There is a triangle $ABC$. A line bisects the angle at the vertex $A$ and cuts the side $BC$ in point $D$ such that $BD=9$ and $DC=12$. If $O$ is the center of the circle that is inscribed inside the triangle $ABC$ and $AO:OD=4:3$, find the perimeter of the triangle $ABC$.

This seems very hard to me.

edit: I even drew one bad picture of this problem:



closed as off-topic by Namaste, Saad, Leucippus, Will Fisher, Xander Henderson Jun 21 '18 at 2:49

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  • $\begingroup$ A picture would be nice $\endgroup$ – lhf Jun 18 '18 at 21:50
  • $\begingroup$ @lhf Here it comes. $\endgroup$ – Hanlon Jun 18 '18 at 21:55
  • $\begingroup$ Your attempts? The bisector theorem gives many informations. $\endgroup$ – Jack D'Aurizio Jun 18 '18 at 21:58
  • $\begingroup$ @lhf Here. I'm sorry, but I couldn't position $O$ at the center of the circle. Just keep that in mind, and otherwise I think it's a good "approximation". $\endgroup$ – Hanlon Jun 18 '18 at 22:03
  • $\begingroup$ @JackD'Aurizio None. I'm not familiar with such a theorem. $\endgroup$ – Hanlon Jun 18 '18 at 22:05

Since $AO$ is a bisector of $\Delta ADC$, we obtain:$$\frac{AC}{DC}=\frac{AO}{OD}$$ or $$\frac{AC}{12}=\frac{4}{3},$$ which gives $AC=16$.

Now, since $AD$ is a bisector of $\Delta ABC$, we obtain: $$\frac{AB}{AC}=\frac{BD}{DC}$$ or $$\frac{AB}{16}=\frac{9}{12},$$ which gives $AB=12$ and the perimeter is $$21+16+12=49.$$

  • $\begingroup$ Hello. Can you please explain why is $\frac{AC}{DC}=\frac{AO}{OD}$ (not in the comments, it would be better to edit the answer)? I thought that I understand it but it turned out that my understanding was wrong. $\endgroup$ – Hanlon Jun 22 '18 at 14:52
  • $\begingroup$ @Hanlon I added something. See now. $\endgroup$ – Michael Rozenberg Jun 22 '18 at 17:44
  • $\begingroup$ I don't see how $AO$ is a bisector of $\triangle ADC$ (at least not from my picture). It's obviously a part of the side $AD$. $\endgroup$ – Hanlon Jun 22 '18 at 18:19
  • $\begingroup$ $O$ is a center of the inscribed circle of $\Delta ABC$,, which is a common point of the triangle bisectors. Yes, your picture is not good. $\endgroup$ – Michael Rozenberg Jun 22 '18 at 18:29
  • $\begingroup$ Thank you, I understand now. $\endgroup$ – Hanlon Jun 22 '18 at 18:33

I am not going to give the complete answer, but enough so that you can solve this yourself.

  1. You already know BC (=BD+DC). So, you need to find AB and AC to find perimeter.

  2. Angle bisector theorem should give you the ratio of AB:AC

  3. The incenter divides AD in the ratio (AB+AC):BC.

(2) and (3) should give you the 2 equations for your to solve the problem.


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