Triangle with inscribed circle

klawesyn28
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P is a point inside triangle ABC. In the triangle there is inscribed
circle which radius is greater than 1. Prove that PA>2, PB>2 or PC>2.


I don't know how to solve it. Could anybody help me?
 
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Did you at least start the problem? If so then post what you have and we will help you from there.
 
Yes I started with this problem and then I don't know how to prove that when PA<=2, PB<=2, PC<=2, the circumradius is smaller than 2. P is point inside the triangle.
 
klawesyn28 said:
P is a point inside triangle ABC. In the triangle there is inscribed
circle which radius is greater than 1. Prove that PA>2, PB>2 or PC>2.
I don't know how to solve it. Could anybody help me?

This is one of those cases where the first step can be tough. You haven't said what kind of class this is for and so it's hard for us to figure out what kind of direction you're supposed to go in.

With that, here are some things you might think about.

(a) It would be really handy if you had a formula for the minimum that you could differentiate. It might be useful for you to know that:

min(A,B) = (|A+B| - |A-B|)/2

(b) Do you think you could do something with the area formula for a triangle as compared to the area formula for the circle? Can you write an area formula that is defined around the point P? Perhaps the areas of the three triangles PAB, PBC, PCA?

Carl
 
It's task from russian books with task from geometry. I don't see how your hints can help me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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