B3. Dana is a square grid with $7 \times 7$ points, whose vertices are marked as $A, B, C$ and $D$ (see the figure). The sides of the grid are segments $A B, B C, C D$ and $D A$. In how many ways can we choose 3 different points on the grid such that at least 1 chosen point lies on each side of the grid?

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-05.jpg?height=388&width=394&top_left_y=2122&top_left_x=1502)

67th Mathematical Competition for High School Students of Slovenia National Competition, April 22, 2023

# Solutions to problems for 1st year 

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-06.jpg?height=143&width=234&top_left_y=474&top_left_x=928)
B3. When we choose 3 points, and the grid has 4 sides, at least one of the chosen points must lie in one of the grid's corners. Let's consider several cases.

1) If exactly 1 corner of the grid contains one of the chosen points (4 possibilities), then the remaining two points must lie on the interior of each of the sides not adjacent to this corner (5 possibilities for each). There are $4 \cdot 5 \cdot 5=100$ such choices.
2) If exactly 2 corners of the grid contain one of the chosen points and these two corners are adjacent (4 possibilities), then the third point lies on the interior of the opposite side (5 possibilities). There are $4 \cdot 5=20$ such choices.
3) If exactly 2 corners of the grid contain one of the chosen points and these two corners are opposite (2 possibilities), then the third chosen point lies anywhere on the grid except in the corners ( $49-4=45$ possibilities). There are $2 \cdot 45=90$ such choices.
4) If exactly 3 corners of the grid contain one of the chosen points, then all three points are already chosen. There are 4 such choices.

The total number of valid choices of 3 points on the grid is therefore $100+20+90+4=214$.

Conclusion that at least one point lies in a corner of the grid ......................................................................... Calculation of the number of possibilities (100) for the case where exactly one point is in a corner ............. 1 point Calculation of the number of possibilities (20) for the case where exactly two points are in adjacent corners . 1 point Calculation of the number of possibilities (90) for the case where exactly two points are in opposite corners 1 point Calculation of the number of possibilities (4) for the case where all three points are in corners ..................... 1 point Adding the obtained possibilities (correct use of the addition rule) ............................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-19.jpg?height=60&width=1739&top_left_y=2083&top_left_x=176)