Let , , , Z and let A( , ),B(1,0),C( , ) and D(1,2) be the vertices of a parallelogram ABC…

Question

Let $\alpha ,\beta ,\gamma ,\delta \in Z$ 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

Accepted Answer

Since $ABCD$ is a parallelogram, the diagonals $AC$ and $BD$ bisect each other. Midpoint of $BD$: $\left(\dfrac{1+1}2,\dfrac{0+2}2\right)=(1,1)$, so midpoint of $AC$ is also $(1,1)$: $\alpha+\gamma=2,\ \beta+\delta=2$.

$A(\alpha,\beta)$ lies on $3y=2x+1$, and with $AB=\sqrt{10}$: $(\alpha-1)^2+\beta^2=10$. Combined with $3\beta=2\alpha+1$, and requiring integer solutions, gives $\alpha=-2,\beta=-1$ (or symmetric choice); then $\gamma=4,\delta=3$.

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

Solutions · 1