Understanding the Triangle and In-circle Properties

The problem provides information about a triangle PQR, specifically about its largest angle P, the value of \(\cos P\), and the lengths of the segments formed by the in-circle touching the sides.

The in-circle of a triangle touches the sides at certain points. A key property is that the lengths of tangents from a vertex to the in-circle are equal. In triangle PQR, the in-circle touches sides PQ, QR, and RP at points N, L, and M respectively.

• Tangent segments from vertex P: PN and PM. Therefore, PN = PM.
• Tangent segments from vertex Q: QN and QL. Therefore, QN = QL.
• Tangent segments from vertex R: RL and RM. Therefore, RL = RM.

We are given the lengths of some of these segments:

• PN = n
• QL = n + 2
• RM = n + 4

Using the property that tangent segments from a vertex are equal, we have:

• PM = PN = n
• QN = QL = n + 2
• RL = RM = n + 4

Now we can express the lengths of the sides of the triangle PQR in terms of n:

• Side PQ = PN + NQ = PN + QN = \(n + (n+2) = 2n + 2\)
• Side QR = QL + LR = QL + RL = \((n+2) + (n+4) = 2n + 6\)
• Side RP = RM + MP = RM + PN = \((n+4) + n = 2n + 4\)

So, the side lengths are:

Applying the Law of Cosines

We are given that P is the largest angle and \(\cos P = \frac{1}{3}\). The Law of Cosines relates the sides of a triangle to one of its angles. For angle P and the side opposite it (QR), the Law of Cosines states:

\[QR^2 = PQ^2 + RP^2 - 2(PQ)(RP)\cos P\]

Substitute the expressions for the side lengths and the given value of \(\cos P\):

\[(2n + 6)^2 = (2n + 2)^2 + (2n + 4)^2 - 2(2n + 2)(2n + 4)\left(\frac{1}{3}\right)\]

Solving the Equation for n

Now, we expand and simplify the equation to find the value of n. Expand the squared terms and the product terms:

• \((2n + 6)^2 = (2n)^2 + 2(2n)(6) + 6^2 = 4n^2 + 24n + 36\)
• \((2n + 2)^2 = (2n)^2 + 2(2n)(2) + 2^2 = 4n^2 + 8n + 4\)
• \((2n + 4)^2 = (2n)^2 + 2(2n)(4) + 4^2 = 4n^2 + 16n + 16\)
• \((2n + 2)(2n + 4) = 4n^2 + 8n + 4n + 8 = 4n^2 + 12n + 8\)

Substitute these back into the Law of Cosines equation:

\[4n^2 + 24n + 36 = (4n^2 + 8n + 4) + (4n^2 + 16n + 16) - \frac{2}{3}(4n^2 + 12n + 8)\]

Combine the terms on the right side:

\[4n^2 + 24n + 36 = 8n^2 + 24n + 20 - \frac{8}{3}n^2 - \frac{24}{3}n - \frac{16}{3}\]

\[4n^2 + 24n + 36 = 8n^2 + 24n + 20 - \frac{8}{3}n^2 - 8n - \frac{16}{3}\]

To eliminate the fraction, multiply the entire equation by 3:

\[3(4n^2 + 24n + 36) = 3(8n^2 + 24n + 20) - 3\left(\frac{8}{3}n^2 + 8n + \frac{16}{3}\right)\]

\[12n^2 + 72n + 108 = 24n^2 + 72n + 60 - (8n^2 + 24n + 16)\]

\[12n^2 + 72n + 108 = 24n^2 + 72n + 60 - 8n^2 - 24n - 16\]

Combine like terms on the right side:

\[12n^2 + 72n + 108 = (24n^2 - 8n^2) + (72n - 24n) + (60 - 16)\]

\[12n^2 + 72n + 108 = 16n^2 + 48n + 44\]

Move all terms to one side to form a quadratic equation:

\[0 = 16n^2 - 12n^2 + 48n - 72n + 44 - 108\]

\[0 = 4n^2 - 24n - 64\]

Divide the entire equation by 4:

\[0 = n^2 - 6n - 16\]

This is a quadratic equation in the form \(an^2 + bn + c = 0\). We can solve it by factoring. We need two numbers that multiply to -16 and add up to -6. These numbers are -8 and +2.

\[n^2 - 8n + 2n - 16 = 0\]

Factor by grouping:

\[n(n - 8) + 2(n - 8) = 0\]

\[(n - 8)(n + 2) = 0\]

The possible values for n are \(n - 8 = 0\) or \(n + 2 = 0\), which gives \(n = 8\) or \(n = -2\).

Validating the Value of n

The problem states that n is an integer and represents a length (PN). Lengths must be positive, so n must be positive. Also, the side lengths of a triangle must be positive:

• PQ = \(2n + 2 > 0 \implies 2n > -2 \implies n > -1\)
• QR = \(2n + 6 > 0 \implies 2n > -6 \implies n > -3\)
• RP = \(2n + 4 > 0 \implies 2n > -4 \implies n > -2\)

The condition \(n > -1\) is the most restrictive. Since n must be an integer, n must be 0 or a positive integer.

The solution \(n = -2\) is not valid because it is not greater than -1.

The solution \(n = 8\) is a positive integer and satisfies the condition \(n > -1\).

Let's check the side lengths when n = 8:

• PQ = \(2(8) + 2 = 16 + 2 = 18\)
• QR = \(2(8) + 6 = 16 + 6 = 22\)
• RP = \(2(8) + 4 = 16 + 4 = 20\)

The side lengths are 18, 22, and 20. These are all positive, so a triangle can be formed.

We also need to verify the condition that P is the largest angle. In a triangle, the largest angle is opposite the longest side. The side lengths are 18, 22, 20. The longest side is QR, which has length 22. The side QR is opposite angle P. Thus, P is indeed the largest angle, consistent with the problem statement.

Therefore, the only valid value for the integer n is 8.

Revision Table: Key Concepts Review

Additional Information: In-circle and Ex-circles

The in-circle is the largest circle that can be inscribed inside the triangle. Its center is the incenter, which is the intersection of the angle bisectors. The radius of the in-circle is called the inradius, often denoted by r.

Besides the in-circle, a triangle also has three ex-circles. Each ex-circle is tangent to one side of the triangle and the extensions of the other two sides. The center of an ex-circle is an excenter, which is the intersection of the bisector of one internal angle and the bisectors of the two external angles.

The points where the in-circle touches the sides (N, L, M in this problem) are important for various triangle properties and formulas, including the semi-perimeter and area.

In this problem, the segments PN, QL, and RM are related to the distances from the vertices to the in-circle tangency points on the adjacent sides. If s is the semi-perimeter of the triangle \((s = \frac{PQ+QR+RP}{2})\), then PN = PM = s - QR, QN = QL = s - RP, and RL = RM = s - PQ. Let's verify this with our side lengths for n=8:

• PQ = 18, QR = 22, RP = 20
• Semi-perimeter \(s = \frac{18 + 22 + 20}{2} = \frac{60}{2} = 30\)
• PN = s - QR = 30 - 22 = 8 (Matches given PN = n = 8)
• QL = s - RP = 30 - 20 = 10 (Matches given QL = n + 2 = 8 + 2 = 10)
• RM = s - PQ = 30 - 18 = 12 (Matches given RM = n + 4 = 8 + 4 = 12)

The values match, confirming our value of n=8 is consistent with these in-circle properties as well.