Let A, B, C respresent the vertices of a triangle, where A is the origin and B and C have position b and c respectively. Points M, N and P are taken on sides AB, BC and CA respectively, such that `(AM)/(AB)=(BN)/(BC)=(CP)/(CA)=alpha Q. AN+BP+CM is

Options

A. a. `3alpha(b+c)`
B. b. `alpha(b+c)`
C. c. `(1-alpha)(b+c)`
D. d. `0`

Answer

To solve the problem, we need to find the expression for \( \vec{AN} + \vec{BP} + \vec{CM} \) given the conditions about the points \( M, N, P \) on the triangle formed by vertices \( A, B, C \).

### Step 1: Define the Position Vectors
Let:
- \( \vec{A} = \vec{0} \) (the origin),
- \( \vec{B} = \vec{b} \),
- \( \vec{C} = \vec{c} \).

### Step 2: Find the Position Vector of Point M
Since \( M \) is on side \( AB \) such that \( \frac{AM}{AB} = \alpha \), we can express the position vector of \( M \) as:
\[ \vec{M} = (1 - \alpha) \vec{A} + \alpha \vec{B} = (1 - \alpha) \vec{0} + \alpha \vec{b} = \alpha \vec{b}. \]

### Step 3: Find the Position Vector of Point N
Since \( N \) is on side \( BC \) such that \( \frac{BN}{BC} = \alpha \), we can express the position vector of \( N \) as:
\[ \vec{N} = (1 - \alpha) \vec{B} + \alpha \vec{C} = (1 - \alpha) \vec{b} + \alpha \vec{c}. \]

### Step 4: Find the Position Vector of Point P
Since \( P \) is on side \( CA \) such that \( \frac{CP}{CA} = \alpha \), we can express the position vector of \( P \) as:
\[ \vec{P} = (1 - \alpha) \vec{C} + \alpha \vec{A} = (1 - \alpha) \vec{c} + \alpha \vec{0} = (1 - \alpha) \vec{c}. \]

### Step 5: Calculate \( \vec{AN} + \vec{BP} + \vec{CM} \)
Now we can compute each vector:
- \( \vec{AN} = \vec{N} - \vec{A} = \vec{N} = (1 - \alpha) \vec{b} + \alpha \vec{c} \),
- \( \vec{BP} = \vec{P} - \vec{B} = (1 - \alpha) \vec{c} - \vec{b} \),
- \( \vec{CM} = \vec{M} - \vec{C} = \alpha \vec{b} - \vec{c} \).

Now, we add these vectors together:
\[ \vec{AN} + \vec{BP} + \vec{CM} = [(1 - \alpha) \vec{b} + \alpha \vec{c}] + [(1 - \alpha) \vec{c} - \vec{b}] + [\alpha \vec{b} - \vec{c}]. \]

### Step 6: Simplify the Expression
Combining the terms:
\[ = (1 - \alpha) \vec{b} + \alpha \vec{c} + (1 - \alpha) \vec{c} - \vec{b} + \alpha \vec{b} - \vec{c}. \]
Grouping similar terms:
\[ = [(1 - \alpha) - 1 + \alpha] \vec{b} + [\alpha + (1 - \alpha) - 1] \vec{c}. \]
This simplifies to:
\[ = 0 \cdot \vec{b} + 0 \cdot \vec{c} = \vec{0}. \]

### Conclusion
Thus, we find that:
\[ \vec{AN} + \vec{BP} + \vec{CM} = \vec{0}. \]