Problem Analysis #21

To solve the problem, we begin by analyzing the structure of the lattice points and the number of ways to choose three distinct points.

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Step 1: Total Number of Ways to Choose Three Points

We are given 15 lattice points in a 3×5 grid (i.e., x $$\in$$ \{1, 2, 3\}, y $$\in$$ \{1, 2, 3, 4, 5\}). The total number of ways to choose 3 distinct points from these is:

$$\binom{15}{3} = \frac{15 \cdot 14 \cdot 13}{6} = 455$$

This will be the denominator for our probability.

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Step 2: Counting Collinear Triples

A triangle is formed if and only if the three chosen points are not collinear. Therefore, the number of favorable outcomes (triangles) is:

$$\text{Number of triangles} = \text{Total triples} - \text{Collinear triples}$$

We now count all sets of three collinear points in the grid.

(a) Vertical Lines

There are 3 vertical lines (x = 1, 2, 3), each with 5 points. For each such line, the number of collinear triples is:

$$\binom{5}{3} = 10$$

So total vertical collinear triples:

$$3 \times 10 = 30$$

(b) Horizontal Lines

There are 5 horizontal lines (y = 1 to 5), each with 3 points. Each gives:

$$\binom{3}{3} = 1$$

Total horizontal collinear triples:

$$5 \times 1 = 5$$

(c) Diagonal Lines with Slope 1

These lines have the form $y = x + c$. To stay within the grid, $c$ must satisfy 1 $$\leq$$ y $$\leq$$ 5, so valid values are $c = 0, 1, 2$. Each such line contains exactly 3 points. Thus:

$$3 \times 1 = 3 \text{ collinear triples}$$

(d) Diagonal Lines with Slope -1

These lines have the form $y = -x + c$. Valid values of $c$ are $4, 5, 6$, giving again 3 lines with 3 points each:

$$3 \times 1 = 3 \text{ collinear triples}$$

(e) Diagonal Lines with Slope 2

Only one such line exists: $y = 2x - 1$, passing through (1,1), (2,3), (3,5). It has 3 points:

$$1 \text{ collinear triple}$$

(f) Diagonal Lines with Slope -2

Only one such line exists: $y = -2x + 7$, passing through (1,5), (2,3), (3,1). Again, 3 points:

$$1 \text{ collinear triple}$$

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Step 3: Total Collinear Triples

Summing up all collinear triples:

$$30 \text{ (vertical)} + 5 \text{ (horizontal)} + 3 \text{ (slope 1)} + 3 \text{ (slope -1)} + 1 \text{ (slope 2)} + 1 \text{ (slope -2)} = 43$$

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Step 4: Number of Triangles

$$\text{Triangles} = 455 - 43 = 412$$

Thus, the probability that the three selected points form a triangle is:

$$p_5 = \frac{412}{455}$$

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Step 5: Simplifying the Fraction

We check whether $\gcd(412, 455) = 1$:

• Prime factorization:
  • 455 = 5 $$\times$$ 7 $$\times$$ 13
  • 412 = 4 $$\times$$ 103

Since 103 is a prime number and does not divide 5, 7, or 13, the fraction is in lowest terms.

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Final Answer

The problem asks for $p + q$, where $p/q$ is the reduced form of the probability.

$$p = 412,\quad q = 455 \Rightarrow p + q = 412 + 455 = \boxed{867}$$