Solve for the area of the triangle formed by the lines $x+y=2005, \frac{x}{2005}+\frac{y}{2006}=1, \frac{x}{2006}+\frac{y}{2005}=1$.
Solution. Let 2005 be denoted by $a$ and 2006 by $b$. Then the equations are:

$$
\begin{aligned}
& y=-x+a \\
& \frac{x}{a}+\frac{y}{b}=1 \\
& \frac{x}{b}+\frac{y}{a}=1
\end{aligned}
$$

Calculate the coordinates of the intersection points of the lines.

From equations (1) and (2):

$$
\begin{aligned}
& \frac{x}{a}+\frac{-x+a}{b}=1, \quad \text { hence } \\
& b x-a x+a^{2}=a b, \\
& x(b-a)=a b-a^{2} \\
& x=\frac{a(b-a)}{b-a}=a \quad \text { and } \quad y=0
\end{aligned}
$$

The intersection point of the two lines is $P_{1}(a ; 0)$.

Let the intersection point of lines (1) and (3) be $P_{2}$. From the symmetry of the equations, it follows that $P_{2}(0 ; a)$.

To find the coordinates of the intersection point $P_{3}$ of lines (2) and (3), solve the system of equations. From equation (2): $x=\frac{a b-a y}{b}$. Substitute this into (3):

$$
\begin{gathered}
\frac{a b-a y}{b^{2}}+\frac{y}{a}=1 \\
a^{2} b-a^{2} y+y b^{2}=a b^{2}, \quad \text { hence } \quad y=\frac{a b}{a+b}
\end{gathered}
$$

By the symmetry of the equations, the $x$-coordinate of $P_{3}$ is the same, thus

$$
P_{3}\left(\frac{a b}{a+b} ; \frac{a b}{a+b}\right)
$$

The area of the triangle $P_{1} P_{2} P_{3}$ needs to be determined.

![](https://cdn.mathpix.com/cropped/2024_05_02_852be0e42f0402ef96f9g-1.jpg?height=338&width=455&top_left_y=1628&top_left_x=824)

It is easy to show that the triangle $P_{1} P_{2} P_{3}$ is isosceles. Its base is $\overline{P_{1} P_{2}}$, and the height corresponding to this base starts from $P_{3}$ and bisects the segment $\overline{P_{1} P_{2}}$. The midpoint of $\overline{P_{1} P_{2}}$ is $F$, with coordinates $\left(\frac{a}{2} ; \frac{a}{2}\right)$.

$$
\overline{P_{1} P_{2}}=a \sqrt{2}, \quad \overline{P_{3} F}=\sqrt{2\left(\frac{a b}{a+b}-\frac{a}{2}\right)^{2}}=\left(\frac{a b}{a+b}-\frac{a}{2}\right) \sqrt{2}
$$

The area:

$$
T=\frac{a \sqrt{2}}{2} \cdot \frac{a(b-a)}{2(a+b)} \cdot \sqrt{2}=\frac{a^{2}(b-a)}{2(a+b)}
$$

Substitute the numerical values of $a$ and $b$, we get:

$$
T=\frac{2005^{2}}{2(2005+2006)} \approx 501.13 \text { square units. }
$$