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The Triangle Inequality: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Classify three attempted frames

Classify each positive side triple as a nondegenerate triangle, a flat boundary arrangement, or an impossible assembly. All lengths use the same unit. Explain each classification.

Case A: 6, 9, and 14.

Case B: 4, 11, and 15.

Case C: 2, 6, and 10.
1. Hint 1
  Compare the two shorter lengths with the longest.
2. Hint 2
  A sum equal to the longest can meet only on a line; a smaller sum cannot reach.
Answer

Case A: nondegenerate triangle. Case B: flat boundary arrangement. Case C: impossible assembly.

Full solution

Case A: The shorter lengths satisfy
$6+9>14.$
The three positive lengths make a nondegenerate triangle.

Case B: The comparison is
$4+11=15.$
The shorter rods span the longest only when laid straight.

Case C: Even the stretched rods give
$2+6<10.$
They leave a gap and cannot close.

Compare a bend, an exact flat meeting, and a gap.

Answer

Case A: nondegenerate triangle. Case B: flat boundary arrangement. Case C: impossible assembly.

Key idea

For positive lengths, the comparison with the longest side distinguishes the three closure cases.
Problem 2 Find every integer third side

Two sides of a nondegenerate triangle measure 11 and 20 centimeters. Find the full range of the third side x, then count the possible integer values. Explain what happens at each endpoint.

The drawing shows only the stated givens.

Text description of this figure

Triangle ABC has AB of length 20 centimeters, AC of length 11 centimeters, and unknown BC labeled x. No computed bound is drawn.
1. Hint 1
  The third side must be long enough to span the difference and short enough to stay below the sum.
2. Hint 2
  After finding the strict interval, start and end at integers inside it.
Answer

$9<x<31.$
There are 21 integer values, from 10 through 30 centimeters. Each endpoint gives a flat arrangement.

Full solution

The difference gives the lower bound:
$x>20-11.$
Thus x is greater than 9.

The sum gives the upper bound:
$x<20+11.$
Thus x is less than 31.

The full range is
$9<x<31.$
The integers are 10 through 30, inclusive.

Their count is
$30-10+1=21.$

At the lower endpoint, 9 and 11 add to 20, so the arrangement is flat.

At the upper endpoint, 11 and 20 add to 31, so the arrangement is flat.

The boundaries are excluded; integer lengths lie inside.

Answer

$9<x<31.$
There are 21 integer values, from 10 through 30 centimeters. Each endpoint gives a flat arrangement.

Key idea

Strict geometric endpoints must be removed before counting integer values.
Problem 3 Three expressions must fit together

Proposed side lengths are $2s+5$ , $3s-4$ , and $16$ units, where s is real. Find every value of s for which they form a nondegenerate triangle. Check that your interval makes all lengths positive.
1. Hint 1
  The longest expression can change, so compare each side with the other two together.
2. Hint 2
  Combine the three strict inequalities, then check positivity throughout the remaining interval.
Answer

$3<s<25.$
All three lengths are positive throughout this interval.

Full solution

Comparing the first two lengths with 16 gives
$(2s+5)+(3s-4)>16.$
Hence
$s>3.$

Comparing the first length and 16 with the second gives
$(2s+5)+16>3s-4.$
Hence
$s<25.$

The remaining comparison is
$(3s-4)+16>2s+5.$
It simplifies to
$s>-7.$

All comparisons therefore hold precisely when
$3<s<25.$

In this interval, the first expression exceeds 11 and the second exceeds 5, while the last side is 16.
All are positive.

At s equal to 3, lengths 11 and 5 add to 16.
This lower boundary is flat.

At s equal to 25, lengths 55 and 16 add to 71.
This upper boundary is also flat.

Answer

$3<s<25.$
All three lengths are positive throughout this interval.

Key idea

Keep all pair comparisons when you cannot fix the longest side in advance.
Problem 4 Recover all side triples from a perimeter

A nondegenerate triangle has positive integer side lengths and perimeter 25 units. Its shortest side has length 4 units. Find every unordered side triple that is possible, and explain why your list is complete.
1. Hint 1
  Name the larger of the two remaining sides and express the smaller one using the perimeter.
2. Hint 2
  The largest side must be less than the other two combined; its size also reflects the order of the remaining sides.
Answer

The unordered triples are 4, 9, 12 and 4, 10, 11, in units.

Full solution

Let the remaining sides be a and b, with a no larger than b.
The perimeter gives
$a+b=21.$

Since b is at least as large as a, b is at least 10.5.
As an integer, b is at least 11.

The triangle inequality for the largest side gives
$4+(21-b)>b.$
Thus
$b<12.5.$

The only integer choices for b are 11 and 12.

Case b of 11: The remaining side a is 10.

Case b of 12: The remaining side a is 9.

Both pairs leave 4 as the shortest side, and their shorter lengths exceed the longest when added.
The two triples are therefore valid and exhaustive.

One of the two triples found by the solution: 4, 10, 11 units.

Answer

The unordered triples are 4, 9, 12 and 4, 10, 11, in units.

Key idea

A perimeter equation and a strict longest-side bound can make a complete finite list.
Problem 5 How much distance can a shortcut save?

A straight walkway from entrance A to fountain P measures 9 meters. Another straight walkway from P to gate B measures 14 meters. P is off line AB.

A new straight walkway from A to B would save s meters compared with the route through P. Find the possible range of s. Decide whether a saving of exactly 18 meters is possible.

The two existing legs meet off the direct line.

Text description of this figure

A triangular walkway has entrance A, gate B, and fountain P off line AB. AP is 9 meters, PB is 14 meters, and the direct segment AB has no numerical label.
1. Hint 1
  First bound the direct distance using the two existing walkway lengths.
2. Hint 2
  Savings equal the old route length minus the direct length; examine what happens at the lower distance boundary.
Answer

$0<s<18.$
A saving of exactly 18 meters is impossible.

Full solution

Let d be the direct distance from A to B.
The two fixed lengths require
$5<d<23.$

The existing route measures
$9+14=23.$
Savings therefore satisfy
$s=23-d.$

Because d is less than 23, the saving is positive.
Because d is greater than 5, the saving is less than 18.
Thus
$0<s<18.$

A saving of exactly 18 meters would force the direct distance to be 5 meters.
That is a flat boundary for lengths 9 and 14, contradicting P being off line AB.

Answer

$0<s<18.$
A saving of exactly 18 meters is impossible.

Key idea

A strict range for a distance becomes a strict range for the amount saved.
Problem 6 Check a third-side range

Triangle ABC has AB of length 10 units, AC of length 7 units, and BC of length x units. Inspect the four numbered lines. Identify the first incorrect line and repair the final range. Explain the lower boundary.

Line 1:
$x<17.$
Line 2:
$7+x>10.$
Line 3:
$x>3.$
Line 4:
$3\le x<17.$

The drawing shows only the stated givens.

Text description of this figure

Triangle ABC has two given sides 10 and 7 units and unknown third side x. No solved endpoint is marked.
1. Hint 1
  Check whether each new line preserves the strict comparison from the line before it.
2. Hint 2
  The difference boundary corresponds to a flat arrangement and must stay excluded.
Answer

Line 4 is the first incorrect line. The repaired range is
$3<x<17.$
The lower boundary is flat and excluded.

Full solution

Line 1 correctly keeps x below the sum of the fixed lengths.

Line 2 is one of the strict triangle inequalities.
Subtracting 7 correctly gives Line 3.

Line 4 incorrectly changes a strict lower bound into an inclusive one.
It should read
$3<x<17.$

At the lower boundary,
$3+7=10.$
The three lengths lie flat, so that boundary cannot belong to the range.

Answer

Line 4 is the first incorrect line. The repaired range is
$3<x<17.$
The lower boundary is flat and excluded.

Key idea

Algebraic simplification must preserve the strictness that geometry requires.
Problem 7 Three positions for the intermediate point

A and B are 18 units apart. In each case below, find the length of the route A to P to B. Explain which routes tie the direct distance and which are longer.

Case A: P is between A and B, and AP measures 5 units.

Case B: P is beyond B on ray AB, and BP measures 5 units.

Case C: P is beyond A on ray BA, and AP measures 5 units.

The cases differ by point order, even though all points are collinear.

Text description of this figure

Three separate rows display the given point orders: A, P, B; A, B, P; and P, A, B. No derived segment length or route total is labeled.
1. Hint 1
  Use the stated order to decide whether a segment is a whole or a part.
2. Hint 2
  A route that overshoots must travel the extra segment twice.
Answer

Case A: 18 units, equal to the direct distance. Case B: 28 units, longer. Case C: 28 units, longer.

Full solution

Case A: P lies on segment AB, so segment addition gives
$AP+PB=18.$
The route ties the direct distance.

Case B: The distance AP is
$AP=18+5.$
It is 23 units.
Returning from P to B adds 5, making the route
$23+5=28.$

Case C: The distance PB is
$PB=5+18.$
It is 23 units.
The route first goes 5 units to P, then to B, totaling
$5+23=28.$

Only Case A travels forward along the direct segment.
The other cases backtrack, each adding twice the 5-unit overshoot.

Answer

Case A: 18 units, equal to the direct distance. Case B: 28 units, longer. Case C: 28 units, longer.

Key idea

Equality needs the correct point order as well as collinearity.
Problem 8 What does a polygon bound settle?

A nondegenerate pentagon ABCDE has consecutive side lengths AB of 3 units, BC of 5 units, CD of 4 units, and DE of 6 units.

Decide whether the polygon inequality rules out each claimed length for EA. Do not assume that passing one bound proves a complete shape exists.

Case A: EA is claimed to be 18 units.

Case B: EA is claimed to be 17 units.

Use the route along the four given sides.

Text description of this figure

Pentagon ABCDE displays the four given consecutive lengths 3, 5, 4, and 6 units. The closing side EA has no length label.
1. Hint 1
  Compare the closing side with the combined length of the other four sides.
2. Hint 2
  A necessary strict upper bound can reject a value without determining a unique length.
Answer

Case A is ruled out. Case B is not ruled out by this polygon bound, which requires
$EA<18.$
This bound alone does not determine EA.

Full solution

The route from A through B, C, and D to E has total length
$3+5+4+6=18.$

Because the polygon is nondegenerate, the direct side must satisfy
$EA<18.$

Case A: A length of 18 would require the alternate route to lie on the direct segment in forward order.
That is a flat arrangement, so it is excluded.

Case B: The value 17 satisfies the strict upper bound.
This comparison therefore does not rule it out; it does not prove the full configuration or determine EA.

Answer

Case A is ruled out. Case B is not ruled out by this polygon bound, which requires
$EA<18.$
This bound alone does not determine EA.

Key idea

A necessary bound can rule out a proposal without constructing or measuring the whole figure.
Problem 9 Combine two route comparisons

A and B are distinct points. C and D lie off line AB on opposite sides. Let the lengths of the two broken routes be L and M.

The route through C has length
$L=AC+CB.$
The route through D has length
$M=AD+DB.$
Prove that
$L+M>2AB.$
Then suppose C and D are instead distinct points strictly inside segment AB. Decide whether equality can occur.

Each route can be compared with the same direct segment.

Text description of this figure

Two nondegenerate triangles share base AB, with C and D on opposite sides of its line. The upper and lower broken routes are visually distinct, without marks asserting equal side lengths.
1. Hint 1
  Apply the two-leg path comparison separately to the routes through C and D.
2. Hint 2
  Add the two strict comparisons. For the last question, replace each route by segment addition.
Answer

For the off-line points,
$L+M>2AB.$
If C and D are strictly inside segment AB, equality does occur.

Full solution

Since C is off line AB, triangle ACB is nondegenerate.
Thus
$L>AB.$

Likewise, triangle ADB is nondegenerate, so
$M>AB.$

Adding the two strict comparisons gives
$L+M>AB+AB.$
Hence
$L+M>2AB.$

If C lies strictly inside segment AB, segment addition makes L equal to AB.
If D also lies strictly inside, M equals AB too.

Their total is then
$L+M=2AB.$
The relative order of C and D does not matter, since they belong to separate routes.

Answer

For the off-line points,
$L+M>2AB.$
If C and D are strictly inside segment AB, equality does occur.

Key idea

Two comparisons with the same reference length can be added, with equality checked separately.
Problem 10 Fit one length under three constraints

A positive integer length x must be the third side of two nondegenerate triangles. All lengths are in units.

Triangle A has fixed side lengths 8 and 17.

Triangle B has fixed side lengths 13 and 24. Its perimeter must be strictly less than 53.

Find every possible x. Explain why the integer just below your smallest value and the integer just above your largest value are excluded for different reasons.

The same length must satisfy both triangle tests and the perimeter condition.

Text description of this figure

Two separate triangles use the same unknown x. Triangle A is labeled with fixed lengths 8 and 17; Triangle B with 13 and 24. No solved interval or final integer value is marked.
1. Hint 1
  Write a third-side interval for each triangle before using the perimeter condition.
2. Hint 2
  The second perimeter includes x and the sum of its two fixed lengths. Keep the overlap of all three strict restrictions.
3. Hint 3
  For the two excluded values, test whether the obstruction comes from a flat triangle or the perimeter cap.
Answer

The possible lengths are 12, 13, 14, and 15 units. The value 11 makes Triangle B flat; 16 makes its perimeter 53, which is excluded.

Full solution

Triangle A requires
$9<x<25.$

Triangle B requires
$11<x<37.$

The perimeter condition for Triangle B gives
$x+37<53.$
Thus
$x<16.$

Combining all three restrictions leaves
$11<x<16.$
The integer values are 12, 13, 14, and 15.

At x equal to 11, the shorter sides of Triangle B add to its longest:
$11+13=24.$
It is flat.

At x equal to 16, both triangle inequalities pass, but Triangle B has perimeter
$16+13+24=53.$
The strict perimeter condition excludes it.

Answer

The possible lengths are 12, 13, 14, and 15 units. The value 11 makes Triangle B flat; 16 makes its perimeter 53, which is excluded.

Key idea

The feasible set is the overlap of every condition, and excluded endpoints can have different causes.

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