3. In the square, three parts are shaded as shown in the figure. The common point of the shaded figures is equally distant from all sides of the square, and the total area of the shaded figures is one third of the area of the square. What fraction of the perimeter of the square is the sum of the lengths of the sides of the shaded figures $a+b+c+d$? Justify your answer.

![](https://cdn.mathpix.com/cropped/2024_06_03_c10e5f238c827120e685g-1.jpg?height=586&width=600&top_left_y=1660&top_left_x=748)
3. 

![](https://cdn.mathpix.com/cropped/2024_05_30_4ecdfa2bc6a4546834fdg-2.jpg?height=474&width=460&top_left_y=317&top_left_x=798)

Let the common point of the shaded parts be $S$, and the length of the side of the square be $x$.

Since $S$ is equidistant from all sides of the given square, the heights of the triangles that form the shaded parts are equal to half the length of the side of the square, i.e., $\frac{x}{2}$.

1 POINT

The area of the shaded part is equal to one-third of the area of the square, so we have:

$\frac{1}{2} a \cdot \frac{x}{2} + \frac{1}{2} b \cdot \frac{x}{2} + \frac{1}{2} c \cdot \frac{x}{2} + \frac{1}{2} d \cdot \frac{x}{2} = \frac{1}{3} x^{2}$

$\frac{a x}{4} + \frac{b x}{4} + \frac{c x}{4} + \frac{d x}{4} = \frac{1}{3} x^{2}$

1 POINT

$x(a + b + c + d) = \frac{4 x^{2}}{3}$

$a + b + c + d = \frac{4 x}{3}$

$a + b + c + d = \frac{o}{3}$.