Example 1: Using the Centroid to Find Segment Lengths in Triangle RST.

Problem Analysis:

The problem involves using the centroid of a triangle to find segment lengths. The centroid of a triangle is the point where the three medians intersect, and it divides each median into a ratio of 2:1, with the longer segment being closer to the vertex.

Given the diagram and the problem statements, we will solve each part step by step.

---

Part A: SP = 16. What is SM?

#### Step 1: Understand the relationship
- The centroid \( P \) divides the median \( SM \) in the ratio \( 2:1 \).
- This means \( SP : PM = 2 : 1 \).

#### Step 2: Express the segments
- Let \( SP = 2x \) and \( PM = x \).
- Given \( SP = 16 \), we have:
\[
2x = 16 \implies x = 8
\]

#### Step 3: Find \( SM \)
- The total length of \( SM \) is the sum of \( SP \) and \( PM \):
\[
SM = SP + PM = 2x + x = 3x
\]
- Substituting \( x = 8 \):
\[
SM = 3 \times 8 = 24
\]

#### Final Answer for Part A:
\[
\boxed{24}
\]

---

Part B: LP = 9. What is LT?

#### Step 1: Understand the relationship
- The centroid \( P \) divides the median \( LT \) in the ratio \( 2:1 \).
- This means \( LP : PT = 2 : 1 \).

#### Step 2: Express the segments
- Let \( LP = 2y \) and \( PT = y \).
- Given \( LP = 9 \), we have:
\[
2y = 9 \implies y = \frac{9}{2} = 4.5
\]

#### Step 3: Find \( LT \)
- The total length of \( LT \) is the sum of \( LP \) and \( PT \):
\[
LT = LP + PT = 2y + y = 3y
\]
- Substituting \( y = 4.5 \):
\[
LT = 3 \times 4.5 = 13.5
\]

#### Final Answer for Part B:
\[
\boxed{13.5}
\]

---

Part C: RN = 33. What is PN?

#### Step 1: Understand the relationship
- The centroid \( P \) divides the median \( RN \) in the ratio \( 2:1 \).
- This means \( RP : PN = 2 : 1 \).

#### Step 2: Express the segments
- Let \( RP = 2z \) and \( PN = z \).
- The total length of \( RN \) is the sum of \( RP \) and \( PN \):
\[
RN = RP + PN = 2z + z = 3z
\]
- Given \( RN = 33 \), we have:
\[
3z = 33 \implies z = \frac{33}{3} = 11
\]

#### Step 3: Find \( PN \)
- From the expression \( PN = z \):
\[
PN = 11
\]

#### Final Answer for Part C:
\[
\boxed{11}
\]

---

Summary of Answers:

1. \( SM = 24 \)
2. \( LT = 13.5 \)
3. \( PN = 11 \)

\[
\boxed{24, 13.5, 11}
\]