Key Concepts and Formulas

• Midpoint Theorem and Vertex Recovery: Given midpoints $D, E, F$ of sides $BC, CA, AB$ respectively, the vertices of $\triangle ABC$ are $A = E + F - D$, $B = D + F - E$, $C = D + E - F$.
• Distance Formula: The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Incenter Formula: The incenter $I(x_I, y_I)$ of a triangle with vertices $A(x_A, y_A)$, $B(x_B, y_B)$, $C(x_C, y_C)$ and side lengths $a, b, c$ (opposite to $A, B, C$ respectively) is given by: $I = \left( \frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c} \right)$

Step-by-Step Solution

Step 1: Determine the Vertices of the Original Triangle

We are given the midpoints $D(0, 1)$, $E(1, 1)$, and $F(1, 0)$. We want to find the vertices $A$, $B$, and $C$ of the original triangle. We use the formulas $A = E + F - D$, $B = D + F - E$, and $C = D + E - F$.

• Vertex A: $A = (1, 1) + (1, 0) - (0, 1) = (1+1-0, 1+0-1) = (2, 0)$.
• Vertex B: $B = (0, 1) + (1, 0) - (1, 1) = (0+1-1, 1+0-1) = (0, 0)$.
• Vertex C: $C = (0, 1) + (1, 1) - (1, 0) = (0+1-1, 1+1-0) = (0, 2)$.

Thus, the vertices are $A(2, 0)$, $B(0, 0)$, and $C(0, 2)$.

Step 2: Calculate the Side Lengths of the Triangle

We need to find the lengths of the sides $a$, $b$, and $c$ opposite to vertices $A$, $B$, and $C$, respectively. We use the distance formula.

• Side a (BC): $a = \sqrt{(0-0)^2 + (2-0)^2} = \sqrt{0 + 4} = 2$.
• Side b (AC): $b = \sqrt{(0-2)^2 + (2-0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
• Side c (AB): $c = \sqrt{(0-2)^2 + (0-0)^2} = \sqrt{4 + 0} = 2$.

So, $a = 2$, $b = 2\sqrt{2}$, and $c = 2$.

Step 3: Calculate the x-coordinate of the Incenter

We use the incenter formula to find the $x$-coordinate of the incenter: $x_I = \frac{ax_A + bx_B + cx_C}{a+b+c}$

Substituting the values: $x_I = \frac{(2)(2) + (2\sqrt{2})(0) + (2)(0)}{2 + 2\sqrt{2} + 2} = \frac{4}{4 + 2\sqrt{2}} = \frac{2}{2 + \sqrt{2}}$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate $2 - \sqrt{2}$: $x_I = \frac{2(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{4 - 2\sqrt{2}}{4 - 2} = \frac{4 - 2\sqrt{2}}{2} = 2 - \sqrt{2}$

Thus, the $x$-coordinate of the incenter is $2 - \sqrt{2}$.

Common Mistakes & Tips:

• Midpoint Formula Mix-up: Ensure you correctly apply the midpoint formula when finding the vertices from the midpoints. Double-check your calculations to avoid errors.
• Rationalization: Always rationalize the denominator to simplify the expression.
• Incenter Formula: Remember the incenter formula correctly. The side lengths are weights for the opposite vertices.

Summary

We used the given midpoints to find the vertices of the triangle. Then, we calculated the side lengths and applied the incenter formula to find the x-coordinate of the incenter. The final answer is $2 - \sqrt 2$.

Final Answer The final answer is $\boxed{2 - \sqrt 2}$, which corresponds to option (B).