The distances of vertices of an equilateral triangle

[tsimone75]

Homework Statement

The distances from the vertices of an equilateral triangle to an interior point P are √a, √b, and √c respectively,where a, b, and c are positive integers. Find the minimum and the maximum values of the sum a + b + c if the side of a triangle is 13.

Homework Equations

the altitude(height) of triangle is √3/2 (a), where a = the side of the triangle which is 13.
we know the altitude is a straight line from each vertex perpendicular to the opposite side and that there are three. We know that three triangles are formed. We also know that the sum of the distances from any interior point to the sides of the equilateral triangle equal the altitude

The Attempt at a Solution

I attempted one case where P is at the intersection where all the altitudes intersect. Therefore if the vertices are A, B, and C respectively, then T, S, and R are points on the opposite sides where the altitude bisects. Therefore AT= BS =CR = 13√3/2 since a = 13. Since the distance from the vertices to point P = √a, √b, and √c respectively, then PT + PS + PR = 13√3/2 or AT-√a + BS -√b + CR -√c =13√3/2. Don't know where to go from here.

 

[lippyka]

Aint:We have $AT^2=BS^2=CR^2=\frac{169}{4}$So $AT^2+BS^2+CR^2=\frac{507}{4}=a+b+c$On the other hand, consider the triangle $\Delta ABC$. The sum of its angles is $180^{\circ}$.Now you can use the cosine formula to get the maximum and minimum value for $a+b+c$.