1
$\begingroup$

I'm trying to solve following geometry question, but it is quite challenging.(at least for me!) Thanks for your help in advance.

On plane, there is some triangle ABC. Also, there is a point P satisfying

(1) triangles ABP,BCP and CAP have same area

(2) triangles ABP,BCP and CAP have same perimeter

Q: then, how does ABC look like?

I strongly believe that ABC must be an equilateral triangle or right-angle triangle. but proving this fact seems not so obvious. How can we approach this?

Let me write down my progress until now. we can show that every triangle ABC can have exactly for possible positions for point P satisfying condition (1) (one of them is, of course, center of mass) but problem is condition (2). how can we use same perimeter condition? we know that B,C are on some ellipse with focuses A and P. but does this mean something?

$\endgroup$
  • $\begingroup$ P needn't be inside. Doesn that mean P can be in a different plane? $\endgroup$ – Win Vineeth Feb 16 '16 at 0:54
  • $\begingroup$ @Win Vineeth no. all of them are on the same plane. $\endgroup$ – KGEO Feb 16 '16 at 1:02
  • $\begingroup$ @Win Vineeth I deleted that sentence. it is redundant sentence, I think. $\endgroup$ – KGEO Feb 16 '16 at 1:08
0
$\begingroup$

You are going in the right direction. Fact (1) implies that $P$ is the centroid of $\triangle ABC$. We can then make use of the median length formular: suppose that the edges of $\triangle ABC$ are ordered $$a\ge b\ge c,\tag{1}$$ then $$m_a\le m_b \le m_c.\tag{2}$$ Now, from (2), $\triangle ABP$ and $\triangle BPC$ has the same perimeter, that is, $$AP+AB = CP + BC,$$ or $$\frac{2m_a}3+c = \frac{2m_c}3+a.$$ From (1) and (2), it follows that $a=c$ and $m_a=m_c$. Showing $a=b$ is similar.

$\endgroup$
  • $\begingroup$ Thank you so much! why I didn't try inequality... $\endgroup$ – KGEO Feb 16 '16 at 7:08
  • $\begingroup$ That's the case where $P$ is inside $\triangle ABC$. There's a case where $P$ is outside $\triangle ABC$ which makes $\triangle ABC$ a right triangle. $\endgroup$ – Quang Hoang Feb 16 '16 at 7:15
  • $\begingroup$ Yes, I realized that and solved that case. really appreciate your help $\endgroup$ – KGEO Feb 16 '16 at 7:18

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.