8,9,10,11 |

Author: S $\underline{\text { Saghafian M. }}$.

In the plane, five points are marked. Find the maximum possible number of similar triangles with vertices at these points.
Example. Vertices and the center of a square.

Evaluation. Let's describe all configurations of four points forming four similar triangles. Let $\$ \mathrm{~A} \$$, $\$ \mathrm{~B} \$$, \$C \$ \$ \$

1. Point $\$ \mathrm{~A} \$$ lies inside triangle $\$ B C D \$$. Let $\$ B \$$ be the largest angle of triangle $\$ B C D \$$. Then angle \$CAD is similar to

2.1. Diagonal \$AC\$ bisects angle \$ \$ \$ABCD\$ - a rhombus, and since the triangles \$

2.2. There will be a vertex for which the diagonal is not a bisector, let this vertex be \$A\$. Then angles \$BAC\$, \$CAD\$, and \$BAD\$ are pairwise different. Therefore, they are the three angles of each of the four similar triangles, i.e., their sum is $\$ 180 \wedge\{\backslash$ circ $\} \$$ and \$ \$angle BAD=90^\{|circ $\} \$$.

Among the angles formed by the given points, there are four right angles and eight acute angles. Angles $\$ \mathrm{ABD} \$$, \$CBD\$ and so on cannot be right angles, as then one of the angles of the quadrilateral would be obtuse. Therefore, the right angles are \$BAD\$, \$ABC\$, \$BCD\$, and \$ADC\$.

Thus, in both subcases, $\$ \mathrm{ABCD} \$$ is a rectangle.

Returning to the problem. Let's call a triangle bad if it is formed by any three of the given points and does not belong to the set of similar triangles. (In particular, a bad triangle is formed by three collinear points). We will show that there are at least two bad triangles.

Note that five points can form no more than one rectangle. As shown above, each of the other quadruples contains at least one bad triangle. Each bad triangle is defined by two quadruples, so the total number of bad triangles is no less than $4 / 2=2$.

## Answer