orl wrote: Determine the least real number $ a$ greater than $ 1$ such that for any point $ P$ in the interior of the square $ ABCD$, the area ratio between two of the triangles $ PAB$, $ PBC$, $ PCD$, $ PDA$ lies in the interval $ \left[\frac {1}{a},a\right]$. Assume triangle $ PAB$ has the largest area, let say $ x$. Also assume triangle $ PBC$ has a greater area than triangle $ PAD$, which is $ y$ Then area of the triangle $ PCD$ and $ PAD$ are $ 2-x$ and $ 2-y$ respectively. With the condition $ 2 \ge x \ge y \ge 1$ So taking two triangles each time, we have six pairs of ratio, and we consider the one not less than $ 1$. Those are $ \frac{x}{y}, \frac{2-y}{2-x}, \frac{y}{2-x}, \frac{x}{2-y}, \frac{x}{2-x}, \frac{y}{2-y}$. We are going to consider the smallest one among these six ratios. Obviously we have $ x \ge 2-x, y \ge 2-y, 2-y \ge 2-x, x+y \ge 2$, so $ \frac{x}{y} \le \frac{x}{2-x} \le \frac{x}{2-y}$ and $ \frac{y}{2-y} \le \frac{y}{2-x}$ Also $ \frac{x}{y} \le \frac{2-y}{2-x}$ Now we have $ \frac{x}{y}$ and $ \frac{y}{2-y}$ left, and they are both not less than $ 1$. So the maximum value of the smaller one among them is actually the required $ a$. If you assume $ \frac{x}{y}=\frac{y}{2-y}$ and try to maximize them, you will get the golden ratio. Now we will prove that at least one of the $ \frac{x}{y}, \frac{y}{2-y}$ is not greater than $ \frac{1+\sqrt{5}}{2}$. If it is not true, then we will get two inequalities $ y > \sqrt{5}-1$ and $ x > \frac{1+\sqrt{5}}{2}y$, which gives $ x > 2$, hence a contradiction. It can easily checked that when $ a=\frac{1+\sqrt{5}}{2}$, the equality can hold. Therefore the answer is $ \frac{1+\sqrt{5}}{2}$