## Task A-1.1.

Marko drew a rectangle with two blue sides of length 24 and two red sides of length 36. He colored each point inside the rectangle with the color of the nearest side. Points that are equally distant from a blue and a red side were colored black. Determine the area of the red part of the rectangle.
## First Solution.

Let $ABCD$ be a rectangle such that $|AB|=|CD|=36$ and $|BC|=|DA|=24$.

Let point $E$ be the intersection of the angle bisector of $\angle BAD$ and the angle bisector of $\angle ADC$, and point $F$ be the intersection of the angle bisector of $\angle ABC$ and the angle bisector of $\angle BCD$.

Points on segment $\overline{AE}$ are equidistant from segments $\overline{AB}$ and $\overline{AD}$, so all of them are black. Similarly, points on segments $\overline{BF}$, $\overline{CF}$, and $\overline{DE}$ are all black.

The closest side to points inside triangle $AED$ is side $\overline{AD}$, so all these points are blue. Similarly, all points inside triangle $BCF$ are blue.

Points inside trapezoid $ABFE$ are closest to side $\overline{AB}$, so all of them are red. Similarly, all points inside trapezoid $EFCD$ are red.

![](https://cdn.mathpix.com/cropped/2024_05_30_9a2ee33c697c48e6349cg-01.jpg?height=537&width=743&top_left_y=1670&top_left_x=588)

The desired area is the area of rectangle $ABCD$ minus the areas of triangles $AED$ and $BCF$.

Notice that triangles $AED$ and $BCF$ are right triangles and that $|AE|=|ED|=|CF|=|FB|$. Let the length of side $|AE|$ be $a$.

The Pythagorean theorem in triangle $AED$ tells us that $a \sqrt{2}=24$.

Therefore, the area of each of the triangles $AED$ and $BCF$ is $\frac{a^{2}}{2}=\frac{24^{2}}{2 \cdot 2}=144$.

1 point

Finally, the desired area is $36 \cdot 24 - 2 \cdot 144 = 576$.

1 point