The Triangle Inequality: Practice

Comparing the first two sides with 17 gives

Hence

The other pair comparisons simplify to 15 greater than 3 and 20 greater than negative 2, which hold. All lengths are positive for v greater than 8, so there is no extra upper bound.

The pair of variable sides must exceed 16:

This gives u greater than 4. The pair involving u plus 1 and 16 gives

This gives u less than 14. The third comparison is weaker, and positivity holds throughout:

The endpoints make flat side triples.

The base must be positive:

The equal sides together must exceed the base:

These give

The comparisons of an equal side with the other side plus the base hold precisely because the base is positive.

The first triangle requires

The second requires

The common interval is the first one. Its positive multiples of 4 are 8, 12, and 16, so there are 3.

The comparison with the x side gives

Thus x is less than 20. The comparison with the 27 minus x side gives

Thus x is greater than 7. The third comparison is already true. The integers are 8 through 19, giving

These values also make all lengths positive.

Because Q, P, and S are in order,

The copied length replaces PS by PR. Thus the proved comparison with QR is

The other inequalities can be proved by relabeling, but they are not this construction's conclusion.

The three shorter sides total

A polygon side must be strictly shorter than the route around its other sides. A 13-unit closing side violates that requirement:

The proposed quadrilateral cannot exist.

The first bend makes the entire broken route strictly longer than the direct segment:

Simplify:

Hence

This is a necessary restriction from the path comparison. It does not determine the route's angles.

Write the longest side as c. The other two sides sum to p minus c, so the needed comparison is

Rearrange:

For positive lengths, checking the longest side suffices. Equality at half the perimeter is a flat boundary.

Since Q is between A and P,

Thus AQ is 5 units. Segment addition then gives

The direct distance is 16 units. The specified route has length

It travels the segment QP twice extra, adding 8 units.

Let b be the base. Each equal side has length

Integer equal sides require b to be even. The two equal sides together must exceed the base:

Thus

The even integers are 2, 4, 6, 8, 10, 12, and 14: seven possibilities. Each gives positive equal sides and satisfies the other inequalities.

The route has total length

The direct distance cannot exceed this, so 20 is impossible. The other choices occur on a number line.

Distance 19: Put A, P, Q, and B at successive positions 0, 4, 10, and 19.

Distance 11: Put them at 0, negative 4, 2, and 11.

Distance 7: Put them at 0, 4, negative 2, and 7.

The order of travel can include reversals, so collinearity alone does not force distance 19.