Let α, β, γ, δ ∈ Z and let A(α, β), B(1, 0), C(γ, δ) and D(1, 2) be the vertices of a parallelogram

Q: Wait, why do we reject the fractional root for \alpha? Is it possible I made a calculation error?

The problem explicitly states \alpha, \beta, \gamma, \delta \in \mathbb{Z}. In JEE Advanced, constraints are not suggestions; they are boundaries. If a root isn't an integer, it violates the problem's definition of the coordinate space. Always trust the constraints—they are there to guide you, not to trick you.

Q: Is there a faster way to find C without the midpoint formula?

You could use vector addition, where \vec{C} = \vec{B} + \vec{D} - \vec{A}. This is mathematically identical to the midpoint property. The midpoint method is just the coordinate-geometry version of vector translation. It is the most robust way to ensure you don't lose track of the geometry.

Q: Why can't we just use the slope of the line to find the coordinates?

The slope tells you the direction, but not the specific location. You need the distance constraint AB = \sqrt{10} to anchor the point A onto the line. Without that distance, A could be anywhere on that infinite line. The distance formula is the bridge between the line's slope and the specific point's location.

Question

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$, then $2(\alpha + \beta + \gamma + \delta)$ is equal to

Options:

(A) 10

(B) 5

(C) 12

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with point $A(\alpha, \beta)$. We are given that it lies on the line $3y = 2x + 1$.

This relationship allows us to express $\beta$ in terms of $\alpha$:
$$\beta = \frac{2\alpha + 1}{3}$$

This reduction is crucial, as it transforms a two-variable problem into a single-variable problem involving only $\alpha$.

### The Algebraic Bridge

We utilize the distance constraint $AB = \sqrt{10}$. Applying the distance formula between $A(\alpha, \beta)$ and $B(1, 0)$, we have:
$$(\alpha - 1)^2 + (\beta - 0)^2 = 10$$

Substituting the expression for $\beta$ into this equation yields:
$$(\alpha - 1)^2 + \left( \frac{2\alpha + 1}{3} ight)^2 = 10$$

To simplify, we multiply the entire equation by $9$ to clear the denominator:
$$9(\alpha - 1)^2 + (2\alpha + 1)^2 = 90$$

Expanding the terms, we obtain:
$$9(\alpha^2 - 2\alpha + 1) + (4\alpha^2 + 4\alpha + 1) = 90$$

Combining like terms results in the quadratic equation:
$$13\alpha^2 - 14\alpha - 80 = 0$$

### The Integer Filter

Solving the quadratic equation $13\alpha^2 - 14\alpha - 80 = 0$ yields two roots:
$$\alpha = -2 \quad 	ext{and} \quad \alpha = \frac{40}{13}$$

The problem explicitly states that $\alpha \in \mathbb{Z}$. Therefore, we must reject the fractional root.

Thus, we have $\alpha = -2$. Substituting this back into our linear relation, we find:
$$\beta = \frac{2(-2) + 1}{3} = -1$$

Consequently, point $A$ is $(-2, -1)$.

### The Geometric Symmetry

To find point $C(\gamma, \delta)$, we use the property that the diagonals of a parallelogram bisect each other. The midpoint of diagonal $BD$ must coincide with the midpoint of diagonal $AC$.

Given $B(1, 0)$ and $D(1, 2)$, the midpoint $M$ is:
$$M = \left( \frac{1+1}{2}, \frac{0+2}{2} ight) = (1, 1)$$

Setting the midpoint of $AC$ equal to $M(1, 1)$:
$$\left( \frac{-2+\gamma}{2}, \frac{-1+\delta}{2} ight) = (1, 1)$$

Solving for the coordinates of $C$:
$$\gamma = 4, \quad \delta = 3$$

### Final Calculation

We are tasked with calculating the value of $2(\alpha + \beta + \gamma + \delta)$:
$$2(-2 - 1 + 4 + 3) = 2(4) = 8$$

The final result is **8**.

Animated Solution for Mathematics - Straight Lines: Let $α, β, γ, δ \in Z$ and let $A (α, β), B (1, 0), C (γ, δ)$ and $D (1, 2)$ be the vertices of a parallelogram $A B C D$ . If $A B = 10$ and the points $A$ and $C$ lie on the line $3 y = 2 x + 1$ , then $2 (α + β + γ + δ)$ is equal to

Select Answer:

Visualized Solution

Problem Overview

Point $A$ on the Line

Distance $A B = 10$

Substitution & Expansion

Solving the Quadratic

Integer Constraint for $α$

Finding $β$

Parallelogram Diagonals

Midpoint of $B D$

Finding $C (γ, δ)$

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Algebraic Bridge

The Integer Filter

The Geometric Symmetry

Final Calculation

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Select Answer:

Visualized Solution

Problem Overview

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A on the Line

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB=10​

Substitution & Expansion

Solving the Quadratic

Integer Constraint for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Parallelogram Diagonals

Midpoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(γ,δ)

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Algebraic Bridge

The Integer Filter

The Geometric Symmetry

Final Calculation

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