1
$\begingroup$

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:

picture

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Saad, Leucippus, Will Fisher, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
1
$\begingroup$

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 $$21+16+12=49.$$

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

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.

$\endgroup$

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