JEE Advanced 2023 Paper 1Question 1
Let $X=\left\{(x,y)\in\mathbb{Z}\times\mathbb{Z}: \frac{x^2}{8}+\frac{y^2}{20}<1 \text{ and } y^2<5x\right\}$. Three distinct points $P$, $Q$ and $R$ are randomly chosen from $X$. Then the probability that $P$, $Q$ and $R$ form a triangle whose area is a positive integer, is
Probability
Solution
Integer points with $\frac{x^2}8+\frac{y^2}{20}<1$ and $y^2<5x$: $x=1,\ y\in\{-2..2\}$ and $x=2,\ y\in\{-3..3\}$, giving $|X|=12$ points hence $220$ triples. Counting triples with non-zero, even doubled area (integer area) gives $73$, so probability $\frac{73}{220}$.