The triangle inequality and geometric transformations

Often the triangle to which we must apply the triangle inequality does not appear in the diagram for the problem. In these cases, a suitable choices of geometric transformation can help. The following series of problems illustrate the use of symmetry together with the triangle inequality.

Problem 10. Point C lies inside a given right angle, and points A and B lie on its sides. Prove that the perimeter of triangle ABC is not less than twice the distance OC, where O is the vertex of the given right angle.

  

Figure 4: Problem 10

Proof. Reflect point C in lines OA and OB, to obtain C' and C'' (see Figure 4). It is easy to see that point O lies on straight line C'C''. Then we can replace the perimeter of triangle ABC with the sum of the lengths of segments C'A, AB and BC''. The triangle inequality tells us that this sum is no less than the length of C'C''. This, in turn, is equal to 2OC, since it is the hypotenuse of a right triangle of which OC is the median.

For teachers. It is important to solve the last r problem carefully, getting students to give a logical exposition of the solution, and not just an intuitive explanation. First we can remind the students that line reflection does not change distances. Then we can point out the common idea in such kind of problems: to transform the required path so that the length doe not change, and so that the problem becomes one of the connecting two points with the shortest path possible.

Problem 11. A fly sits on one vertex of a wooden cube. What is the shortest path it can follow to the opposite vertex.

Hint. Unfold the cube and use the fact that the shortest distance between two points is the straight line. Folding the cube back will give you the answer. Try to make a paper model of this problem.

Problem 12. Prove that the sum of the distances from point O to the vertices of a given triangle is less than the perimeter, if point O lies inside the triangle. What if point O is outside the triangle?

Hint. It is enough to prove that AO+OC<AB+BC. To prove it extend segment AO to intersect with the side BC at point D and use the triangle inequalities AB+BD>AD and OD+DC>OC.