To find the incenter of the triangle given the midpoints of its sides, we can follow these steps:
### Step 1: Identify the midpoints and label the triangle vertices
The midpoints given are:
- \( M_A = (0, \frac{1}{2}) \) (midpoint of BC)
- \( M_B = (\frac{1}{2}, \frac{1}{2}) \) (midpoint of AC)
- \( M_C = (\frac{1}{2}, 0) \) (midpoint of AB)
Let's label the vertices of the triangle as \( A(x, y) \), \( B(0, 1) \), and \( C(1, 0) \).
### Step 2: Use the midpoint formula to find the vertices of the triangle
The midpoint formula states that if \( M \) is the midpoint of points \( (x_1, y_1) \) and \( (x_2, y_2) \), then:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Using this formula, we can find the coordinates of the vertices \( A \), \( B \), and \( C \).
1. For \( M_A = (0, \frac{1}{2}) \):
\[
\frac{B_x + C_x}{2} = 0 \implies B_x + C_x = 0 \implies 0 + 1 = 0 \quad \text{(not valid)}
\]
\[
\frac{B_y + C_y}{2} = \frac{1}{2} \implies 1 + 0 = 1 \quad \text{(valid)}
\]
2. For \( M_B = (\frac{1}{2}, \frac{1}{2}) \):
\[
\frac{A_x + C_x}{2} = \frac{1}{2} \implies A_x + C_x = 1 \quad \text{(valid)}
\]
\[
\frac{A_y + C_y}{2} = \frac{1}{2} \implies A_y + C_y = 1 \quad \text{(valid)}
\]
3. For \( M_C = (\frac{1}{2}, 0) \):
\[
\frac{A_x + B_x}{2} = \frac{1}{2} \implies A_x + 0 = 1 \quad \text{(valid)}
\]
\[
\frac{A_y + B_y}{2} = 0 \implies A_y + 1 = 0 \quad \text{(valid)}
\]
### Step 3: Solve for the coordinates of vertices
From the equations:
- From \( M_B \): \( A_x + C_x = 1 \) and \( A_y + C_y = 1 \)
- From \( M_C \): \( A_x = 1 \) and \( A_y = -1 \)
Thus, we find:
- \( A(0, 0) \)
- \( B(0, 1) \)
- \( C(1, 0) \)
### Step 4: Calculate the lengths of the sides
Using the distance formula:
- \( a = BC = \sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2} \)
- \( b = AC = \sqrt{(1-0)^2 + (0-0)^2} = 1 \)
- \( c = AB = \sqrt{(0-0)^2 + (1-0)^2} = 1 \)
### Step 5: Use the incenter formula
The coordinates of the incenter \( I \) can be calculated using the formula:
\[
I_x = \frac{aA_x + bB_x + cC_x}{a + b + c}
\]
\[
I_y = \frac{aA_y + bB_y + cC_y}{a + b + c}
\]
Substituting the values:
\[
I_x = \frac{\sqrt{2} \cdot 0 + 1 \cdot 0 + 1 \cdot 1}{\sqrt{2} + 1 + 1} = \frac{1}{\sqrt{2} + 2}
\]
\[
I_y = \frac{\sqrt{2} \cdot 0 + 1 \cdot 1 + 1 \cdot 0}{\sqrt{2} + 1 + 1} = \frac{1}{\sqrt{2} + 2}
\]
### Step 6: Rationalize the incenter coordinates
To rationalize:
\[
I_x = I_y = \frac{1}{\sqrt{2} + 2} \cdot \frac{\sqrt{2} - 2}{\sqrt{2} - 2} = \frac{\sqrt{2} - 2}{2 - 2} = \frac{\sqrt{2} - 2}{2}
\]
### Final Answer
The incenter of the triangle is:
\[
I\left(\frac{\sqrt{2} - 2}{2}, \frac{\sqrt{2} - 2}{2}\right)
\]
