Let X=\(x, y) \in \mathbbZ \times \mathbbZ: \fracx^28+\fracy^220\lt 1. and .y^2\lt 5 x\ . 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 PYQ — JEE Advanced 2023 · Shift 1 · Q23 with Solution

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Correct answer: B

Find all integer points in $X=\left\{(x,y)\in \mathbb Z^2: \frac{x^2}{8}+\frac{y^2}{20}<1,\; y^2<5x\right\}.$

We need integer lattice points satisfying both inequalities.

Use the ellipse condition $\frac{x^2}{8}+\frac{y^2}{20}<1.$ Since $\dfrac{x^2}{8}<1$ , we get $x^2<8$ , so $x\in\{-2,-1,0,1,2\}.$ Also $\dfrac{y^2}{20}<1$ , so $y^2<20$ , hence $y\in\{-4,-3,-2,-1,0,1,2,3,4\}.$

Now combine with the second condition: $y^2<5x.$ Since $y^2\ge 0$ , this forces $x>0$ . Thus only $x=1,2$ are possible.

Check possible points

For $x=1$

Then $y^2<5 \implies y=0,\pm1,\pm2.$ But ellipse gives

\implies \frac{y^2}{20}<\frac78 \implies y^2<\frac{35}{2}=17.5,$$ which is true for $y=0,\pm1,\pm2$. So points are $$(1,0),(1,1),(1,-1),(1,2),(1,-2).$$ ### For $x=2$ Then $$y^2<10 \implies y=0,\pm1,\pm2,\pm3.$$ But ellipse gives $$\frac{4}{8}+\frac{y^2}{20}<1 \implies \frac12+\frac{y^2}{20}<1 \implies y^2<10,$$ so same possibilities hold. Thus points are $$(2,0),(2,1),(2,-1),(2,2),(2,-2),(2,3),(2,-3).$$ Hence total number of points in $X$ is $$5+7=12.$$ --- 4. **Total number of ways to choose 3 distinct points** $$\binom{12}{3}=220.$$ We must count triples forming a triangle whose area is a **positive integer**. --- 5. **Structure of the points** All points lie on the two vertical lines $x=1$ and $x=2$. So any non-collinear triangle must have: - two points from one vertical line, and - one point from the other vertical line. If all three are on the same vertical line, area is $0$. Since the horizontal distance between the lines is $1$, if the two points on one line have $y$-coordinates $a,b$, then the area is $$\frac12 |a-b|.$$ So the area is a positive integer iff $|a-b|$ is an even positive integer. Thus we count all valid triples where the pair taken from the same vertical line has **even difference in $y$**. --- 6. **Case 1: choose 2 points from $x=1$ and 1 point from $x=2$** On $x=1$, the available $y$-values are $$\{-2,-1,0,1,2\}.$$ We need pairs with even difference. This happens when both $y$'s have same parity. - Even values: $-2,0,2$ → number of pairs $=\binom{3}{2}=3$ - Odd values: $-1,1$ → number of pairs $=\binom{2}{2}=1$ So valid pairs on $x=1$: $$3+1=4.$$ For each such pair, the third point can be any of the 7 points on $x=2$. Hence count in this case: $$4\times 7=28.$$ --- 7. **Case 2: choose 2 points from $x=2$ and 1 point from $x=1$** On $x=2$, the available $y$-values are $$\{-3,-2,-1,0,1,2,3\}.$$ Again need even difference, i.e. same parity. - Even values: $-2,0,2$ → pairs $=\binom{3}{2}=3$ - Odd values: $-3,-1,1,3$ → pairs $=\binom{4}{2}=6$ So valid pairs on $x=2$: $$3+6=9.$$ For each such pair, choose any of the 5 points on $x=1$ as third point. Hence count in this case: $$9\times 5=45.$$ --- 8. **Total favorable triples** $$28+45=73.$$ Therefore the required probability is $$\frac{73}{220}.$$ --- 9. **Match with options** This is $$\boxed{\frac{73}{220}},$$ which is **Option B**. --- 10. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

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View written solutionFree

Correct answer: B

Find all integer points in $X=\left\{(x,y)\in \mathbb Z^2: \frac{x^2}{8}+\frac{y^2}{20}<1,\; y^2<5x\right\}.$

We need integer lattice points satisfying both inequalities.

Use the ellipse condition $\frac{x^2}{8}+\frac{y^2}{20}<1.$ Since $\dfrac{x^2}{8}<1$ , we get $x^2<8$ , so $x\in\{-2,-1,0,1,2\}.$ Also $\dfrac{y^2}{20}<1$ , so $y^2<20$ , hence $y\in\{-4,-3,-2,-1,0,1,2,3,4\}.$

Now combine with the second condition: $y^2<5x.$ Since $y^2\ge 0$ , this forces $x>0$ . Thus only $x=1,2$ are possible.

Check possible points

For $x=1$

Then $y^2<5 \implies y=0,\pm1,\pm2.$ But ellipse gives

\implies \frac{y^2}{20}<\frac78 \implies y^2<\frac{35}{2}=17.5,$$ which is true for $y=0,\pm1,\pm2$. So points are $$(1,0),(1,1),(1,-1),(1,2),(1,-2).$$ ### For $x=2$ Then $$y^2<10 \implies y=0,\pm1,\pm2,\pm3.$$ But ellipse gives $$\frac{4}{8}+\frac{y^2}{20}<1 \implies \frac12+\frac{y^2}{20}<1 \implies y^2<10,$$ so same possibilities hold. Thus points are $$(2,0),(2,1),(2,-1),(2,2),(2,-2),(2,3),(2,-3).$$ Hence total number of points in $X$ is $$5+7=12.$$ --- 4. **Total number of ways to choose 3 distinct points** $$\binom{12}{3}=220.$$ We must count triples forming a triangle whose area is a **positive integer**. --- 5. **Structure of the points** All points lie on the two vertical lines $x=1$ and $x=2$. So any non-collinear triangle must have: - two points from one vertical line, and - one point from the other vertical line. If all three are on the same vertical line, area is $0$. Since the horizontal distance between the lines is $1$, if the two points on one line have $y$-coordinates $a,b$, then the area is $$\frac12 |a-b|.$$ So the area is a positive integer iff $|a-b|$ is an even positive integer. Thus we count all valid triples where the pair taken from the same vertical line has **even difference in $y$**. --- 6. **Case 1: choose 2 points from $x=1$ and 1 point from $x=2$** On $x=1$, the available $y$-values are $$\{-2,-1,0,1,2\}.$$ We need pairs with even difference. This happens when both $y$'s have same parity. - Even values: $-2,0,2$ → number of pairs $=\binom{3}{2}=3$ - Odd values: $-1,1$ → number of pairs $=\binom{2}{2}=1$ So valid pairs on $x=1$: $$3+1=4.$$ For each such pair, the third point can be any of the 7 points on $x=2$. Hence count in this case: $$4\times 7=28.$$ --- 7. **Case 2: choose 2 points from $x=2$ and 1 point from $x=1$** On $x=2$, the available $y$-values are $$\{-3,-2,-1,0,1,2,3\}.$$ Again need even difference, i.e. same parity. - Even values: $-2,0,2$ → pairs $=\binom{3}{2}=3$ - Odd values: $-3,-1,1,3$ → pairs $=\binom{4}{2}=6$ So valid pairs on $x=2$: $$3+6=9.$$ For each such pair, choose any of the 5 points on $x=1$ as third point. Hence count in this case: $$9\times 5=45.$$ --- 8. **Total favorable triples** $$28+45=73.$$ Therefore the required probability is $$\frac{73}{220}.$$ --- 9. **Match with options** This is $$\boxed{\frac{73}{220}},$$ which is **Option B**. --- 10. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

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