Knop K.A.

Given a triangle with no equal angles. Petya and Vasya play the following game: on each turn, Petya marks a point on the plane, and Vasya paints it either red or blue at his discretion. Petya wins if any three of the points he has marked and painted by Vasya form a monochromatic triangle similar to the original one. What is the minimum number of moves Petya can guarantee to win (regardless of the original triangle)?

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In 4 moves, Petya cannot guarantee a win: if Vasya paints two points red and two blue, there will be no monochromatic triangles.

We will show how to win in 5 moves. In the first three moves, Petya marks three points on the plane, which are the vertices of a triangle \(ABC\) similar to the original one. If Vasya paints all the marked points in one color, then Petya has already won.

Suppose vertices \(A\) and \(B\) are painted red, and \(C\) is painted blue. Then, in the next two moves, Petya marks points \(P\) and \(Q\) in the same half-plane relative to the line \(AB\) as point \(C\), such that triangles \(ABC\), \(PAB\), and \(BQA\) are similar (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-48.jpg?height=563&width=397&top_left_y=27&top_left_x=819)

If Vasya paints at least one of the points \(P\) and \(Q\) red, then a triangle similar to the original one will be formed, with all vertices red. If Vasya paints both these vertices blue, then a "blue" triangle \(CPQ\) will be formed, which is also similar to the original one. Let's prove this.

First method. \(\angle CAQ = \angle CAB - \angle QAB = \angle CAB - \angle ACB = \angle QBA - \angle PBA = \angle PBQ\), and also \(\frac{AC}{AQ} = \frac{AC}{AB} \cdot \frac{AB}{AQ} = \frac{AC}{AB} \cdot \frac{AC}{CB} = \frac{BA}{BQ} \cdot \frac{BP}{BA} = \frac{BP}{BQ}\). Therefore, triangles \(CAQ\) and \(PBQ\) are similar, with the similarity ratio \(\frac{QB}{QA} = \frac{BA}{BC}\).

Thus, \(\angle CQP = \angle CQA + \angle AQP = \angle CQA + \angle AQB - \angle PQB = \angle AQB = \angle CBA\) and \(\frac{QP}{QC} = \frac{BA}{BC}\), from which the similarity of triangles \(CQP\) and \(CBA\) follows.

Second method. Mark points \(C'\), \(P'\), and \(Q'\), which are symmetric to points \(C\), \(P\), and \(Q\) relative to the perpendicular bisector of \(A\) (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-48.jpg?height=592&width=401&top_left_y=1305&top_left_x=840)

From the equality of angles, it is clear that each triplet of points \(A, C, Q'\); \(A, P', Q\); \(A, P, C'\); \(B, P', C\); \(B, P, Q'\); \(B\), \(C', Q\) lies on the same line.

\(\angle ACB = \angle QAB = \angle AQ Q'\), so quadrilateral \(CQ'QP'\) is cyclic. The perpendicular bisector of \(QQ'\) coincides with the perpendicular bisector of \(AB\) and passes through the center of the circumcircle of quadrilateral \(CQ'QP'\). Therefore, this circle is symmetric with respect to it, and thus points \(C'\) and \(P\) also lie on it.

From the cyclic nature of hexagon \(Q'Q C' P P' C\), it follows that \(\angle ACB = \angle Q'BA = \angle BQ'Q = \angle PCQ\) and \(\angle CBA = \angle BQ'A = \angle CQP\). We obtain the similarity of triangles \(ABC\) and \(PQC\) by two angles.

## Answer

In 5 moves.