Problem
Let X = {(x, y) ∈ Z x Z: x²/8 + y²/20 < 1 and y² < 5x}. Three distinct points P, Q, and R are randomly chosen from X. Then find the probability that P, Q, and R form a triangle whose area is a positive integer.
Solution
Let’s begin by sketching out the region bounded by the two inequalities in X. Then, we’ll find the integral points within that region.
The equation x²/8 + y²/20 = 1 represents an ellipse and the inequality represent the region inside it.
Similarly, y² = 5x is a parabola and the given inequality represents the region inside it. Take a look.
Now, let’s mark all integral points inside it.
The number of ways to select any 3 points from these 12 is ¹²C₃ or 220.
Now, a triangle will only be formed from the three points if two of them (forming the base) are selected from the left ‘column’ of points, and one from the right.
Or, vice versa. That is, select two points (forming the base) from the right column, and one from the left.
The height of each of these triangles will always be 1 unit. So, to form a triangle with an integer area, the base must be even.
I’ll discuss two methods of counting such triangles.
Method 1
Let’s start with the base being chosen from the left column, and the opposite vertex from the right.
Since there are only 5 points in the left column, the base length can only be either 2 or 4 units.
Let’s again with bases of length 2. In how many ways can we select the base and the opposite vertex? Take a look.
Drag the first slider to observe all the possible bases, and the second one to see all possible opposite vertices.
Since there are 3 possible bases and 7 possible vertices, the number of such triangles will be 3 x 7 or 21.
Similarly, if the base is 4 units, there will be only 1 possible base, and 7 possible opposite vertices again.
So, 7 more triangles, giving a total of 21 + 7 or 28 triangles.
Now, lets flip the triangles – base from the right column, and the opposite vertex from the left.
This time, there are 7 points in the right column. So, the base length can be 2, 4, or 6 units.
Let’s start again with bases of length 2. In how many ways can we select the base and the opposite vertex? Take a look.
Drag the first slider to observe all the possible bases, and the second one to see all possible opposite vertices.
Since there are 5 possible bases and 5 possible vertices, the number of such triangles will be 5 x 5 or 25.
If the base length is 4 units, there will be 3 possible bases, and 5 possible opposite vertices again. So, 3 x 5 or 15 more triangles (i.e. with base length as 4).
Finally, if the base length is 6 units, there can be only 1 possible base (joining the extreme points). Since there are still 5 possible opposite vertices, there will be 1 x 5 or 5 such triangles (i.e. with base length as 6).
So, the total number of triangles formed will be 25 + 15 + 5 or 45.
Combining this with the previous 28, we get a total of 45 + 28 or 73 triangles.
So, the required probability will be 73/220. Phew!
Method 2
For the base to be even, the y-coordinates of the selected points, from a given column, must be both odd or both even.
This is because the base equals the difference of the y-coordinates of the selected points.
And, the difference of two odd or two even numbers is always even, implying that the base length is even!
So, here are the points from the left column: {(1, -2), (1, -1), (1, 0), (1, 1), (1, 2)}
We can select points with even y-coordinates in ³C₂ ways, and those with odd y-coordinates in ²C₂ ways.
So, the total number of selections is ³C₂ + ²C₂, i.e. 3 + 1 or 4.
Now, given the base, the opposite vertex from the right column can be selected in 7 ways, just like before.
So, the number of triangles will be 4 x 7 or 28.
Similarly, here are the points from the left column: {(2, -3), (2, -2), (2, -1), (2, 0), (2, 1), (2, 2), (2, 3)}
This time, we can select points with even y-coordinates in ³C₂ ways, and those with odd y-coordinates in ⁴C₂ ways.
So, the total number of selections is ³C₂ + ⁴C₂, i.e. 3 + 6 or 9.
For these bases, the opposite vertex, from the left column, can be selected in 5 ways.
So, the number of triangles will be 9 x 5 or 45.
Finally, the total number of triangles with integer area equals 28 + 45 or 73, and the required probability is 73/220.
Comments
A tedious problem overall – involved drawing out the regions and figuring out the integral points. And then, it became a counting / probability problem. Should’ve been worth more marks, e.g. by being a two-part paragraph style problem.