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Problem 79 states that every square by its intersection lines can be divided into equal rectangles, one example is given on figure as you see square by the 3 line is divided into 6 equal rectangles,we are asked to say what is the possible number of equal rectangles if the square is intersected by the 10 lines. This is one task from a national exam test and because I have an exam on the 14th of July, I need to know answers and every trick of such tasks; my basic attitude is to find a recurrence relation if it exist here.

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So your question is "what are the answers and tricks for every such task" ??? (-1) –  The Chaz 2.0 Jul 12 '11 at 12:51
@THE Chaz question seems clear please read again –  dato datuashvili Jul 12 '11 at 13:01
Both of the answers refer to the ambiguity/unreadability of the question. It's interesting, but might benefit from a little refinement for posterity's sake. –  The Chaz 2.0 Jul 12 '11 at 14:17
@user3196, Thee question is quite not clear, really. Are you asking for "answers and every trick of such tasks"? Do you want to know how to solve the particular problem you describe? Usually, when people ask for clarifications it is safe to assume that clarifications are needed: even if the question is clear (to you), that someone asks for clarification is a clear indication that further clarifications might be helpful, don't you think? –  Mariano Suárez-Alvarez Jul 12 '11 at 19:43

2 Answers 2

up vote 8 down vote accepted

Too bad that your exam is so soon. With more time I might learn to read Georgian.

Let me guess what the problem asks for. We draw a total of $10$ lines, some (perhaps $0$) horizontal, the rest vertical. The question (I am guessing) now says that these lines divide the square into "equal" (presumably congruent) rectangles which are not squares. What is the maximum possible number of such rectangles?

Using the notation of Ross Millikan, let us see what happens if we draw $m$ horizontal lines and $n$ vertical lines, where $m+n=10$.

This procedure divides the square into $(m+1)(n+1)$ "rectangles".

We want to maximize $(m+1)(n+1)$, subject to the condition that $m+n=10$ and $m \ne n$. (If $m=n$, we will have divided the square into squares.)

To put it in other terms, we want to maximize $xy$, subject to the following conditions:

$1$) $x$ and $y$ are unequal positive integers

$2$) $x+y=12$

There is general theory around this: $xy$ is maximized when $x$ and $y$ are as nearly equal as possible. But in an exam context, you can just quickly examine various possibilities in your mind. Best choice is $x=5$, $y=7$ (or the other way around) for a total of $35$.

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thanks very much you are great person and user thanks ones again –  dato datuashvili Jul 12 '11 at 14:14
It depends upon the reading of the problem: squares are rectangles, so maybe $m=n$ is permitted, in which case the answer would be $36$. –  Ross Millikan Jul 12 '11 at 17:46
@Ross Millikan: My answer, if I were asked, would be $36$. However, I went to the link, it was a multiple choice question, with $35$ as one of the answers, and $36$ not. None of the other choices was an answer to any question that could be reasonably connected to the post. Hence the inference that the Georgian text specified non-square rectangle. –  André Nicolas Jul 12 '11 at 18:08
@user6312: good point about multiple choice problems. Often the choices lead the way. –  Ross Millikan Jul 12 '11 at 18:14

Hint: for this one, if there are $n$ horizontal lines and $m$ vertical lines, how many rows and columns of rectangles are there?

I can't read the rest of the problems.

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