Frankin B.R.

On three red and three blue cards, six positive numbers are written, all of them are different. It is known that on the cards of one color, the pairwise sums of some three numbers are written, and on the cards of the other color - the pairwise products of the same three numbers. Can these three numbers always be guaranteed to be determined?

#
All the sought numbers are distinct, otherwise some numbers on the cards would be equal. Since their pairwise products are positive, all numbers have the same sign, and since their pairwise sums are positive, this sign is positive.

Let the sought numbers be \( x < y < z \).

First method. For the cards of each color, consider the ratio of the largest number on them to the smallest. In one case, this is \(\frac{y+z}{x+y}\), in the other \(\frac{x+y}{x}\). Since the first ratio is less than the second, it is clear on which cards the sums are written and on which the products.

Knowledge of the pairwise sums of three numbers determines these numbers, for example, \( x = \frac{1}{2}((x+y) + (x+z) - (y+z)) \).

Second method. Notice that \( x+y < x+z < y+z \) and \( xy < xz < yz \). Let \( a < b < c \) be the numbers on the cards of one color, and \( A < B < C \) be the numbers on the cards of the other color. Then \( x = \sqrt{\frac{(a+A+1)(b+B+1)}{c+C+1}} - 1 \). Similarly, \( y \) and \( z \) can be calculated.

## Answer

Always.

Author: \(\underline{\text{Mukhin}}\) D.G.

On the graph of the function \( y = \frac{1}{x} \), Misha marked all points with abscissas \( 1, 2, 3, \ldots \) until he got tired. Then Masha came and colored all rectangles, one of whose vertices is a marked point, another is the origin, and the other two lie on the axes (the figure shows which rectangle Masha would have colored for the marked point \( P \)). Then the teacher asked the students to calculate the area of the figure consisting of all points colored exactly once. What did they get?

![](https://cdn.mathpix.com/cropped/2024_05_06_0ffd3d08f157224b5b77g-59.jpg?height=552&width=551&top_left_y=453&top_left_x=-1)

## Solution

Suppose Misha got tired after marking the point with abscissa \( n \). Let's see how the figure consisting of all points colored exactly once is structured. On the segment of abscissas \([i-1, i]\), where \( i = 1, \ldots, n-1 \), this is a rectangle of width \( 1 \) and height \( h_i = \left| \frac{1}{i} - \frac{1}{i+1} \right| \). On the segment \([n-1, n]\), this is a rectangle of width \( 1 \) and height \( h_n = \frac{1}{n} \). Then the area of the figure \( S = 1 \cdot h_1 + 1 \cdot h_2 + \cdots + 1 \cdot h_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right) + \frac{1}{n} = 1 \).

## Comments

1. One can conduct a proof by induction. Suppose Masha colors the rectangle immediately after Misha marks a point. When Misha marks the point \( (1, 1) \), Masha colors a rectangle of area \( 1 \). Next, when Misha marks a point with abscissa \( n \), Masha colors a rectangle of area \( \frac{1}{n} \) for the first time and a rectangle of area \( \frac{1}{n} \) for the second time (the rest will be points already colored more than once). Thus, the area of the figure consisting of all points colored exactly once does not change.
2. This fact can be seen visually by "shifting" all rectangles to the y-axis.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ffd3d08f157224b5b77g-59.jpg?height=491&width=823&top_left_y=1701&top_left_x=0)

## Answer

Submit a comment