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.
Ok, here's what to do.
First, draw up a triangle, like this one:
After viewing the picture, we can show that 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
After that, just add the three inequalities together, and simplify.
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?
What do you mean by "more formal way"?
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?
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.
I think that there's an easy way to prove that , using the law of cosines. But I don't know. Is there a theorem that states something along these lines?
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 .