Convex Quadratic?

The equation, as can be easily found by using the given points is $x^2+x+1$

Now, observe that area of triangle $ABD$ is fixed, hence all we have to do is maximize area of triangle $BCD$ . Since base $BD$ is fixed, all that's left to do is to maximize perpendicular distance of $C$ from $BD$ . This happens when the tangent at point $C$ has same value as the slope of line connecting $B$ and $D$ . Since in a parabola, the slope at a given point is proportional to its $x$ coordinate, the coordinate $\alpha$ that we are looking for will be the A.M of $B$ and $D$ . $\Rightarrow \alpha=1/2$

Hence answer is $\boxed{11}$

On an interesting note, we could have found out $\alpha$ without finding the equation of the curve! We had to find it only because of $a,b,c$ !

We get the eq $y=x^2+x+1$ by the given three points.

Now for C( $\alpha,\beta$ ) $\equiv (\alpha,\alpha^2+\alpha+1)$

Now writing area of $\Delta BCD$ in coordinates form

we obtain $A=\frac{-3x^2+3x+6}{2}$ Differentiating and equating to zero

$x=\frac{1}{2} =\alpha$

$\beta=7/4$

Hence

$a+b+c+2\alpha+4\beta= 1+1+1+2(\frac{1}{2})+4(\frac{7}{4})=11$

Actually I think the 'Level 5' label and the 39% correct rate doesn't match the real difficulty of this question...

First, it has friendly give out three points on the parabola, so we have: $\begin{cases} 4a-2b+c=3\\ a-b+c=1\\ 4a+2b+c=7 \end{cases}$

Easy to know that $a=b=c=1$ , so the equation of the parabola is $y=x^2+x+1$ .

For $A, B, D$ are fixed points, $S_{\triangle ABD}$ is a fixed value. So we can just consider when $S_{\triangle BDC}$ has maximum value.

Think the edge $BD$ as the base of $\triangle BDC$ , which is fixed, and when does the height has maximun value? The answer is when $BD\parallel\text{Tangent of the parabola through point C}$ .

So we have, $\displaystyle{Slope_{BD}=\frac{dy}{dx}\Big|_{x=\alpha}}$ , which is $2=2\alpha+1$ .

It is obivious that $\alpha=1/2$ , and $\beta=7/4$ . Substitute $a,b,c,\alpha,\beta$ into the origin fomula and we'll have $\boxed{11}$ .