Let ABC be a triangle. Consider four points p₁, p₂, p₃, p₄ on the side AB, five points p₅, p₆, p₇, p

Q: Wait, why can't we just pick any 5 points out of 13? Why do we care about the sides?

That is a great question! If we just picked any 5 points out of 13, we would be calculating \binom{13}{5}, which is 1287. However, that includes cases where 3 or more points are collinear. A polygon cannot have three consecutive vertices on a straight line. Since all points on a single side of the triangle are collinear, picking 3 or more from one side would result in a line segment, not a pentagon. We must exclude those cases.

Q: How do I know I have found all the possible distributions of points?

We need to distribute 5 points across 3 sides (AB, BC, AC) such that no side has more than 2 points. Let the number of points on each side be n1, n2, and n3. We have the constraints: n1 + n2 + n3 = 5, where 0 \le n1 \le 4, 0 \le n2 \le 5, 0 \le n3 \le 4, and crucially, each ni \le 2. The only integer partitions of 5 into three parts, where each part is at most 2, are permutations of (1, 2, 2). These are (1, 2, 2), (2, 1, 2), and (2, 2, 1). Any other combination would either sum to less than 5 or violate the 'at most 2' rule.

Q: What if I choose 0 points from one of the sides?

If you choose 0 points from one side, you would need to choose 5 points from the remaining two sides. But each side can contribute at most 2 points. So, the maximum number of points you could get from two sides is 2 + 2 = 4. Since 4 is less than 5, it is impossible to form a pentagon if you choose 0 points from any side. Therefore, every side must contribute at least 1 point.

Question

Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, \dots, p_{13}$, is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Constraint

Imagine a triangle $ABC$ with thirteen distinct points distributed along its edges. We must select exactly five points to form a pentagon.

The fundamental constraint is that no three points can be collinear. Since points on the same side of the triangle are collinear, we can select **at most two points** from any single side.

### The Art of Partitioning

We have thirteen points total: four on $AB$, five on $BC$, and four on $AC$. Let $n_1$, $n_2$, and $n_3$ be the number of points chosen from sides $AB$, $BC$, and $AC$, respectively.

We must satisfy the equation:
$$n_1 + n_2 + n_3 = 5$$

Subject to the constraint $n_i \le 2$ for all $i \in \{1, 2, 3\}$. The only integer partition of 5 that satisfies this constraint is the set $\{1, 2, 2\}$.

### Calculating the Possibilities

We now evaluate the permutations of the distribution $(1, 2, 2)$ to determine the total number of valid pentagons.

**Case 1: The (1, 2, 2) Distribution**
We choose one point from $AB$, two from $BC$, and two from $AC$:
$$\binom{4}{1} 	imes \binom{5}{2} 	imes \binom{4}{2} = 4 	imes 10 	imes 6 = 240$$

**Case 2: The (2, 1, 2) Distribution**
We choose two points from $AB$, one from $BC$, and two from $AC$:
$$\binom{4}{2} 	imes \binom{5}{1} 	imes \binom{4}{2} = 6 	imes 5 	imes 6 = 180$$

**Case 3: The (2, 2, 1) Distribution**
We choose two points from $AB$, two from $BC$, and one from $AC$:
$$\binom{4}{2} 	imes \binom{5}{2} 	imes \binom{4}{1} = 6 	imes 10 	imes 4 = 240$$

### The Final Synthesis

These three cases are mutually exclusive and cover all possible ways to select five points without violating the collinearity constraint.

To find the total number of unique pentagons, we sum the results:
$$240 + 180 + 240 = 660$$

The total number of valid pentagons that can be formed is **660**.

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle and Points

The Pentagon Constraint

Identifying Valid Distributions

Case 1: $(1, 2, 2)$ Distribution

Case 2: $(2, 1, 2)$ Distribution

Case 3: $(2, 2, 1)$ Distribution

Final Summation

The Sigma Insight: Combinations and Selection

Analyzing the Geometric Constraint

The Art of Partitioning

Calculating the Possibilities

The Final Synthesis

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