Finding Side Lengths of Triangle Given Cosine and Incircle Tangent Segments

To find the possible side lengths of triangle $PQR$, let us analyze the given information:

1. Let the lengths of the tangent segments from the vertices $P, Q, R$ to the incircle be:

   • Tangent from $P$: $PN = PM = x$
   • Tangent from $Q$: $QL = QN = y$
   • Tangent from $R$: $RM = RL = z$

2. The side lengths of triangle $PQR$ are then:

   • $p = QR = QL + RL = y + z$
   • $q = RP = RM + PM = z + x$
   • $r = PQ = PN + QN = x + y$

3. We are given that the segment lengths $PN, QL, RM$ (i.e., $x, y, z$) are consecutive even integers. Since $P$ is the largest angle of $\triangle PQR$, the side opposite to $P$, which is $p = QR = y + z$, must be the largest side of the triangle. Since $p = y + z$, $q = x + z$, and $r = x + y$, for $p$ to be the largest side, we must have: $y+z > x+z \implies y > x$ $y+z > x+y \implies z > x$ Thus, $x$ must be the smallest among $x, y, z$. Since $x, y, z$ are consecutive even integers and $x$ is the smallest, the set $\{y, z\}$ must consist of $x+2$ and $x+4$. Without loss of generality for the side lengths, let: $x = 2k, \quad y = 2k+2, \quad z = 2k+4 \quad (\text{or } y = 2k+4, z = 2k+2)$ where $k$ is a positive integer.

4. The side lengths of the triangle are:

   • $p = QR = y + z = (2k+2) + (2k+4) = 4k + 6$
   • $q = RP = z + x = (2k+4) + 2k = 4k + 4$
   • $r = PQ = x + y = 2k + (2k+2) = 4k + 2$

5. We are given that $\cos P = \frac{1}{3}$. Using the Law of Cosines for angle $P$: $\cos P = \frac{q^2 + r^2 - p^2}{2qr}$ Substitute $\cos P = \frac{1}{3}$ and the expressions for $p, q, r$: $\frac{1}{3} = \frac{(4k+4)^2 + (4k+2)^2 - (4k+6)^2}{2(4k+4)(4k+2)}$

   Dividing all terms by $4$ (since each term in parentheses has a factor of 2, so their squares have a factor of 4): Let $k$ terms be simplified by dividing inside the terms by 2: $p = 2(2k+3), \quad q = 2(2k+2), \quad r = 2(2k+1)$ $\frac{1}{3} = \frac{(2k+2)^2 + (2k+1)^2 - (2k+3)^2}{2(2k+2)(2k+1)}$

   Now simplify the numerator: $(2k+2)^2 + (2k+1)^2 - (2k+3)^2 = (4k^2 + 8k + 4) + (4k^2 + 4k + 1) - (4k^2 + 12k + 9)$ $= 4k^2 - 4$ $= 4(k^2 - 1)$

   Now set up the equation: $\frac{1}{3} = \frac{4(k^2 - 1)}{2(2k+2)(2k+1)} = \frac{4(k-1)(k+1)}{4(k+1)(2k+1)} = \frac{k-1}{2k+1}$

6. Solve for $k$: $2k + 1 = 3(k - 1)$ $2k + 1 = 3k - 3$ $k = 4$

7. Now calculate the lengths of the sides of the triangle:

   • $p = 4k + 6 = 4(4) + 6 = 22$
   • $q = 4k + 4 = 4(4) + 4 = 20$
   • $r = 4k + 2 = 4(4) + 2 = 18$

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

Comparing these side lengths with the given options:

• Option A: $16$
• Option B: $18$ (Correct)
• Option C: $24$
• Option D: $22$ (Correct)

Therefore, the possible lengths of the sides of triangle $PQR$ are $18$ and $22$.

Correct Options: B, D