Triangle inequality

Although it is easily motivated, triangle inequality is extremely useful. There are many non-trivial applications involving the triangle inequality. The inequality itself states that for any triangle ABC we have the following three inequalities

AB<AC+BC, AC<BC+AB, BC<AB+AC,

showing that any side of the triangle is less than the sum of two others. Also if we have an equality for one of them (for example AC=BC+AB) then the points A, B and C are on the same line.

Problem 4. Prove that the length of any side of a triangle is not more than half of its perimeter.

Solution. Straitforward application of the triangle inequality. If the sides of the triangle are a, b and c, then then we have that b+c>a. Adding a to each side, we find a+b+c>2a.

Problem 5. Let A, B, C and D are four points. If we have that the distance between A and B is 6, between A and C is 3, between B and D is 1 and between C and D is 2, find the distance between A and D

Solution. Although such a problem can't be solved in the general case, in this case we have that AC+CD+DB=AB. But then all the points are on the same line - just use the fact that the sum of any three sides of a quadrilateral is greater than the fourth side (use two times the triangle inequality). So AD=AC+CD=5

Note that in solving Problem 5, we used a slight generalization of the triangle inequality (quadrilateral inequality). In fact, for any polygon, the sum of all but one of the sides is greater than the remaining side. If you are a teacher, you can ask the students to give a formal proof of this fact, using induction.

Problem 6. Point O is given on the plane of the square ABCD. Prove that the distance from O to one of the vertices of the square is not greater than the sum of the distances from O to the other three vertices.

Solution. Lets show that OB+OC+OD>OA (see Figure 1). Adding the triangle inequalities AC+OC>OA and OB+OD>BD, we find that AC+OB+OC+OD>OA+BD=OA+OC. Is there any difference if the point O is inside or outside the square?

  

Figure 1: Problem 6

Problem 7. Prove that the sum of the diagonals of a convex quadrilateral is less than the perimeter but more than half the perimeter.

Solution. Suppose the diagonals of the quadrilateral ABCD intersect at O (see Figure 2). Then AB+BC>AC, BC+CD>BD, CD+AC>AC and AD+AB>BD. Adding, we obtain 2(AB+BC+CD+DA)>2(AC+BD), so the first part is proved. For the second part see that OA+OB>AB, OB+OC>BC, OC+OD>CD and OD+OA>AD. So 2(OA+OB+OC+OD)=2(AC+BD)>AB+BC+CD+DA.

  

Figure 2: Problem 7

Problem 8. Prove that the distance between any two points inside a triangle is not greater than half the perimeter of the triangle.

Solution. If the points are X and Y, we extend the segment connecting them in both directions, until it intersects the sides AB and AC in E and F respectively (see Figure 3). Then EF<EA+AF and EF<EB+BC+CF. Adding, we find that EF is less than half the perimeter of the triangle. Since XY<EF, XY is also less than half the perimeter.

  

Figure 3: Problem 8

Problem 9. Point A, inside an acute angle, is reflected in either side of the angle to obtain points B and C. Line segment BC intersects the sides of the angle at D and E. Show that BC/2>DE

Hint. Use Problem 4 and the fact that BC=AD+DE+EA