0
$\begingroup$

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 $t_1$?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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}$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.