This problem is a comprehensive test of coordinate geometry, specifically involving triangles and their special points, and conic sections, particularly parabolas. We will systematically determine the vertices of the triangle, its incentre, the parameter of the given parabola, and finally, the required expression.

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1. Finding the Vertices of the Triangle

Key Concept: The midpoint formula states that if a point $M(x_m, y_m)$ is the midpoint of a line segment connecting $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, then $x_m = \frac{x_1+x_2}{2}$ and $y_m = \frac{y_1+y_2}{2}$.

Explanation: We are given the midpoints A(0,1), B(1,1), and C(1,0) of the sides of a triangle. Let the vertices of the triangle be $V_1(x_1, y_1)$, $V_2(x_2, y_2)$, and $V_3(x_3, y_3)$. Let A be the midpoint of $V_1V_2$, B be the midpoint of $V_2V_3$, and C be the midpoint of $V_3V_1$.

From the midpoint formula, we have the following system of equations:

1. $\frac{V_1 + V_2}{2} = A \implies V_1 + V_2 = 2A$
2. $\frac{V_2 + V_3}{2} = B \implies V_2 + V_3 = 2B$
3. $\frac{V_3 + V_1}{2} = C \implies V_3 + V_1 = 2C$

To find $V_1$, we can add (1) and (3) and subtract (2): $(V_1 + V_2) + (V_3 + V_1) - (V_2 + V_3) = 2A + 2C - 2B$ $2V_1 = 2(A - B + C)$ $V_1 = A - B + C$

Similarly, we can find $V_2$ and $V_3$ using cyclic permutations: $V_2 = A + B - C$ $V_3 = B + C - A$

Step-by-step Calculation: Given midpoints: $A=(0,1)$, $B=(1,1)$, $C=(1,0)$.

• For $V_1$: $V_1 = A - B + C = (0,1) - (1,1) + (1,0)$ $V_1 = (0-1+1, 1-1+0) = (0,0)$

• For $V_2$: $V_2 = A + B - C = (0,1) + (1,1) - (1,0)$ $V_2 = (0+1-1, 1+1-0) = (0,2)$

• For $V_3$: $V_3 = B + C - A = (1,1) + (1,0) - (0,1)$ $V_3 = (1+1-0, 1+0-1) = (2,0)$

So, the vertices of the triangle are $V_1(0,0)$, $V_2(0,2)$, and $V_3(2,0)$. Tip: This is a right-angled isosceles triangle with the right angle at $V_1(0,0)$. This observation can simplify some calculations later.

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2. Finding the Incentre D of the Triangle

Key Concept: The incentre $D$ of a triangle with vertices $V_1(x_1, y_1)$, $V_2(x_2, y_2)$, $V_3(x_3, y_3)$ and opposite side lengths $v_1, v_2, v_3$ (where $v_1$ is the length of side $V_2V_3$, $v_2$ is the length of side $V_1V_3$, and $v_3$ is the length of side $V_1V_2$) is given by the formula: $D = \left(\frac{v_1x_1 + v_2x_2 + v_3x_3}{v_1 + v_2 + v_3}, \frac{v_1y_1 + v_2y_2 + v_3y_3}{v_1 + v_2 + v_3}\right)$

Explanation: First, we need to calculate the lengths of the sides of the triangle using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Step-by-step Calculation: Vertices are $V_1(0,0)$, $V_2(0,2)$, $V_3(2,0)$.

• Side $v_1$ (opposite $V_1$, length of $V_2V_3$): $v_1 = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$

• Side $v_2$ (opposite $V_2$, length of $V_1V_3$): $v_2 = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$

• Side $v_3$ (opposite $V_3$, length of $V_1V_2$): $v_3 = \sqrt{(0-0)^2 + (2-0)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$

Now, substitute the side lengths and vertices into the incentre formula: $D_x = \frac{v_1x_1 + v_2x_2 + v_3x_3}{v_1 + v_2 + v_3} = \frac{(2\sqrt{2})(0) + (2)(0) + (2)(2)}{2\sqrt{2} + 2 + 2} = \frac{4}{4 + 2\sqrt{2}}$ $D_y = \frac{v_1y_1 + v_2y_2 + v_3y_3}{v_1 + v_2 + v_3} = \frac{(2\sqrt{2})(0) + (2)(2) + (2)(0)}{2\sqrt{2} + 2 + 2} = \frac{4}{4 + 2\sqrt{2}}$

So, $D = \left(\frac{4}{4 + 2\sqrt{2}}, \frac{4}{4 + 2\sqrt{2}}\right)$. Let's simplify the coordinates by rationalizing the denominator: $\frac{4}{4 + 2\sqrt{2}} = \frac{4}{2(2 + \sqrt{2})} = \frac{2}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} = \frac{2(2 - \sqrt{2})}{4 - 2} = \frac{2(2 - \sqrt{2})}{2} = 2 - \sqrt{2}$.

Therefore, the incentre is $D(2 - \sqrt{2}, 2 - \sqrt{2})$. Common Mistake: Forgetting to rationalize the denominator or making arithmetic errors during rationalization.

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3. Finding the Parabola's Parameter

Key Concept: The equation of a parabola with its vertex at the origin and symmetric about the x-axis is $y^2 = 4ax$, where 'a' is the focal length (distance from the vertex to the focus).

Explanation: We are given that the parabola $y^2 = 4ax$ passes through the incentre $D$. This means the coordinates of $D$ must satisfy the parabola's equation. We will substitute the coordinates of $D$ into the equation to find the value of the parameter 'a' (note: this 'a' is different from the side length 'a' used in the incentre formula, which we denoted as $v_1$).

Step-by-step Calculation: The parabola is $y^2 = 4ax$. Point $D(2 - \sqrt{2}, 2 - \sqrt{2})$ lies on the parabola. Substitute $x = 2 - \sqrt{2}$ and $y = 2 - \sqrt{2}$ into the equation: $(2 - \sqrt{2})^2 = 4a(2 - \sqrt{2})$

Since $2 - \sqrt{2} \neq 0$, we can divide both sides by $(2 - \sqrt{2})$: $2 - \sqrt{2} = 4a$ $a = \frac{2 - \sqrt{2}}{4}$

So, the parameter of the parabola is $a = \frac{2 - \sqrt{2}}{4}$.

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4. Determining the Focus of the Parabola

Key Concept: For a parabola of the form $y^2 = 4ax$, the focus is located at the point $(a, 0)$.

Explanation: We have found the parameter 'a' of the parabola. Now we can directly state the coordinates of its focus. The problem specifies that this focus is $(\alpha + \beta \sqrt{2}, 0)$, allowing us to equate the two forms and solve for $\alpha$ and $\beta$.

Step-by-step Calculation: The focus of the parabola $y^2 = 4ax$ is $(a, 0)$. Using the value of 'a' we found: Focus $= \left(\frac{2 - \sqrt{2}}{4}, 0\right)$ Focus $= \left(\frac{2}{4} - \frac{\sqrt{2}}{4}, 0\right)$ Focus $= \left(\frac{1}{2} - \frac{1}{4}\sqrt{2}, 0\right)$

We are given that the focus is $(\alpha + \beta \sqrt{2}, 0)$. Comparing the x-coordinates: $\alpha + \beta \sqrt{2} = \frac{1}{2} - \frac{1}{4}\sqrt{2}$

Since $\alpha$ and $\beta$ are rational numbers, we can equate the rational and irrational parts: $\alpha = \frac{1}{2}$ $\beta = -\frac{1}{4}$

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5. Calculating the Final Expression

Explanation: Now that we have the values of $\alpha$ and $\beta$, we can substitute them into the required expression $\frac{\alpha}{\beta^2}$ and simplify.

Step-by-step Calculation: $\alpha = \frac{1}{2}$ $\beta = -\frac{1}{4}$

$\beta^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}$

Now, calculate $\frac{\alpha}{\beta^2}$: $\frac{\alpha}{\beta^2} = \frac{\frac{1}{2}}{\frac{1}{16}}$ $\frac{\alpha}{\beta^2} = \frac{1}{2} \times 16$ $\frac{\alpha}{\beta^2} = 8$

The final answer is 8.

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Summary and Key Takeaways

This problem effectively combined multiple concepts from coordinate geometry and conic sections.

1. Triangle Properties: The relationship between midpoints of sides and vertices of a triangle (using vector addition/subtraction or solving a system of equations) is a crucial shortcut.
2. Incentre Calculation: Remember the formula for the incentre and the need to calculate side lengths accurately. Rationalizing denominators is often necessary for simplification.
3. Parabola Basics: Understanding the standard form of a parabola ($y^2=4ax$) and the meaning of its parameter 'a' for finding the focus is fundamental.
4. Careful with Notation: Pay attention when the same letter (like 'a' here) is used for different quantities (side length vs. parabola parameter).
5. Rational and Irrational Parts: When equating expressions involving rational and irrational numbers, separate the rational and irrational components to solve for unknown rational variables.

The final result is 8. This corresponds to option (D). *Self