Question

Determine all values of α for which the point (α, α²) lies inside the triangle formed by the lines 2x + 3y − 1 = 0 ; x + 2y − 3 = 0 ; 5x − 6y − 1 = 0.

Answer 1

Problem.

Determine all values of $\alpha$ for which the point $(\alpha, \alpha^2)$ lies inside the triangle formed by the lines $2x + 3y - 1 = 0; x + 2y - 3 = 0; 5x - 6y - 1 = 0$.

Solution

The given triangle is defined by the system of inequalities (see Figure)

$$\left\{ \begin{array}{l} 2x + 3y - 1 \geq 0 \\ x + 2y - 3 \leq 0 \\ 5x - 6y - 1 \leq 0 \end{array} \right.$$

The point $(t, t^2)$ will be the inner point of this triangle when

$$\left\{ \begin{array}{l} 2t + 3t^2 - 1 > 0 \\ t + 2t^2 - 3 < 0 \\ 5t - 6t^2 - 1 < 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} (t + 1)(t - \frac{1}{3}) > 0 \\ (t + \frac{3}{2})(t - 1) < 0 \\ (t - \frac{1}{3})(t - \frac{1}{2}) > 0 \end{array} \right.$$

We will find a solution to this system of inequalities by intervals (see Figure)

$$t \in \left(- \frac{3}{2}, -1\right) \cup \left(\frac{1}{2}, 1\right)$$

**Answer.** The point $(\alpha, \alpha^2)$ lies inside the given triangle when $\alpha$ belongs to the set $\left(-\frac{3}{2}, -1\right) \cup \left(\frac{1}{2}, 1\right)$.

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