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Prove that the sum of the distances of any point within a triangle , from its vertices, is smaller than the perimeter of the triangle .(As shown in the following figure )Prove that : AP + PC + PB < AB + BC + AC . ( Point P lying anywhere within the triangle ).

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6 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										Dear student,
Lets take a different view of this problem muh more interesting and relevant as far as JEE is concerned...
Fermat point: Another center is the Fermat point 		(also called the isogonic center or the Rorricelli point). This is the point 		which minimizes the sum of the distances from the three vertices. This problem 		is called Fermat's problem or Steiner's problem. The three angles at this 		point, between the vertices, are all 120 degrees. This point has many practical 		uses, as it is often a good idea to minimize distances, so save money. 		Telephone lines between three cities, using the minimum amount of wire, will be 		three lines which meet at the Fermat point. More points result in connections 		between Fermat points. The Fermat point can be constructed by constructing 		equilateral triangles on each side of the triangle, and connecting their 		farthest vertices with the opposite vertex of the original triangle.

All the best.
Win exciting gifts by                                                      answering the     questions    on            Discussion        Forum.    So      help              discuss         any                query   on        askiitians      forum    and         become  an       Elite            Expert      League                askiitian.

Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
6 years ago
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