On the sides $ BC,CA,AB $ of a triangle $ABC$ semicircles $c_1,c_2,c_3$ are described externally.
If $t_1,t_2,t_3$ are the lengths of common tangents of $c_2,c_3;\;c_3,c_1$ and $c_1,c_2$ then $t_1t_2t_3$ in terms of semiperimeter and area of triangle is?

If we are able to find $t_1$ in terms of sides then $t_2 ,t_3$ will also be found similarly. We can then use relation $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$.
But how to find $t_1$?

  • $\begingroup$ Hint: Connect the endpoints of $t_1$ to the midpoints of the adjacent sides of the triangle, and you'll have three sides of a right trapezoid. $\endgroup$ – Blue Jul 19 at 6:22
  • $\begingroup$ Hint: The common tangent between 2 circles whose centers are distance D apart is .... $\endgroup$ – Calvin Lin Jul 19 at 6:26

From the hints,consider evaluating $t_1$,I found the right angled triangle with sides $t_1, \frac{(b-c)}{2}$ and hypotenuse $\frac{a}{2}$. so by pythagoras' theorem

$(t_1)^2=(\frac{a}{2})^2-(\frac{(b-c)}{2})^2$ . so $t_1=\frac{(a-b+c)(a+b-c)}{4}=(s-b)(s-c)$.

Similarly $(t_2)^2=(s-c)(s-a)$,


so $t_1t_2t_3=(s-a)(s-b)(s-c)$ $\Longleftrightarrow$ $ \frac {\Delta^2} {s}$

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