5. On each side of the rectangle, 5 points are highlighted which are not the vertices of the rectangle. How many quadrilaterals have vertices that are some of these 20 highlighted points?
5. On each side of the rectangle, 5 points are highlighted that are not the vertices of the rectangle. How many quadrilaterals have vertices among these 20 highlighted points?

## Solution.

The following cases are possible:

(i) All four vertices of the quadrilateral are on different sides of the rectangle $ABCD$.

(ii) Two vertices of the quadrilateral are on one side, and two are on another side of the rectangle $ABCD$.

(iii) Two vertices of the quadrilateral are on one side, one on another, and one on a third side of the rectangle $ABCD$.

1 POINT

Case (i): On each side of the rectangle, we choose one of the five possible vertices, so the total number of possibilities is $5 \cdot 5 \cdot 5 \cdot 5=5^{4}$,

1 POINT or 625.

Case (ii): We can choose two sides of the rectangle $ABCD$ on which the vertices of the quadrilateral will be in $\frac{4 \cdot 3}{2}$ ways.

1 POINT

We can choose two vertices on one fixed side of the rectangle $ABCD$ in $\frac{5 \cdot 4}{2}$ ways, so we can choose two vertices on one fixed side of the rectangle $ABCD$ and two on another in $\left(\frac{5 \cdot 4}{2}\right)^{2}$ ways.

1 POINT

Therefore, the number of quadrilaterals with two vertices on one side and two on another side of the rectangle $ABCD$ is $\frac{4 \cdot 3}{2} \cdot\left(\frac{5 \cdot 4}{2}\right)^{2}$

1 POINT* or 600.

Case (iii): We can choose one side of the rectangle $ABCD$ on which there will be two vertices of the quadrilateral in 4 ways, and for each such choice, we can choose two sides of the rectangle $ABCD$ on which there will be one vertex of the quadrilateral in $\frac{3 \cdot 2}{2}=3$ ways. Thus, we can choose the sides on which the vertices of the quadrilateral are located in $4 \cdot 3$ ways.

1 POINT

We can choose two vertices on one fixed side of the rectangle $ABCD$ in $\frac{5 \cdot 4}{2}$ ways, and one vertex on each of the two remaining fixed sides of the rectangle in $5 \cdot 5=5^{2}$ ways, so the number of possible vertex selections for the quadrilateral is $\frac{5 \cdot 4}{2} \cdot 5^{2}$.

1 POINT

Therefore, the number of quadrilaterals with two vertices on one side, one on another, and one on a third side of the rectangle $ABCD$ is $4 \cdot 3 \cdot \frac{5 \cdot 4}{2} \cdot 5^{2}$,

1 POINT** or 3000.

In total, the number of quadrilaterals is $625+600+3000=4225$.

2 POINTS

TOTAL 10 POINTS

Alternatively, in case (iii), we can count as follows:

We can choose one side of the rectangle $ABCD$ on which there will be two vertices of the quadrilateral in 4 ways, and two vertices on it determine $\frac{5 \cdot 4}{2}=10$ different segments (sides of the quadrilateral).

1 POINT

We can choose two sides of the rectangle $ABCD$ on which there will be one vertex of the quadrilateral in $\frac{3 \cdot 2}{2}=3$ ways, and on each of them, a vertex of the quadrilateral in 5 ways.

1 POINT

Thus, we can choose the sides on which the vertices of the quadrilateral are located in $4 \cdot 10 \cdot 3 \cdot 5 \cdot 5=3000$ ways.

1 POINT**

## Notes: