The question asks for a specific number that connects the area of a triangle, its perimeter, and the radius of its in-circle.

Triangle Area Formula

Let's recall the formula for the area of a triangle ($\text{Area}$) in terms of its semi-perimeter ($s$) and the radius of its in-circle ($r$). The formula is:

\begin{equation*} \text{Area} = s \times r \end{equation*}

Here, $s$ represents the semi-perimeter of the triangle. The semi-perimeter is half of the total perimeter ($P$) of the triangle. So, we can write:

\begin{equation*} s = \frac{P}{2} \end{equation*}

Now, substitute the expression for $s$ into the area formula:

\begin{equation*} \text{Area} = \left(\frac{P}{2}\right) \times r \end{equation*}

This formula can be rearranged slightly:

\begin{equation*} \text{Area} = \frac{1}{2} \times P \times r \end{equation*}

So, the area of a triangle is half the product of its perimeter and the in-radius.

Perimeter, In-radius, and Area Relation

The question states that the product of the perimeter of a triangle, the radius of its in-circle, and a number gives the area of the triangle. Let the unknown number be $N$. According to the question, we have the following relationship:

\begin{equation*} \text{Perimeter} \times \text{In-radius} \times N = \text{Area} \end{equation*}

Using the symbols $P$ for perimeter, $r$ for in-radius, and $\text{Area}$ for area, this equation is:

\begin{equation*} P \times r \times N = \text{Area} \end{equation*}

Comparing Formulas

We have two formulas for the area of the triangle:

1. From the standard geometric relation: $\text{Area} = \frac{1}{2} \times P \times r$
2. From the question's statement: $\text{Area} = P \times r \times N$

By comparing these two equations, we can find the value of the number $N$:

\begin{equation*} P \times r \times N = \frac{1}{2} \times P \times r \end{equation*}

Assuming the perimeter $P$ and the in-radius $r$ are not zero (which is true for any valid triangle), we can divide both sides of the equation by $P \times r$:

\begin{equation*} N = \frac{1}{2} \end{equation*}

Thus, the number that satisfies the condition is $1/2$. This number is a constant value for any triangle when relating its area, perimeter, and in-radius.