To find the vertices of the triangle given the midpoints of its sides, we can follow these steps:
### Step 1: Define the vertices and midpoints
Let the vertices of the triangle be \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \). The midpoints of the sides are given as:
- Midpoint of \( AB \): \( M_1(3, 2, \frac{3}{2}) \)
- Midpoint of \( BC \): \( M_2(1, \frac{3}{2}, 3) \)
- Midpoint of \( CA \): \( M_3(2, \frac{5}{2}, \frac{5}{2}) \)
### Step 2: Set up equations for the midpoints
Using the midpoint formula, we can set up the following equations:
1. For midpoint \( M_1 \):
\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = (3, 2, \frac{3}{2})
\]
This gives us three equations:
\[
\frac{x_1 + x_2}{2} = 3 \quad \Rightarrow \quad x_1 + x_2 = 6 \quad \text{(Equation 1)}
\]
\[
\frac{y_1 + y_2}{2} = 2 \quad \Rightarrow \quad y_1 + y_2 = 4 \quad \text{(Equation 2)}
\]
\[
\frac{z_1 + z_2}{2} = \frac{3}{2} \quad \Rightarrow \quad z_1 + z_2 = 3 \quad \text{(Equation 3)}
\]
2. For midpoint \( M_2 \):
\[
\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right) = (1, \frac{3}{2}, 3)
\]
This gives us:
\[
\frac{x_2 + x_3}{2} = 1 \quad \Rightarrow \quad x_2 + x_3 = 2 \quad \text{(Equation 4)}
\]
\[
\frac{y_2 + y_3}{2} = \frac{3}{2} \quad \Rightarrow \quad y_2 + y_3 = 3 \quad \text{(Equation 5)}
\]
\[
\frac{z_2 + z_3}{2} = 3 \quad \Rightarrow \quad z_2 + z_3 = 6 \quad \text{(Equation 6)}
\]
3. For midpoint \( M_3 \):
\[
\left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2}, \frac{z_3 + z_1}{2} \right) = (2, \frac{5}{2}, \frac{5}{2})
\]
This gives us:
\[
\frac{x_3 + x_1}{2} = 2 \quad \Rightarrow \quad x_3 + x_1 = 4 \quad \text{(Equation 7)}
\]
\[
\frac{y_3 + y_1}{2} = \frac{5}{2} \quad \Rightarrow \quad y_3 + y_1 = 5 \quad \text{(Equation 8)}
\]
\[
\frac{z_3 + z_1}{2} = \frac{5}{2} \quad \Rightarrow \quad z_3 + z_1 = 5 \quad \text{(Equation 9)}
\]
### Step 3: Solve the equations for \( x_1, x_2, x_3 \)
From Equations 1, 4, and 7:
1. From Equation 1: \( x_1 + x_2 = 6 \)
2. From Equation 4: \( x_2 + x_3 = 2 \)
3. From Equation 7: \( x_3 + x_1 = 4 \)
Now, we can solve these equations:
- Subtract Equation 4 from Equation 1:
\[
(x_1 + x_2) - (x_2 + x_3) = 6 - 2 \quad \Rightarrow \quad x_1 - x_3 = 4 \quad \text{(Equation 10)}
\]
- Add Equation 10 to Equation 7:
\[
(x_3 + x_1) + (x_1 - x_3) = 4 + 4 \quad \Rightarrow \quad 2x_1 = 8 \quad \Rightarrow \quad x_1 = 4
\]
- Substitute \( x_1 = 4 \) into Equation 1:
\[
4 + x_2 = 6 \quad \Rightarrow \quad x_2 = 2
\]
- Substitute \( x_2 = 2 \) into Equation 4:
\[
2 + x_3 = 2 \quad \Rightarrow \quad x_3 = 0
\]
### Step 4: Solve the equations for \( y_1, y_2, y_3 \)
Using Equations 2, 5, and 8:
1. From Equation 2: \( y_1 + y_2 = 4 \)
2. From Equation 5: \( y_2 + y_3 = 3 \)
3. From Equation 8: \( y_3 + y_1 = 5 \)
- Subtract Equation 5 from Equation 2:
\[
(y_1 + y_2) - (y_2 + y_3) = 4 - 3 \quad \Rightarrow \quad y_1 - y_3 = 1 \quad \text{(Equation 11)}
\]
- Add Equation 11 to Equation 8:
\[
(y_3 + y_1) + (y_1 - y_3) = 5 + 1 \quad \Rightarrow \quad 2y_1 = 6 \quad \Rightarrow \quad y_1 = 3
\]
- Substitute \( y_1 = 3 \) into Equation 2:
\[
3 + y_2 = 4 \quad \Rightarrow \quad y_2 = 1
\]
- Substitute \( y_2 = 1 \) into Equation 5:
\[
1 + y_3 = 3 \quad \Rightarrow \quad y_3 = 2
\]
### Step 5: Solve the equations for \( z_1, z_2, z_3 \)
Using Equations 3, 6, and 9:
1. From Equation 3: \( z_1 + z_2 = 3 \)
2. From Equation 6: \( z_2 + z_3 = 6 \)
3. From Equation 9: \( z_3 + z_1 = 5 \)
- Subtract Equation 3 from Equation 6:
\[
(z_2 + z_3) - (z_1 + z_2) = 6 - 3 \quad \Rightarrow \quad z_3 - z_1 = 3 \quad \text{(Equation 12)}
\]
- Add Equation 12 to Equation 9:
\[
(z_3 + z_1) + (z_3 - z_1) = 5 + 3 \quad \Rightarrow \quad 2z_3 = 8 \quad \Rightarrow \quad z_3 = 4
\]
- Substitute \( z_3 = 4 \) into Equation 6:
\[
z_2 + 4 = 6 \quad \Rightarrow \quad z_2 = 2
\]
- Substitute \( z_2 = 2 \) into Equation 3:
\[
z_1 + 2 = 3 \quad \Rightarrow \quad z_1 = 1
\]
### Final Result
The vertices of the triangle are:
- \( A(4, 3, 1) \)
- \( B(2, 1, 2) \)
- \( C(0, 2, 4) \)
