The problem describes a triangle PQR with specific properties related to its incircle and angles. We are given that P is the largest angle, and \( \cos P = \frac{1}{3} \). The incircle touches the sides at points N, L, and M on PQ, QR, and RP respectively. The lengths of the tangent segments from the vertices are given as PN = \(n\), QL = \(n+2\), and RM = \(n+4\), where \(n\) is an integer.

Understanding Incircle Tangent Properties

A key property of the incircle is that the lengths of tangents from a vertex to the incircle are equal. Therefore:

• Tangent segments from P: PN = PM = \(n\)
• Tangent segments from Q: QN = QL = \(n+2\)
• Tangent segments from R: RL = RM = \(n+4\)

Calculating Side Lengths

The sides of the triangle are formed by the sum of these tangent segments:

• Side PQ (opposite angle R) = PN + NQ = \(n + (n+2) = 2n+2\)
• Side QR (opposite angle P) = QL + LR = \((n+2) + (n+4) = 2n+6\)
• Side RP (opposite angle Q) = RM + MP = \((n+4) + n = 2n+4\)

Let the side lengths be \(r = PQ = 2n+2\), \(p = QR = 2n+6\), and \(q = RP = 2n+4\).

Identifying Smallest and Largest Sides

We are given that P is the largest angle. In a triangle, the largest angle is opposite the largest side. The side opposite angle P is QR, which has length \(p = 2n+6\). Comparing the side lengths \(2n+2\), \(2n+4\), and \(2n+6\), it is clear that \(2n+2 < 2n+4 < 2n+6\). Thus, the side lengths are \(2n+2\), \(2n+4\), and \(2n+6\) in increasing order. The smallest side has length \(2n+2\).

Using the Law of Cosines

We can relate the side lengths and the cosine of angle P using the Law of Cosines: \[p^2 = q^2 + r^2 - 2qr \cos P\] Substitute the expressions for \(p\), \(q\), \(r\) and the given value \( \cos P = \frac{1}{3} \): \[(2n+6)^2 = (2n+4)^2 + (2n+2)^2 - 2(2n+4)(2n+2)\left(\frac{1}{3}\right)\]

Solving for n

Expand and simplify the equation: \[(4n^2 + 24n + 36) = (4n^2 + 16n + 16) + (4n^2 + 8n + 4) - \frac{2}{3}(4n^2 + 8n + 4n + 8)\] \[4n^2 + 24n + 36 = 8n^2 + 24n + 20 - \frac{2}{3}(4n^2 + 12n + 8)\] Multiply by 3 to eliminate the fraction: \[3(4n^2 + 24n + 36) = 3(8n^2 + 24n + 20) - 2(4n^2 + 12n + 8)\] \[12n^2 + 72n + 108 = 24n^2 + 72n + 60 - 8n^2 - 24n - 16\] Combine like terms: \[12n^2 + 72n + 108 = (24n^2 - 8n^2) + (72n - 24n) + (60 - 16)\] \[12n^2 + 72n + 108 = 16n^2 + 48n + 44\] Rearrange into a quadratic equation: \[(16n^2 - 12n^2) + (48n - 72n) + (44 - 108) = 0\] \[4n^2 - 24n - 64 = 0\] Divide by 4: \[n^2 - 6n - 16 = 0\] Factor the quadratic equation: \[(n-8)(n+2) = 0\] This gives two possible solutions for \(n\): \(n=8\) or \(n=-2\). Since \(n\) represents a length (part of a side), it must be a positive value. Also, the problem states \(n\) is an integer. Therefore, the valid value is \(n=8\).

Calculating Side Lengths with n=8

Substitute \(n=8\) back into the expressions for the side lengths:

• Smallest side: \(2n+2 = 2(8) + 2 = 16 + 2 = 18\)
• Middle side: \(2n+4 = 2(8) + 4 = 16 + 4 = 20\)
• Largest side: \(2n+6 = 2(8) + 6 = 16 + 6 = 22\)

The lengths of the sides of the triangle are 18, 20, and 22.

Finding the Smallest Side

From the calculated side lengths (18, 20, 22), the smallest side is 18.

The final answer is 18.