Kondakov G.V.

Divide the segment $[-1,1]$ into black and white segments so that the integrals of any a) linear function;

b) quadratic trinomial over the white and black segments are equal.

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Since the integral of a polynomial over an interval is equal to the increment of its antiderivative, our task is a special case of problem $\underline{98268}$.

## Answer

For example, a) intervals $[-1,-1 / 2],[1 / 2,1]$ - black, $[-1 / 2,1 / 2]-$ white.

b) intervals $[-1,-3 / 4],[-1 / 4,0],[1 / 4,3 / 4]$ - white, $[-3 / 4,-1 / 4],[0,1 / 4],[3 / 4,1]$ - black.

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-14.jpg?height=286&width=2067&top_left_y=106&top_left_x=-2)

Author: Shestakov S.A.

In a convex quadrilateral $A B C D$, points $E$ and $F$ are the midpoints of sides $B C$ and $C D$ respectively.

The segments $A E, A F$, and $E F$ divide the quadrilateral into four triangles, the areas of which are (in some order) consecutive natural numbers. What is the greatest possible value of the area of triangle $A B D$?

## Solution

Author: $\underline{\text { Shestakov S.A. }}$

Let the areas of the triangles be $n, n+1, n+2$, and $n+3$. Then $S_{A B C D}=4 n+6$. Since $E F$ is the midline of triangle $B C D$, we have $S_{B C D}=4 S_{E C F} \geq 4 n$. Therefore, $S_{A B D}=S_{A B C D}-S_{B C D} \leq 6$.

We will show that this area can be equal to 6. Consider an isosceles trapezoid $A B C D$ with bases $A D=6, B C=4$ and height 2. Then

$S_{C F E}=1, S_{A B E}=2, S_{A D F}=3, S_{A E F}=S_{A B C D}-1-2-3=4$, and in this case $S_{A B D}=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1cbe8f655c608884328cg-14.jpg?height=214&width=526&top_left_y=1048&top_left_x=797)

## Answer

6.