2
$\begingroup$

If $I_1$ and $I_2$ and $I_3$ be the centres of the escribed circles of $\triangle ABC$ and if $R_1$, $R_2$ and $R_3$ are radius of the circles inscribed in the triangles $\triangle BI_1C , \triangle AI_2C, \triangle AI_3B $, then prove that

$$R_1 : R_2 : R_3 =\sin(A/2):\sin(B/2):\sin(C/2).$$

I got $$ R_3 = 8R\sin(C/2) \cos(A/2)\cos(B/2)\cos(C/2) $$ But I am not sure whether it is correct or not.

Can anyone help me?

$\endgroup$
  • $\begingroup$ I tried to post it using latex syntax but failed to do that...can anyone tell me where is the latex option?? $\endgroup$ – Vamsi Spidy Oct 14 '15 at 10:13
  • $\begingroup$ Here is a tutorial for writing math on the site. $\endgroup$ – G-man Oct 14 '15 at 12:35
  • $\begingroup$ In your edit you replaced S with C, so what role does the circumcentre now play? $\endgroup$ – G-man Oct 14 '15 at 17:00
  • $\begingroup$ @G-man no role does the circumcentre play.......the question is printed wrong in sl loney book.....i corrected it $\endgroup$ – Vamsi Spidy Oct 15 '15 at 4:21
0
$\begingroup$

I think there is a mistake in the question. The given result does not hold for the inscribed circles in those triangles but for the circumcircles.

It's really simple actually. Just draw $\triangle ABC$ and all its exterior and interior angle bisectors. Let the exterior ones meet at $I_1,I_2$ and $I_3$ . Doing some angle chasing you'll find that $\angle BI_1C=90^{\circ}-\frac A2$. Now if $R_1$ is the circumradius of $\triangle BI_1C$ then according to the sine law:$$R_1=\frac{BC}{2\sin\angle BI_1C}=\frac{a}{2\cos \frac A2}$$.

Also we know that $a=2R\sin A$ so just apply that in that to finally get $R_1=2R\sin \frac A 2$. After that there's not much left to do.

$\endgroup$

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.