Let $X=\left\{\right(x,y)\in Z\times Z:\frac{x^2}{8}+\frac{y^2}{20}<1$ and $y^2<5x$}. Three distinct points $P,Q,R$ are randomly chosen from $X$. What is the probability that $P,Q,R$ form a triangle whose area is a positive integer, is ?

The correct option is $\frac{73}{220}$.

Step 1: Determine the points belonging to set $X$

The set $X$ is given by integer coordinates $(x,y)\in Z\times Z$ satisfying two inequalities:

$\frac{x^2}{8}+\frac{y^2}{20}<1$

and

$y^2<5x$

From $y^2<5x$, since $y^2\ge <!--\; \ge \; -->0$, we must have $x>0$. So $x$ can take positive integer values $1,2,3,\dots$

From $\frac{x^2}{8}+\frac{y^2}{20}<1$, we observe that $x^2<8$, which implies $x$ can only be $1$ or $2$.

Let's find the valid integer values for $y$ corresponding to each possible value of $x$:

1. For $x=1$:

$y^2<5\left(1\right)=5$ and $\frac{1}{8}+\frac{y^2}{20}<1\Rightarrow <!--\; \Rightarrow \; -->y^2<20\times \frac{7}{8}=17.5$

Combining both conditions, $y^2<5\Rightarrow <!--\; \Rightarrow \; -->y\in <!--\; \in \; -->\{-2,-1,0,1,2\}$ (5 points).

2. For $x=2$:

$y^2<5\left(2\right)=10$ and $\frac{4}{8}+\frac{y^2}{20}<1\Rightarrow <!--\; \Rightarrow \; -->\frac{y^2}{20}<\frac{1}{2}\Rightarrow <!--\; \Rightarrow \; -->y^2<10$

This gives $y\in <!--\; \in \; -->\{-3,-2,-1,0,1,2,3\}$ (7 points).

Thus, the total number of points in set $X$ is:

$n\left(X\right)=5+7=12$

Step 2: Total number of ways to choose 3 distinct points

The total number of ways to select 3 distinct points from set $X$ is:

$N=\left(\begin{array}{}123\end{array}\right)=\frac{12\times 11\times 10}{3\times 2\times 1}=220$

Step 3: Condition for a triangle to have a positive integer area

All points in $X$ lie on only two vertical lines: $x=1$ and $x=2$.

Any 3 chosen points lying on the same line are collinear and form an area of $0$.

Therefore, to form a non-degenerate triangle, we must select either:

• 2 points from $x=1$ and 1 point from $x=2$, OR
• 1 point from $x=1$ and 2 points from $x=2$.

If two points $(x_1,y_1)$ and $(x_1,y_2)$ lie on the same vertical line and the third point lies on the other vertical line, the height of the triangle perpendicular to the vertical line is $h=|2-1|=1$.

The area of the triangle is given by:

$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}|y_1-y_2|\times 1=\frac{|{\mathrm{ystyle="border-style:none;"}}_{}{y}_1-y_2|}{2}$

For the area to be a positive integer, $|y_1-y_2|$ must be a positive even integer. This means $y_1$ and $y_2$ must have the same parity (both even or both odd).

Step 4: Count the favorable outcomes

Case 1: 2 points on $x=1$ and 1 point on $x=2$

On line $x=1$, $y\in <!--\; \in \; -->\{-2,-1,0,1,2\}$:

• Even values: $\{-2,0,2\}$ (3 values) $\Rightarrow <!--\; \Rightarrow \; -->\left(\begin{array}{}32\end{array}\right)=3$ pairs
• Odd values: $\{-1,1\}$ (2 values) $\Rightarrow <!--\; \Rightarrow \; -->\left(\begin{array}{}22\end{array}\right)=1$ pair

Total pairs of same parity on $x=1$ is $3+1=4$.

For each pair, we can choose any of the 7 points on $x=2$:

$n_1=4\times 7=28$

Case 2: 1 point on $x=1$ and 2 points on $x=2$

On line $x=2$, $y\in <!--\; \in \; -->\{-3,-2,-1,0,1,2,3\}$:

• Even values: $\{-2,0,2\}$ (3 values) $\Rightarrow <!--\; \Rightarrow \; -->\left(\begin{array}{}32\end{array}\right)=3$ pairs
• Odd values: $\{-3,-1,1,3\}$ (4 values) $\Rightarrow <!--\; \Rightarrow \; -->\left(\begin{array}{}42\end{array}\right)=6$ pairs

Total pairs of same parity on $x=2$ is $3+6=9$.

For each pair, we can choose any of the 5 points on $x=1$:

$n_2=9\times 5=45$

Total number of favorable triangles:

$\text{Favorable outcomes}=28+45=73$

Step 5: Calculate the required probability

The required probability is:

$P=\frac{73}{220}$