# Triangle Inequality

## Definition

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 $$a, b$$ and $$c$$ are side lengths of a triangle, then $$a+b \geq c$$. 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 $$\cos \theta \leq 1$$.

There are several useful consequences of the triangle inequality:

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

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

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

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

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

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

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

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

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

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

Extend $$CA$$ to $$E$$ such that $$EA=AC$$. Since $$CD=DB$$, by Parallel lines Property D, triangles $$CAD$$ and $$CEB$$ are similar with ratio 2, which gives $$EB = 2 \cdot AD$$. Applying the triangle inequality to $$EAB$$, we get $$BA + AE \geq EB$$. Substituting the previous equations give $$AB+AC = BA+AE \geq EB = 2\cdot AD$$. Equality holds if points $$E, A$$ and $$B$$ are on the same line, which implies that points $$A, B$$ and $$C$$ 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 $$n\geq 6$$ sides will not satisfy the condition. Suppose not, let $$P$$ be such a polygon. Label the vertices $$V_1, V_2, \ldots V_n$$ in (clockwise) order. Then $$V_1 V_3 V_4 V_6$$ form a convex quadrilateral, so by consequence A, $$V_1 V_3 + V_4 V_6 < V_1 V_4 + V_3 V_6$$. But the condition given states that all the internal diagonals have equal length, so $$V_1 V_3 + V_4 V_6 = V_1 V_4 + V_3 V_6$$, which is a contradiction. Hence, $$n\leq 5$$, and we know that 5 sides is attainable.

Note by Arron Kau
4 years, 7 months ago

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