A Small Restriction With A Huge Impact

Let $f(x) = ax^2 - bx +c$

Using properties of quadratic equations we get many inequalities. The following ones will be useful.

$b^2 > 4ac$ $\phantom{aaaaa}$ ( $f(x) = 0$ has $2$ distinct roots)

$\frac{c}{a}< 1$ $\phantom{aaaaaaa}$ (Product of roots of $f(x)$ is between $0$ and $1$ )

$a - b + c > 0$ $\phantom{aa}$ ( $f(1) > 0$ )

$\frac{b}{a} < 2$ $\phantom{aaaaaaa}$ (Sum of roots of $f(x)$ is between $0$ and $2$ )

In addition, it is given that $a, b, c > 0$

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Rearranging the above inequalites gives

$b^2 >4ac .......... (1)$

$c < a ..............(2)$

$b < a + c ...........(3)$

$b < 2a ..............(4)$

Since $(2)$ is true, $(3)$ implies $(4)$ , therefore $(4)$ becomes redundant.

Using the three inequalities, to minimize $a$ , by inspection we can see $c$ should be minimized, therefore $c = 1$

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Now, inequalities $(1)$ and $(3)$ become

$b^2 > 4a$

$b < a + 1$

The maximum value of $b$ for any value of $a$ can be obtained by setting $b = a + 1 - 1 = a$

This gives us the final inequality $a^2 > 4a, \phantom{a}\text{i.e. } a >4$ .

We can see clearly, that $\boxed{a = 5}$ is the minimum integer value that satisfies the inequality.

Small $a$ results in big opening (more spread) parabola and big $a$ results in small opening.

So, if we want the minimum $a$ we should have the roots as far as possible to each one. It will be possible with axis of symmetry passing by $x = \frac{1}{2}$ .

It means $\frac { -(-b) }{ 2a } =\frac { 1 }{ 2 } \therefore b=a$

Because 2 distinct roots, ${ b }^{ 2 }-4ac>0\Rightarrow { b }^{ 2 }>4ac$

plugging $b=a$ in last inequality, we get: ${ a }^{ 2 }>4ac \Rightarrow a > 4c$

So, to minimize $a$ we need minimize $c$ . The minimum (positive and integer) $c$ is $c = 1$ , so we have the minimum $a$ is $\boxed{a = 5}$ .

and we are done.

By Vieta's Formula, the sum of roots is $\frac {b}{a}$ and the product of roots is $\frac {c}{a}$ . Because the distinct real roots are between $0$ and $1$ exclusive. $\frac {c}{a} < 1 \Rightarrow c < a$ and $\frac {b}{a} < 1+1 \Rightarrow b < 2a$ . Noting that $6 \left (x - \frac {1}{2} \right ) \left ( x - \frac {1}{3} \right ) = 6x^2 - 5x + 1$ satisify the condition, we just need to check for $a \leq 6$

• $a = 1$ : there's no solution for $c$

• $a = 2$ : $c = 1$ only, $b = 1,2,3$ , because they are real roots, the determinant must be positive, $b^2 - 4ac > 0 \Rightarrow b = 3$ only. Solve $2x^2-3x+1 =0$ gives $x=1$ as a solution, which is invalid.

• $a = 3$ : apply the same logic above for the determinant, we have $(b,c) = (4,1), (5,1), (5,2)$ . Trial and error shows none of these satisfy the condition.

• $a=4$ : $(b,c) = (5,1),(6,1),(6,2),(7,1),(7,2),(7,3)$ . No solution.

• $a = 5$ : $(b,c) = (5,1), \ldots$ . Turns out the very first pair works out, so $(b,c) = (5,1)$ is valid. The answer is $\boxed{5}$

P.S: Once we conclude that $a=1,2,3$ don't work. And we know that at least one of $a=4,5,6$ must work. Plus, we are given 3 tries....

Here it is my way to resolve it!

$ax^{2} - bx + c = 0 \Longrightarrow x = \frac{b \pm \sqrt{b^{2} - 4ac} }{2a}$ Therefore we need to define two things: the roots of the equation must be real; the roots must be less than $1$ and greater than $0$ . Looking for the equation, we can do that:

For the real roots:
$b^{2} - 4ac > 0 \Longrightarrow a < \frac{b^{2}}{4c}$ ;

For $0<x<1$ :
$2a > b \pm \sqrt{b^{2} - 4ac} \Longrightarrow a > b - c$ ;

We get this:
$b-c < a < \frac{b^{2}}{4c}$ .

Now, we can play with this inequality until we get $5$ .

And there's a lot of sweet information there like when $c = \frac{b}{2}$ , $b-c$ and $\frac{b^{2}}{4c}$ are equal (which I'd like to know how this happens from the "master" since it looks like too difficult for me understand and work the interception of two functions with two variables - just some titles of what I have to read and study it's enough). I almost failed for thinking that $\frac{b^{2}}{4c}$ also must be an integer. What a dumb conclusion.