Let A(0, 1), B(1, 1) and C(1, 0) be the mid-points of the sides of a triangle with incentre at the p

Q: Wait, can't I just find the incenter of the triangle formed by the midpoints?

That is a classic trap! The incenter of the medial triangle (the one formed by the midpoints) is not the same as the incenter of the original triangle. You must find the vertices of the original triangle first. The medial triangle's incenter is actually related to the orthocenter of the original triangle, which is a completely different geometric entity.

Q: Why is the focus of $y^2 = 4ax$ just $(a, 0)$?

That is the standard definition of a parabola in the form $y^2 = 4ax$. The distance from the vertex $(0,0)$ to the focus is $a$ along the axis of symmetry, which is the x-axis here. If the equation were different, the focus would shift, but for this standard form, $(a, 0)$ is the correct coordinate.

Q: Is there a faster way to find the vertices without solving six equations?

You can use the property that the triangle formed by the midpoints is similar to the original triangle and its sides are parallel to the original sides. However, solving the system of equations is the most robust and error-proof method for JEE. It ensures you don't make a geometric assumption that might be slightly off.

Question

Let $A(0, 1)$, $B(1, 1)$ and $C(1, 0)$ be the mid-points of the sides of a triangle with incentre at the point $D$. If the focus of the parabola $y^2 = 4ax$ passing through $D$ is $(\alpha + \beta\sqrt{2}, 0)$, where $\alpha$ and $\beta$ are rational numbers, then $\frac{\alpha}{\beta^2}$ is equal to

Options:

(A) 8

(B) 12

(C) 6

(D) $\frac{9}{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Structures

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden structure.

Imagine you are standing on a coordinate plane, looking at three points: $A(0, 1)$, $B(1, 1)$, and $C(1, 0)$. These are not just random dots; they are the midpoints of a triangle. The triangle itself is a ghost, a shape waiting to be revealed. Our journey begins by reconstructing this ghost.

### Phase 1: The Mystery of the Vertices

We are given the midpoints of the sides of a triangle. Let the vertices of the original triangle be $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$.

The midpoint formula tells us that the midpoint of a segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. Since we know the midpoints, we can set up a system of linear equations.

For the x-coordinates, we have:
$$x_1+x_2 = 0, \quad x_2+x_3 = 2, \quad x_3+x_1 = 2$$

This is a beautiful, symmetric system. If we sum these three equations, we get $2(x_1+x_2+x_3) = 4$, which simplifies to $x_1+x_2+x_3 = 2$.

By subtracting each original equation from this sum, we find $x_3 = 2$, $x_1 = 0$, and $x_2 = 0$. We repeat this for the y-coordinates, and voilà! The vertices of our triangle are $P(0, 0)$, $Q(0, 2)$, and $R(2, 0)$. We have successfully brought the triangle into existence.

### Phase 2: The Incenter Quest

Now that we have the vertices, we need the incenter $D$. The incenter is the center of the circle inscribed within the triangle.

To find it, we need the lengths of the sides opposite to each vertex. Let $p$ be the length of side $QR$, $q$ be the length of $RP$, and $r$ be the length of $PQ$.

Using the distance formula:
$$p = \sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4+4} = 2\sqrt{2}$$
$$q = \sqrt{(2-0)^2 + (0-0)^2} = 2$$
$$r = \sqrt{(0-0)^2 + (2-0)^2} = 2$$

We have an isosceles right-angled triangle! The incenter formula is a powerful tool:
$$D = \left( \frac{px_1+qx_2+rx_3}{p+q+r}, \frac{py_1+qy_2+ry_3}{p+q+r} ight)$$

Substituting our values, the x-coordinate becomes:
$$x_D = \frac{2\sqrt{2}(0) + 2(0) + 2(2)}{2\sqrt{2}+2+2} = \frac{4}{4+2\sqrt{2}}$$

Simplifying this, we get $\frac{2}{2+\sqrt{2}}$. Rationalizing the denominator by multiplying by $\frac{2-\sqrt{2}}{2-\sqrt{2}}$, we find $x_D = 2-\sqrt{2}$.

By symmetry, $y_D$ is also $2-\sqrt{2}$. Our incenter $D$ is at $(2-\sqrt{2}, 2-\sqrt{2})$.

### Phase 3: The Parabola's Dance

We are given a parabola $y^2 = 4ax$ that passes through $D$. This means the coordinates of $D$ must satisfy the equation.

Substituting $x = 2-\sqrt{2}$ and $y = 2-\sqrt{2}$, we get:
$$(2-\sqrt{2})^2 = 4a(2-\sqrt{2})$$

Since $2-\sqrt{2} 
eq 0$, we can divide both sides by it, yielding $2-\sqrt{2} = 4a$. Thus:
$$a = \frac{2-\sqrt{2}}{4} = \frac{1}{2} - \frac{1}{4}\sqrt{2}$$

The focus of the parabola $y^2 = 4ax$ is at $(a, 0)$. Comparing this to the given form $(\alpha + \beta\sqrt{2}, 0)$, we identify $\alpha = \frac{1}{2}$ and $\beta = -\frac{1}{4}$.

### The Final Celebration

We are asked to calculate $\frac{\alpha}{\beta^2}$. Substituting our values:
$$\frac{\alpha}{\beta^2} = \frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = \frac{1}{2} 	imes 16 = 8$$

The elegance of this result is a testament to the beauty of geometry. We started with midpoints and ended with a clean, integer answer.

Remember, in JEE, the path is often as important as the destination. Keep practicing, keep visualizing, and keep falling in love with the physics and math behind the problems. The final answer is **8**.

Select Answer:

Visualized Solution

Plotting the Given Midpoints

Setting Up Vertex Equations

Finding the Triangle Vertices

Calculating Side Lengths

The Incenter Formula

Substituting into Incenter Formula

Simplifying the Incenter Coordinates

Parabola Passing Through Incenter

Calculating Parameter $a$

Finding the Focus

Comparing to Find $α$ and $β$

Evaluating $\frac{α}{β ^{2}}$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of Hidden Structures

Phase 1

Phase 2

Phase 3

The Final Celebration

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