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
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What are your thoughts on this problem? – Sawarnik May 18 '14 at 20:30
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We can't just help you if we don't know how much you've done. – Jason Chen May 18 '14 at 20:47
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1I understand...I was away from the computer. Let me type in what I have. – mepinon May 18 '14 at 20:49
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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. – mepinon May 18 '14 at 20:54
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1@mepinon but the problem asks for an upper bound not a lower bound... – Fermat May 18 '14 at 21:31
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While its true that $a+b>l_a+l_b$, I have a hard time believing that $a+b>l_a+l_b+l_c$. I should play with Geo-Gebra to check if there's any counterexamples. – Sawarnik May 19 '14 at 06:18
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I'm still stuck on this problem...any ideas – mepinon May 25 '14 at 22:56
1 Answers
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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 $$\angle{ABR}\le\angle{ACR}\lt\angle{ACR}+\angle{RAC}=\angle{ARB}.\quad\blacksquare$$
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