The Triangle Inequality

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Learning goals

Prove why two sides of a triangle together exceed its third side
Classify positive side triples, distinguishing triangles from flat boundary arrangements
Determine strict third-side ranges, including restrictions on variable lengths
Compare direct and broken paths using triangle and polygon inequalities

When three sides can close

Two sides of a triangle together are longer than the third side. For lengths $5$ , $6$ , and $8$ , the two shorter rods have a combined length of $11$ , leaving room to bend while their free ends still reach the endpoints of the $8$ -unit base.

Lengths 5, 6, and 8 close in two reflected positions. The dashed lower position has the same lengths as the solid upper position.

Now keep the rods of lengths $5$ and $6$ , but use a base of length $11$ . The rods meet only when laid along the base. This is a degenerate triangle, a flat boundary arrangement with collinear vertices. An ordinary, nondegenerate triangle has three noncollinear vertices and positive side lengths.

At equality, the rods meet on the base. There is no triangular region.

With a base of length $12$ , even the fully stretched rods leave a gap of $1$ . Bending them cannot make them reach farther.

The rods fail to reach even when stretched along the base. This is an attempted assembly, not a triangle.

The triangle inequality says that in every nondegenerate triangle, the sum of any two side lengths is strictly greater than the third. Here $AB$ names the length of segment $AB$ . For triangle $ABC$ :

AB+AC>BC.

Also,

AB+BC>AC.

And,

AC+BC>AB.

Why the inequality is strict

In the previous lesson, a larger angle in one triangle faced a longer side. Equal sides also gave equal opposite angles in an isosceles triangle.

We can turn a sum of two sides into one straight segment by extending a side. That gives a comparison inside one larger triangle.

Two sides together exceed the third side#

First follow a numerical instance. Extend $BA$ beyond $A$ to $D$ , copying $AC$ onto $AD$ , and join $D$ to $C$ . In an instance where angle $ACD$ is $35^\circ$ and angle $BCA$ is $50^\circ$ , the equal sides $AD$ and $AC$ make angle $ADC$ measure $35^\circ$ too.

B, A, and D lie in that order. The single ticks mark the copied lengths AC and AD. The dashed segments are auxiliary additions.

The whole angle at $C$ is

m\angle BCD=50^\circ+35^\circ.

Thus it measures $85^\circ$ , larger than the $35^\circ$ angle at $D$ . In triangle $BCD$ , side $BD$ is therefore longer than $BC$ . But $BD$ consists of $BA$ followed by a copy of $AC$ , giving the desired sum.

Now take any nondegenerate triangle $ABC$ . On ray $BA$ beyond $A$ , choose $D$ so that

AD=AC.

The order is $B,A,D$ , and $C$ is off their line. Join $DC$ .

The general proof retains the copied lengths and point order, with no special angle measures.

Triangle $ACD$ is isosceles, so

m\angle ACD=m\angle ADC.

Ray $DA$ coincides with ray $DB$ , so angle $ADC$ is angle $BDC$ . Also, ray $CA$ lies inside angle $BCD$ : the segment from vertex $C$ to the interior point $A$ of side $BD$ splits that angle. The other part, angle $BCA$ , is positive because $ABC$ is nondegenerate, so

m\angle BCD>m\angle ACD.

Therefore angle $BCD$ is greater than angle $BDC$ . Apply the larger-angle result in triangle $BCD$ :

BD>BC.

Segment addition and the copied length give

BD=BA+AD.

Hence,

BA+AC>BC.

Nothing depended on the numbers $35$ and $50$ . The strict step used a positive angle at $C$ , which every nondegenerate triangle has. Relabeling the vertices proves the other two side-sum inequalities.

Check your understanding

In the extension proof, B, A, and D lie in that order, and AD is a copy of AC. Which fact makes angle BCD strictly larger than angle BDC?

A A is the midpoint of BD B DC is perpendicular to BD C Angle BCD contains angle ACD and a positive extra angle D BC is assumed shorter than BD

Test a proposed set of lengths

Positive lengths make a triangle exactly when each pair adds to more than the remaining length. To see why lengths passing the test can close, join the two shorter rods at a movable joint. We accept the geometric continuity fact that as the joint turns, the distance between the free ends takes every value from their difference when folded back to their sum when straight. Every distance strictly between those endpoints occurs with a bend. Because both rods are positive, their difference is strictly less than the longer rod. The longest requested length is at least as long as that rod, and passing the longest-side test puts it below their sum. It therefore lies strictly between the endpoints and can be fitted between the free ends.

The same 4-unit and 7-unit rods join at J in every row. Their free ends A and B are 3 units apart when folded, 9 in this bent position, and 11 when straight. The folded rods overlap along one line.

For a numerical test, first identify a longest length. You need check only whether the other two add to more than that length. The other sums automatically exceed a smaller length because every length is positive.

Worked example 1 Distinguish a triangle, a flat boundary, and a gap

Problem. Classify each proposed side triple as a nondegenerate triangle, a flat boundary arrangement, or an impossible assembly. All lengths are in the same unit.

Case A: $5$ , $9$ , $12$ .

Case B: $4$ , $6$ , $10$ .

Case C: $3$ , $7$ , $12$ .

Solution. Case A: Compare the sum of the two shorter lengths with the longest:

5+9>12.

The lengths make a nondegenerate triangle.

Case B: The comparison is equality:

4+6=10.

The pieces can only meet flat, so they do not make an ordinary triangle.

Case C: The pieces are too short even when stretched straight:

3+7<12.

They cannot close.

Case A bends, Case B is flat, and Case C leaves a gap. These drawings illustrate the comparisons in the solution.

The positivity condition matters. A zero or negative proposed side fails before any triangle test. Also, checking an arbitrary pair can miss the problem: lengths $2$ , $5$ , and $9$ satisfy the comparison of $5+9$ with $2$ , but the two shorter lengths cannot span $9$ .

Check your understanding

Which positive side triple makes a nondegenerate triangle?

A 4, 8, 12 B 6, 8, 13 C 2, 3, 7 D 5, 5, 11

Find the whole third-side range

The third-side range lists every possible length for a third side when the other two are fixed. With sides $7$ and $12$ , the third side must be less than their sum, $19$ . It must also be greater than their difference, $5$ , so the $7$ -unit side and the proposed third side together exceed the $12$ -unit side.

Worked example 2 Keep both strict bounds

Problem. Two sides of a triangle measure $7$ centimeters and $12$ centimeters. Find the range of the third side $x$ , and list the possible integer values.

Solution. The sum comparison gives the upper bound:

x<7+12.

The inequality involving the longest fixed side gives the lower bound:

7+x>12.

The remaining comparison, $x+12>7$ , holds automatically for every positive $x$ .

Together these give

5<x<19.

The integer lengths are $6$ , $7$ , $8$ , $9$ , $10$ , $11$ , $12$ , $13$ , $14$ , $15$ , $16$ , $17$ , $18$ centimeters. Neither endpoint belongs: $5$ or $19$ would make a flat arrangement.

Open endpoints exclude the two flat boundaries.

For any positive fixed lengths $a$ and $b$ , the same three comparisons reduce to

|a-b|<x<a+b.

The absolute value means the larger fixed length minus the smaller. If the fixed lengths match, the lower bound is zero, still excluded because a side must be positive.

Worked example 3 Let all three lengths depend on a variable

Problem. Proposed side lengths are $2t+1$ , $t+4$ , and $8$ units. Find every real value of $t$ for which they make a nondegenerate triangle.

The proposed sides carry expressions in t. A sketch alone cannot validate every value of t.

Solution. Since the longest expression may change, check all three inequalities.

First comparison:

(2t+1)+(t+4)>8.

This simplifies to

t>1.

Second comparison:

(2t+1)+8>t+4.

This gives

t>-5.

Third comparison:

(t+4)+8>2t+1.

This gives

t<11.

All three hold exactly when

1<t<11.

Throughout this interval, all side expressions are positive.

At the lower endpoint, $1$ , the lengths are $3$ , $5$ , and $8$ , which are flat.

At the upper endpoint, $11$ , they are $23$ , $15$ , and $8$ , also flat.

Check your understanding

Two sides have lengths 9 and 14 units. What is the complete range of possible third-side lengths x?

0<x<23

5\le x\le23

9<x<14

5<x<23

Direct segments and broken paths

A broken path is a route made from consecutive straight segments. Between distinct points $A$ and $B$ , a two-segment route through a point $P$ off line $AB$ is strictly longer than the direct segment:

AP+PB>AB.

This is the triangle inequality in triangle $APB$ .

If $P$ is between $A$ and $B$ , segment addition gives equality.

If $P$ is on their line beyond either endpoint, the path doubles back and is longer. Collinearity by itself does not guarantee equality.

The upper route bends off AB. The middle route travels forward along the segment and ties the direct distance. The lower route passes B and returns from P. Its dashed arrows below the line show travel along that same line.

The same comparison works for any finite number of legs. For instance, if the three legs from $A$ through $B$ and $C$ to $D$ measure $4$ , $6$ , and $8$ , first replace $AB$ and $BC$ by $AC$ . That replacement cannot increase the length. Then replace $AC$ and $CD$ by $AD$ , giving a direct distance at most $18$ .

This bent 4, 6, 8 route illustrates two shortcuts. Solid segments show each replacement; dashed segments show the route being replaced. In the second panel, faint guides retain the original first two legs. The general bound also allows forward straight routes that give equality.

In general, repeated applications of the two-leg result give

AD\le AB+BC+CD.

Equality holds exactly when the intermediate points lie on the direct segment in travel order, without backtracking. A turn off the segment or a reversal makes the path longer. We allow the non-strict symbol here because a path may include straight, forward steps.

A polygon inequality compares one side of a closed polygon with all the other sides together. Traveling around those other sides gives a broken path between the chosen side’s endpoints. A nondegenerate polygon cannot have all its vertices on one line, so that route cannot be the same forward straight segment, and its length strictly exceeds the chosen side.

Worked example 4 Bound a polygon side without measuring it

Problem. A nondegenerate quadrilateral $ABCD$ has three consecutive sides of lengths $4$ , $7$ , and $6$ units, in that order. Give a strict upper bound for side $DA$ . Explain whether a claimed length of $17$ units is possible.

The three labeled sides form a route from A to D. The fourth side closes that route.

Solution. The route around the other three sides has length

4+7+6=17.

The polygon inequality gives

DA<17.

A claimed length of $17$ would require the route from $A$ through $B$ and $C$ to $D$ to lie straight in travel order. That would be flat, so it cannot describe the stated quadrilateral.

Check your understanding

A route goes from A to P to Q to B with positive leg lengths. Which condition makes its total length equal to AB?

A P and Q lie on segment AB in the travel order A, P, Q, B B P and Q lie anywhere on line AB C The first and last legs have equal lengths D P and Q lie on opposite sides of line AB

The Triangle Inequality

Learning goals

When three sides can close

Why the inequality is strict

Test a proposed set of lengths

Find the whole third-side range

Direct segments and broken paths

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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