5. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn, and the radius of each circle is less than the height of the triangle. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; and blue if inside all three. It turns out that the yellow area is 1000, the green area is 100, and the blue area is 1. Find the area of the triangle. (P. D. Mulyenko)

![](https://cdn.mathpix.com/cropped/2024_05_06_ada50bc4f8866d307083g-2.jpg?height=323&width=354&top_left_y=878&top_left_x=1551)
Solution. The sum of the areas of the three circles is $1000+2 \cdot 100+3 \cdot 1=1203$; the sum of the areas of the three "lenses" is $100+3 \cdot 1=103$ (a "lens" is the intersection of two circles).

The area of the triangle is $S_{1}-S_{2}+S_{3}$, where

$S_{1}=1203 / 6-$ the sum of the areas of the three 60-degree sectors,

$S_{2}=103 / 2$ - the sum of the areas of the halves of the three "lenses" lying inside the triangle;

$S_{3}=1-$ the area of the blue region.

Indeed, with this calculation, each yellow region inside the triangle is counted 1 time, each green region: $2-1=1$ time, the blue region: $3-3+1=1$ time.

In total, we get $\frac{1203}{6}-\frac{103}{2}+1=\frac{1203-309+6}{6}=\frac{900}{6}=150$.

Answer: 150.

Criteria. Up to 3 points will be deducted for errors in the inclusion-exclusion formula.