.$$

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