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

• A. 10
• B. 5
• C. 12
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Let $E$ be the midpoint of the diagonals. By the midpoint formula:

$\frac{\alpha + \gamma}{2} = \frac{1 + 1}{2} = 1 \quad \implies \quad \alpha + \gamma = 2$

Similarly:

$\frac{\beta + \delta}{2} = \frac{2 + 0}{2} = 1 \quad \implies \quad \beta + \delta = 2$

Therefore:

$2(\alpha + \beta + \gamma + \delta) = 2(2 + 2) = 8$