# Triangle inequality

• Mar 4th 2011, 12:25 PM
thalesgil1
Triangle inequality
Please, guys, I would aprecciate any help with this problem:
Prove that the sum of the distances of any point within a triangle, to its vertices, is smaller than the triangle perimeter. I know that I have to use the triangle inequality, but I am having a hard time trying to prove the result (which seems quite obvious).
Thanks a lot.
• Mar 4th 2011, 03:16 PM
scounged
Ok, here's what to do.

First, draw up a triangle, like this one:
Attachment 21047

After viewing the picture, we can show that \$\displaystyle b+c>y+z\$ because the point where y and z meet is closer to the side a than the point where b and c meet.

In a similar fashion, we can show that
\$\displaystyle a+c>x+y\$
\$\displaystyle a+b>x+z\$

After that, just add the three inequalities together, and simplify.
• Mar 4th 2011, 04:32 PM
thalesgil1
Hello scounged! Thanks for the reply.
I know that it is quite obvious that b + c > y + z , but is there a more formal way to prove it?
Thanks again.
• Mar 4th 2011, 06:41 PM
scounged
What do you mean by "more formal way"?
• Mar 4th 2011, 07:19 PM
Plato
Quote:

Originally Posted by scounged
What do you mean by "more formal way"?

How do you prove the following. No diagrams allowed.
If P is any point interior to \$\displaystyle \triangle ABC\$, letting
\$\displaystyle d_X\$ be the distance from P to vertex X that \$\displaystyle b+c>d_C+d_B~?\$
That is what you claimed. You said the we could see that.
But how do you prove that? No diagrams allowed.
• Mar 5th 2011, 07:57 AM
thalesgil1
Yes, Plato! that's exactly what I meant by "more formal way". I agree with you, scounged, I can see, by looking the picture, that b + c > y + z , but I am afraid that only telling "because the point where y and z meet is closer to the side a than the point where b and c meet" wouldn't be a proof (maybe I'm wrong). Is there any theorem for that?
• Mar 5th 2011, 04:28 PM
LoblawsLawBlog
What scounged said is Euclid I.21, which should be easy to find online.. I don't know if you can consider any proof of it "rigorous" according to modern standards without going through quite a bit of preliminary effort first. I know in Hartshorne's Geometry:Euclid and Beyond, this is done by using Hilbert's axioms and maybe a couple more. Whether this is absolutely necessary for this problem, I have no idea.

In any case, whether scounged's suggestion is acceptable or not really depends on the class and what theorems you've proved so far.
• Mar 5th 2011, 05:01 PM
scounged
I think that there's an easy way to prove that \$\displaystyle b+c>d_C+d_B\$, using the law of cosines. But I don't know. Is there a theorem that states something along these lines?

\$\displaystyle \mbox{If}~\mid{a}\mid^x+\mid{b}\mid^x>\mid{c}\mid^ x+\mid{d}\mid^x~\mbox{then}~\mid{a}\mid^{x+n}+\mid {b}\mid^{x+n}>\mid{c}\mid^{x+n}+\mid{d}\mid^{x+n}\$
• Mar 6th 2011, 06:40 AM
thalesgil1
Thank you guys so much for the help. I finally found a proof for the Euclid I.21, which directly leads to the proof that the sum of the distances of a point inside a triangle to the vertices is less than the perimeter! I was having nightmares with this problem for days. Here it is: draw the triangle ABC. Let D be a point inside de triangle. Draw BD through toE (where it meats side AC). In triangle ABE, AB + AE > BE . Add EC to each side: AB + AE + EC > BE + EC , so AB + AC > BE + EC (1) . In triangle CED: EC + ED > DC . Add DB to each side: EC + BE > DC + DB (2) . Now, (1) and (2) give us AB + AC > BE + EC > DC + DB . So AB + AC > DC + DB . If we do this with the other sides of the triangle we can show that AD + DB + DC < AB + BC + AC .