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Neznayka draws a rectangle, divides it into 64 smaller rectangles by drawing $7$ straight lines parallel to each of the original rectangle's sides.
After that, Znayka points to $n$ rectangles of the division at the same time, and Neznayka names/gives/reveals the areas of each of these rectangles.

What is the smallest value of $n$ for Znayka to be able to name/give/reveal the areas of all the rectangles in the division?

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Mar 10 at 20:28
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    $\begingroup$ I suggest: take a simpler example. Divide the rectangle into $4$ sub rectangles by drawing $1$ line parallel to each of the original sides. Now, what would you say about the $4$ smaller areas? $\endgroup$
    – lulu
    Commented Mar 10 at 20:38
  • $\begingroup$ @lulu: I believe the lines are carefully drawn and Neznayka knows the length of every segment, so the area of each small rectangle. In your simple case, if I am given the area of three of the small rectangles I can derive the area of the fourth $\endgroup$
    – Ross Millikan
    Commented Mar 10 at 20:40
  • $\begingroup$ @RossMillikan Thanks, I had indeed misunderstood. $\endgroup$
    – lulu
    Commented Mar 10 at 20:43

2 Answers 2

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The answer is $n=15$.

To convince yourself it is true try to prove using induction the more general property: For a partition of the rectangle into $m\times m$ smaller rectangles (using $m-1$ straight lines parallel to each of the original rectangle's sides) the minimal number is $n=2m-1$ (In your case $m=8$).

After this, try and go even further: what about partitions into $m\times k$ smaller rectangles?

Again, use induction to prove $n=m+k-1$

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  • $\begingroup$ And when you are comfortable with the case of two-dimensional rectangle, you can try and go to higher dimensions of this game... $\endgroup$
    – MOMO
    Commented Mar 11 at 5:47
  • $\begingroup$ Do you have any ideas how to formulate an explanation? $\endgroup$
    – Arseniy Akinshin
    Commented Mar 11 at 12:43
  • $\begingroup$ Explanation to what exactly? $\endgroup$
    – MOMO
    Commented Mar 11 at 16:49
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BASIC LIMIT :

There are $8$ unknowns along the length of the rectangle.
There are $8$ unknowns along the width of the rectangle.

We require $8+8=16$ Equations to get those unknowns.
When we know those $16$ unknowns , then all areas are known.

The areas we might choose are :
$8$ Diagonal Elements
$7$ Elements just below the Diagonal
$1$ Element at the top Corner

AREA

UPDATED LIMIT :

Shown in the Diagram , we have $16$ Elements which are selected to highlight that we have to take ratios of neighboring terms which are left-right & which are up-down.
Imagine that we double all widths while halving all lengths : Since $2 \times 1/2=1$ , we will still have same areas for all Parts.
This implies that we have $1$ less unknown , hence we have only $16-1=15$ unknowns & hence we require only $15$ Equations via $15$ Elements.
We can skip the top-right green Element in the Diagram.
Every row & every Column can be obtained via ratios of neighboring terms.
Hence $n=15$ here.

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    $\begingroup$ This suggests that if we draw one line instead of seven, we need four areas. But in fact three will do. $\endgroup$
    – Gerry Myerson
    Commented Mar 10 at 21:06
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    $\begingroup$ This is not correct. In the $2 \times 2$ case you would say we need all four areas. But if $a,b$ are the areas of the top two rectangles and $c$ is the lower left we have $\frac {bc}{a}$ for the area of the fourth $\endgroup$
    – Ross Millikan
    Commented Mar 10 at 21:10
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    $\begingroup$ We are not asked to assess the sides, just the areas of the rectangles. In my calculation I don't know the sides, but I do know all the areas. $\endgroup$
    – Ross Millikan
    Commented Mar 10 at 21:12
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    $\begingroup$ I have rectified my answer , @GerryMyerson , in line with your Observation. Thank you ! $\endgroup$
    – Prem
    Commented Mar 11 at 3:54
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    $\begingroup$ You still need to prove $n=14$ doesn't work. $\endgroup$
    – D S
    Commented Mar 11 at 8:50

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