Triangle Inequality


The triangle inequality is one of the simplest geometric inequalities. It conveys a very simple idea, that in (Euclidean) space, the straight line distance is the shortest path between two points. Specifically, if a,b a, b and c c are side lengths of a triangle, then a+bc a+b \geq c. The inequality is strict if the triangle is non-degenerate (has non-zero area).

There are several proofs of the triangle inequality; one proof uses AM-GM, while another uses cosine rule and the fact that cosθ1 \cos \theta \leq 1.

There are several useful consequences of the triangle inequality:

A. In a (non-degenerate) convex quadrilateral ABCD ABCD, AB+CD<AC+BD. AB+CD < AC+BD.

Proof: Let AC AC intersect BD BD at P P. From the triangle inequality on APB APB and CPD CPD, ABAP+PB AB \leq AP + PB and CDCP+PD CD \leq CP + PD. Thus, AB+CDAP+PB+CP+PD=AC+BD. AB + CD \leq AP + PB + CP + PD = AC + BD. Equality holds if and only if APB APB and PCD PCD are straight lines, which forces ABCD ABCD to be a straight line.


B. If P P is a point in the interior of triangle ABC ABC, then AP+PC<AB+BC. AP+PC < AB+BC.

Proof: Extend AP AP to intersect BC BC at D D. From the triangle inequality on PDC PDC and ADB ADB, we get PCPD+DC PC \leq PD + DC and ADAB+BD AD \leq AB + BD. Then, AP+PCAP+PD+DC=AD+DCAB+BD+DC=AB+BC. AP + PC \leq AP +PD + DC =AD + DC \leq AB+BD+DC = AB+BC. Equality holds if and only if PDC PDC and ADB ADB are straight lines, which forces P=D=B P=D=B.

Worked Examples

1. If the lengths of a non-degenerate triangle are 9 and 15, what are the possible lengths of the third side?.

Let the third side have length x>0 x > 0. The 3 sides must satisfy x<9+15,15<x+9 x < 9 + 15, 15 < x + 9 and 9<x+15 9 < x + 15. The third inequality is trivially true, and the first two give 6<x<24 6 < x < 24 .


2. If P P is a point in the interior of triangle ABC ABC, show that

AB+BC+CA2<AP+BP+CP<AB+BC+CA. \frac {AB+BC+CA}{2} < AP + BP + CP < AB+BC+CA.

Apply the triangle inequality to triangles PAB,PBC PAB, PBC and PCA PCA to obtain AP+PB>AB,BP+PC>BC AP+PB>AB, BP+PC> BC and CP+PA>CA CP+PA>CA. Adding up these 3 inequalities and dividing by 2 gives the inequality on the left.

Apply consequence B thrice to obtain AP+PB<AC+CB,BP+PC<BA+AC AP+PB < AC+CB, BP+PC < BA+AC and CP+PA<CB+BA CP+PA < CB+BA. Adding these 3 inequalities and dividing by 2 gives the inequality on the right.


3. In triangle ABC ABC, D D is the midpoint of side BC BC. Show that 2ADAB+AC. 2\cdot AD \leq AB + AC.

Extend CA CA to E E such that EA=AC EA=AC. Since CD=DB CD=DB, by Parallel lines Property D, triangles CAD CAD and CEB CEB are similar with ratio 2, which gives EB=2AD EB = 2 \cdot AD. Applying the triangle inequality to EAB EAB, we get BA+AEEB BA + AE \geq EB . Substituting the previous equations give AB+AC=BA+AEEB=2AD AB+AC = BA+AE \geq EB = 2\cdot AD. Equality holds if points E,A E, A and B B are on the same line, which implies that points A,B A, B and C C are on the same line, which is a degenerate triangle.


4. All the internal diagonals of a convex polygon have equal lengths. What is the maximum number of sides that the polygon can have?

The regular pentagon clearly satisfies the condition. We will show by contradiction that any convex polygon with n6 n\geq 6 sides will not satisfy the condition. Suppose not, let P P be such a polygon. Label the vertices V1,V2,Vn V_1, V_2, \ldots V_n in (clockwise) order. Then V1V3V4V6 V_1 V_3 V_4 V_6 form a convex quadrilateral, so by consequence A, V1V3+V4V6<V1V4+V3V6 V_1 V_3 + V_4 V_6 < V_1 V_4 + V_3 V_6. But the condition given states that all the internal diagonals have equal length, so V1V3+V4V6=V1V4+V3V6 V_1 V_3 + V_4 V_6 = V_1 V_4 + V_3 V_6, which is a contradiction. Hence, n5 n\leq 5, and we know that 5 sides is attainable.

Note by Arron Kau
5 years, 7 months ago

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