The Cantor Set. A segment of the number line from 0 to 1 is painted green. Then its middle part, the interval $(1 / 3 ; 2 / 3)$, is repainted red. Next, the middle part of each of the remaining green segments is also repainted red, and the same operation is performed on the remaining green segments, and so on to infinity. The points that remain green form the Cantor set.

a) Find the sum of the lengths of the red intervals.

b) Prove that the number $1 / 4$ will remain painted green.

c) From the sum

$$
\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\ldots
$$

arbitrarily chosen terms are erased. Prove that the sum of the remaining terms is a green number.
Numbers from 0 to 1 can conveniently be considered as infinite ternary fractions of the digits 0, 1, and 2. The numbers mentioned in point v) are those numbers whose ternary representation contains no 1.

|  | Game Theory (Other) |
| :---: | :---: |
|  | Similarity Criteria |
|  | [ Auxiliary Similar Triangles |
| Problem $\underline{64723}$ Topics: | [ Symmetry Helps Solve the Problem |
|  | [ Inscribed and Circumscribed Polygons |
|  | Complex Numbers in Geometry |
|  | Estimation + Example |

Author: Knop K.

Given a triangle with no equal angles, Petya and Vasya play the following game: in one move, 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 marked and painted points form a monochromatic triangle similar to the original. What is the minimum number of moves Petya needs to guarantee a win (regardless of the original triangle)?

## Solution

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 points on the plane that are the vertices of a triangle \(ABC\) similar to the original. If Vasya paints all the marked points the same 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 figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_5476ffd8e94d65b212aag-41.jpg?height=563&width=395&top_left_y=1199&top_left_x=823)

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

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

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

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

![](https://cdn.mathpix.com/cropped/2024_05_06_5476ffd8e94d65b212aag-42.jpg?height=597&width=423&top_left_y=-1&top_left_x=823)

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 a straight 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 it also contains points \(C'\) and \(P\).

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.