3
$\begingroup$

Prove that the sum of distances from any point in the interior of a triangle to three vertices of the triangle is less than the sum of two larger sides of the triangle

$\endgroup$
7
  • $\begingroup$ What are your thoughts on this problem? $\endgroup$
    – Sawarnik
    May 18 '14 at 20:30
  • $\begingroup$ We can't just help you if we don't know how much you've done. $\endgroup$
    – Jason Chen
    May 18 '14 at 20:47
  • 1
    $\begingroup$ I understand...I was away from the computer. Let me type in what I have. $\endgroup$
    – mepinon
    May 18 '14 at 20:49
  • $\begingroup$ I have this: Given triangle ABC with point D in the interior then AD+BD+CD > (AB+BC+AC)/2. Then from here, I'm going to say that AB and BC are the two largest sides. But I don't exactly know where to go from here. Still working on this. $\endgroup$
    – mepinon
    May 18 '14 at 20:54
  • 1
    $\begingroup$ @mepinon but the problem asks for an upper bound not a lower bound... $\endgroup$
    – Fermat
    May 18 '14 at 21:31
1
$\begingroup$

Let us prove that $$PA+PB+PC\lt AB+BC$$ always holds for any point $P$ inside $\triangle{ABC}$ satisfying $AB\ge BC\ge CA$.

Proof :

Consider an ellipse whose foci are $A,C$ that passes through $P$. Let $Q,R$ be the intersection point between the ellipse and the side $AB,BC$ respectively. Here, we have $$PA+PC=QA+QC=RA+RC.$$

So, if $P$ is on the ellipse, we have that $PA+PC$ is constant and that $PB$ is max when $P$ is either on $Q$ or on $R$.

Hence, we have $(1)$ or $(2)$ :

$$PA+PB+PC\lt QA+QB+QC=AB+QC\tag1$$ $$PA+PB+PC\lt RA+RB+RC=BC+RA\tag2$$

For $(1)$, we have $QC\lt BC$ because $$\angle{QBC}\le\angle{QAC}\lt\angle{QAC}+\angle{ACQ}=\angle{BQC}.$$

For $(2)$, we have $RA\lt AB$ because $$\angle{ABR}\le\angle{ACR}\lt\angle{ACR}+\angle{RAC}=\angle{ARB}.\quad\blacksquare$$

$\endgroup$

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.