8.5. Given a convex quadrilateral $A B C D$ with side $A D$ equal to 3. Diagonals $A C$ and $B D$ intersect at point $E$, and it is known that the areas of triangles $A B E$ and $D C E$ are both 1. Find the side $B C$, given that the area of $A B C D$ does not exceed 4.
# Answer: 3.

Solution: Triangles $A B D$ and $A C D$ have the same area, as they are composed of the common $A E D$ and equal-area $A B E$ and $D C E$. Since $A B D$ and $A C D$ have the same base $A D$, their heights to this base are equal. Therefore, $B C$ is parallel to $A D$, meaning our quadrilateral is a trapezoid.

Let $B C$ be $x$. Triangles $B E C$ and $A E D$ are similar, so $B E / E D = B C / A D = x / 3$. Additionally, the area of $B E C$ is $B E / E D$ of the area of $B C D$, since these triangles have the same base, and their heights are in the ratio of $B E$ to $E D$. Therefore, the area of $B E C$ is $x / 3$. Similarly, the area of $A E D$ is $3 / x$.

Since the area of the entire quadrilateral does not exceed 4, $x / 3 + 3 / x \leq 2$.

However, the sum of the reciprocals of positive quantities is always at least 2, and equality is achieved only when the terms are equal, i.e., in our case, $x / 3 = 3 / x$, from which $x$ equals 3.

Criteria: Only the answer - 0 points. Proved that $A B C D$ is a trapezoid - plus 2 points. Obtained the equation - plus 1 point. Not explained why the inequality is possible only when the terms are equal - deduct 1 point.

## Criteria for Determining Winners and Prize Winners of the All-Siberian Open Mathematics Olympiad for Schoolchildren (2015-2016 academic year)

According to the Regulations, the winners and prize winners of the Olympiad were determined based on the results of the Final Stage of the Olympiad. The total number of winners and prize winners was 380 out of 1578 participants, which is 24.08%. The number of winners was 85, which is 5.38%.

Based on the overall ranking of participants and considering the noticeable gaps in the scores of the top-ranked groups, the Olympiad jury developed the following criteria for determining winners and prize winners: The maximum possible score is 35 points.

## 11th Grade:

Winners:

Participants who scored more than 77% of the maximum possible points, i.e., from 27 to 35 points; Prize winners:

2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 26 points 3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 21 points

10th Grade:

Winners:

Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 29 points

3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 21 points

9th Grade:

Winners:

Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than 65% of the maximum possible points, i.e., from 24 to 29 points

3rd degree - more than 51% of the maximum possible points, i.e., from 18 to 23 points

8th Grade:

Winners:

Participants who scored more than 82% of the maximum possible points, i.e., from 29 to 35 points;

Prize winners:

2nd degree - more than 62% of the maximum possible points, i.e., from 22 to 28 points 3rd degree - more than 48% of the maximum possible points, i.e., from 17 to 21 points

7th Grade:

Winners:

Participants who scored more than 85% of the maximum possible points, i.e., from 30 to 35 points;

Prize winners:

2nd degree - more than 71% of the maximum possible points, i.e., from 24 to 29 points

3rd degree - more than 42% of the maximum possible points, i.e., from 15 to 21 points

Co-Chair of the Mathematics Jury

![](https://cdn.mathpix.com/cropped/2024_05_06_1e3236413cdb88a8afc4g-4.jpg?height=143&width=314&top_left_y=2367&top_left_x=928)

A.Yu. Avdyushenko