SOLUTION: Prove that the sum of distances from any point in the interior of a triangle to the three vertices a, b, and c is greater than half the perimeter of the triangle.
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Question 413128: Prove that the sum of distances from any point in the interior of a triangle to the three vertices a, b, and c is greater than half the perimeter of the triangle. Found 2 solutions by richard1234, Edwin McCravy:Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! Let our interior point be P, and the three segments be PA, PB, PC, in which we want to prove that
.
By the triangle inequality, . Similarly, and . We have three inequalities:
Adding all three inequalities up, we get . We can divide both sides by 2 to get , as desired.
Let the sides of the large triangle be a, b, and c.
Let the distances to the vertices from an interior point
be x, y, and z.
Half the perimeter of the large triangle is
Use the triangular inequality on each of the three smaller triangles
that make up the large triangle:
x + y > a
x + z > b
y + z > c
Adding those three inequalities term by term:
x + y > a
x + z > b
y + z > c
--------------------
2x + 2y + 2z > a+b+c
Dividing through by 2
x + y + z >
Edwin
