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Step-by-Step Solution
Step 1: Set up the coordinates of the triangle
Let the triangle have vertices at the origin $(0,0)$, on the x-axis $(a,0)$, and
on the y-axis $(0,b)$, where $a$ and $b$ are integers (non-zero).
Step 2: Write the formula for the area of the triangle
The area $A$ of a triangle with vertices at $(0,0)$, $(a,0)$, and $(0,b)$ is given by:
$A = \frac{1}{2} \bigl| a \times b \bigr|.$
Step 3: Apply the given area condition
We are told that the area is $50$ square units, so:
$\frac{1}{2} |a\,b| = 50 \implies |a\,b| = 100.$
This means $a \times b = \pm 100.$
Step 4: Find integer solutions to $a\,b = \pm 100$
First consider $a\,b = 100.$ The number of integer pairs $(a,b)$ satisfying $a\,b = 100$ is
determined by the number of divisors of $100$.
The prime factorization of $100$ is $2^2 \times 5^2$, which has $(2+1)(2+1) = 9$ positive divisors.
Hence, there are 9 positive pairs $(a, b)$ such that $a\,b = 100$.
Each positive pair $(a,b)$ has a corresponding negative pair $(-a,-b)$ that also satisfies $a\,b = 100.$
Thus, there are $9 \times 2 = 18$ integer solutions to $a\,b = 100.$
Similarly, $a\,b = -100$ also has the same count of solutions: 18.
Adding these gives a total of $18 + 18 = 36$ valid integer pairs $(a,b)$ such that
the area is $50$.
Step 5: Conclude the count of such triangles
For each valid pair $(a,b)$, we get a unique triangle in the plane with area $50$.
Therefore, the number of triangles in the set $S$ is $36$.
