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?

  • $\begingroup$ 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

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.


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