Minimizing $\beta$

First our motivation is to provide an integer bound on $\alpha$ and $\beta$ .

$\frac{77}{17} < \frac{\beta}{\alpha} < \frac{197}{43}$ $4 +\frac{9}{17} < \frac{\beta}{\alpha} < 4+\frac{25}{43}$ ---[1]

From this relation we can say that $4 < \frac{\beta}{\alpha} < 5$ . Hence clearly $\beta > 4\alpha$ ,so we can state that $\beta = 4\alpha + \gamma$ , where $0<\gamma < \alpha$ .

Putting this value of $\beta$ in (1) we get, $4 +\frac{9}{17} < \frac{4\alpha + \gamma}{\alpha} < 4 +\frac{25}{43}$ $4 +\frac{9}{17} < 4+\frac{ \gamma}{\alpha} < 4 +\frac{25}{43}$ $\frac{9}{17} < \frac{ \gamma}{\alpha} < \frac{25}{43}$ $\frac{17}{9} < \frac{ \alpha}{\gamma} < \frac{43}{25}$ $\frac{17}{9}\gamma < \alpha < \frac{43}{25}\gamma$

Now to minimize the value of $\beta$ ,we will select the smallest value of $\gamma$ for which $\alpha$ is an integer.

If we put $\gamma = 1$ ,we get $\frac{17}{9} < \alpha < \frac{43}{25}$ hence we have no possible values.

If we put $\gamma = 2$ ,we get $3\frac{11}{25} < \alpha < 3\frac{7}{9}$ hence we have no possible values.

If we put $\gamma = 3$ ,we get $5\frac{4}{9} < \alpha < 5\frac{2}{3}$ hence we have no possible values.

If we put $\gamma = 4$ ,we get $6\frac{12}{25} < \alpha < 7\frac{5}{9}$ hence $\alpha = 7$ is a possible value and hence $\beta = 4\alpha + \gamma = 28 + 4 = 32$ .

We can also see that this is the minimum possible value since if $x \ge 5$ , $\alpha > \frac{43}{25}x \ge \frac{43}{5} > 8$ . hence $\beta > 32$ .

So we can conclude that the minimum possible value of $\beta$ is $\boxed{32}$

Past INMO question