# Problem related with semicircles on sides of triangle and common tangents through semicircles.

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 $$t_1$$?

• 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. – Blue Jul 19 at 6:22
• Hint: The common tangent between 2 circles whose centers are distance D apart is .... – 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 $$(t_2)^2=(s-c)(s-a)$$,
$$(t_3)^2=(s-a)(s-b)$$
so $$t_1t_2t_3=(s-a)(s-b)(s-c)$$ $$\Longleftrightarrow$$ $$\frac {\Delta^2} {s}$$