Let $X = \{(x,y) \in \mathbb{Z} \times \mathbb{Z} : \frac{x^{2}}{8} + \frac{y^{2}}{20} < 1$ and $y^{2} < 5x\}$. 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

Let X = {(x,y) ∈ Z × Z : x²/8 + y²/20 < 1 and y² < 5x}. 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 $X = \{(x,y)\in\mathbb{Z}\times\mathbb{Z} : \frac{x^{2}}{8}+\frac{y^{2}}{20}<1,\ y^{2}<5x\}$ $\text{To find: } \mathbb{P}\left(\text{area of } \triangle PQR \text{ is a positive integer}\right)$ $\text{Plan: } 1.\ \text{points of } X \quad 2.\ \binom{|X|}{3} \quad 3.\ \text{count integer-area triples}$ $\frac{x^{2}}{8}+\frac{y^{2}}{20}=1: \quad rx=\sqrt{8},\ \ ry=\sqrt{20}$ $y^{2}=5x: \quad \text{right-opening parabola}$ $x=1: \frac{1}{8}+\frac{y^{2}}{20}<1,\ y^{2}<5 \Rightarrow |y|\le 2 \Rightarrow 5$ $x=2: \frac{4}{8}+\frac{y^{2}}{20}<1,\ y^{2}<10 \Rightarrow |y|\le 3 \Rightarrow 7$ $x\ge 3: \frac{x^{2}}{8}\ge \frac{9}{8}>1 \Rightarrow \text{none} \quad \therefore |X|=12$ $\binom{12}{3} = \frac{12\cdot 11\cdot 10}{6} = 220$ $D = \left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ $\text{Area} = \frac{D}{2} \in \mathbb{Z}_{>0} \iff D \text{ is a positive even integer}$ $x\in\{1,2\} \Rightarrow \text{two of } P,Q,R \text{ share a column}$ $(1,a),\ (1,a'),\ (2,b): \quad D = |a-a'|$ $\text{Area} = \frac{|a-a'|}{2}$ $|a-a'| \text{ even}: \{-2,0\},\{-2,2\},\{0,2\},\{-1,1\} \Rightarrow 4 \text{ pairs}$ $4 \times 7 = 28$ $(2,b),\ (2,b'),\ (1,a): \quad D = |b-b'|$ $\text{same parity}: \binom{4}{2}+\binom{3}{2} = 6+3 = 9 \text{ pairs}$ $9 \times 5 = 45$ $28 + 45 = 73 \text{ favourable triples}$ $\mathbb{P} = \frac{73}{220} \quad \Rightarrow \quad \text{option (2)}$