Consider a rectangle ABCD having 5,7,6,9 points in the interior of the line segments AB, CD, BC, DA

Q: Wait, why do we reject the negative root here? Or rather, why can't we just pick any 3 points from the total points?

That is a fantastic question! If you simply picked any 3 points from the total, you would accidentally include sets of 3 points that lie on the same side of the rectangle. Because those points are collinear, they cannot form a triangle. By forcing the selection to be from different sides, we automatically guarantee that no three points are collinear, ensuring every selection forms a valid triangle.

Q: Why do we sum the cases for $\alpha$ but multiply for $\beta$?

Think of it as a logical structure. For $\alpha$, the cases (choosing different sets of 3 sides) are mutually exclusive—you cannot have a triangle that uses both 'Side 1, 2, 3' and 'Side 1, 2, 4' simultaneously as its set of sides. Therefore, we add them. For $\beta$, we are performing a single, unified selection process where we must pick one point from each of the four sides to complete the quadrilateral, so we multiply the possibilities.

Q: What if the points were at the corners? Would that change the math?

It absolutely would! If points were at the corners, they would belong to two sides simultaneously. This would make the selection process much more complex because you would have to subtract cases where you accidentally picked the same vertex twice or created collinear sets. The problem specifically places them in the 'interior' to keep the geometry clean and the combinatorial logic elegant.

Question

Consider a rectangle ABCD having 5,7,6,9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta - \alpha)$ is equal to :

Options:

(A) 795

(B) 1173

(C) 1890

(D) 717

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: A Combinatorial Journey

Imagine you are standing before a large rectangle, $ABCD$. Along its edges, there are tiny, distinct markers—points scattered like stars in the night sky.

We have $5$ points on $AB$, $7$ on $CD$, $6$ on $BC$, and $9$ on $DA$. Our goal is to count the number of triangles ($\alpha$) and quadrilaterals ($\beta$) we can form using these points as vertices.

The vertices must come from different sides. Because three points on a single line are collinear and cannot form a triangle, restricting our selection to different sides ensures that no three points are ever collinear. This is the secret to unlocking the problem.

### The Triangle Hunt: Calculating $\alpha$

A triangle requires exactly $3$ vertices. Since we must pick these from different sides of our rectangle, we are essentially choosing $3$ sides out of the $4$ available.

This leads us to $4$ distinct scenarios, or cases, based on which side we choose to leave out:

1. **Case 1 (Excluding $DA$):** We choose sides $AB, CD,$ and $BC$. The number of ways to pick one point from each is the product of their counts: $5 	imes 7 	imes 6 = 210$.

2. **Case 2 (Excluding $BC$):** We choose sides $AB, CD,$ and $DA$. The product is $5 	imes 7 	imes 9 = 315$.

3. **Case 3 (Excluding $CD$):** We choose sides $AB, BC,$ and $DA$. The product is $5 	imes 6 	imes 9 = 270$.

4. **Case 4 (Excluding $AB$):** We choose sides $CD, BC,$ and $DA$. The product is $7 	imes 6 	imes 9 = 378$.

Since these cases are mutually exclusive, we sum them up to find the total value of $\alpha$:

$$\alpha = 210 + 315 + 270 + 378 = 1173$$

### The Quadrilateral Quest: Calculating $\beta$

Now, let us turn our attention to $\beta$, the number of quadrilaterals. A quadrilateral is defined by $4$ vertices.

Since we are constrained to picking vertices from different sides, and we only have $4$ sides, we are forced to pick exactly one point from every single side. The calculation is a direct application of the Fundamental Counting Principle:

$$\beta = n_1 	imes n_2 	imes n_3 	imes n_4$$

Substituting our values, we get:

$$\beta = 5 	imes 7 	imes 6 	imes 9 = 1890$$

### The Final Synthesis

We have navigated the combinatorial landscape. We found that $\alpha = 1173$ and $\beta = 1890$.

The problem asks for the difference, $(\beta - \alpha)$. This final step is the reward for our careful accounting:

$$\beta - \alpha = 1890 - 1173 = 717$$

The final result is **$717$**. By breaking a complex spatial problem into smaller, manageable logical cases, we transformed a daunting counting task into a simple arithmetic exercise.

Select Answer:

Visualized Solution

Visualizing the Rectangle and Points

Logic for Triangles ( $α$ )

Possible Side Combinations

Computing Cases 1 and 2

Computing Cases 3 and 4

Summing up $α$

Logic for Quadrilaterals ( $β$ )

Computing $β$

Finding $β - α$

Final Answer

The Sigma Insight: Combinations and Selection

The Geometry of Constraints

The Triangle Hunt

The Quadrilateral Quest

The Final Synthesis

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