Lesson Math Olympiad level problems on inequalities

Math Olympiad level problems on inequalities

Problem 1

Let the 3 smallest fishes weight a, b and c kilograms.  Then  

    a + b + c = 25 kg     (1)


    Notice, that from equation (1), the weight at least one of the fishes {a,b,c}  MUST BE MORE than  8 kilograms
    (otherwise, their total weight would be not more than 24 kg).

    Let call this fish M8 ("more than 8 kilograms"):  M8 > 8 kg.    <<<---=== note the strong inequality (!)



Let the 3 largest fishes weight x, y and z kilograms.  Then  

    x + y + z = 35 kg     (2)



The sum (1) plus sum (2) is 25 + 35 = 60 kilograms, which  is 40 kilograms less than 100 kilograms.


Hence, it should be at least 4 or more other fishes, distinct from a, b, c, x, y and z, to balance this difference of 40 kilograms.


    Now, if the set of these distinct fishes contain 5 or more fishes, then the weight of at least one of these 5 or more distinct fishes 
    must be LESS than OR equal to 8 kilograms

    (otherwise, the total weight of these 5 distinct fishes would be more than 40 kilograms).

    Let call this fish L8 ("less or equal to 8 kilograms"):  L8 <= 8.



Then, replacing M8 by L8 in the set {a,b,c}, we would obtain the new set of 3 fishes weighing in total LESS THAN 25 kilograms.


It gives us a CONTRADICTION with the statement that the three smallest fishes weight 25 kilograms.


The contradiction means that the set of distinct fishes consists of EXACTLY 4 fishes.


Hence, the total number of fishes is 3 + 4 + 3 = 10.      ANSWER

Problem 2

We consider the set of integer points {(x,y)} in a coordinate plane, 
where -24 <= x <= 24, -24 <= y <= 24, and we want to find the number of points
(x,y) such that 

    xy < x + y.    (1)


This inequality (1) is equivalent to

    xy - x - y < 0,

    1 + xy - x - y < 1,

     (1-x)*(1-y) < 1.      (2)


    +-------------------------------------------+
    |  This inequality (2) is easy to analyze.  |
    +-------------------------------------------+


(a)  First set of solutions in integer numbers is those pairs (x,y),
     where x = 1.  It gives all the pairs (1,y) with y = -24, -23, . . . ,23, 24.
     In all, there are 24+24+1 = 49 such pairs.



(b)  The other set of solutions in integer numbers is those pairs (x,y),
     where y = 1.  It gives all the pairs (x,1) with x = -24, -23, . . . ,23, 24.

     In all, there are 24+24+1 = 49 such pairs.
     But the pair (1,1) was just accounted in (a);  so, it adds only 49-1 = 48 pairs.

     For now, we have 49 + 48 = 97 different pairs.



Thus, we covered cases, when one or both factors in (2) are zero,
so their product is zero.


Now we will consider cases with the negative left side in (2).
Left side of (2) is negative if and only if one factor is positive, while the other factor is negative.



(c)  So, third set of solutions is all pairs (x,y), where x > 1, y < 1.  

     In all, there are (24-1)*(24+1) = 23*25 = 575 such pairs.     
     
     They all are different from pairs found in (a) and (b), so we can add 575 to 97.

     For now, we have 97 + 575 = 672 different pairs.

 


(d)  Fourth set of solutions is all pairs (x,y), where x < 1, y > 1.  ( This set is symmetric to (c) ).

     In all, there are (24-1)*(24+1) = 23*25 = 575 such pairs.     
     
     They all are different from pairs found in (a), (b) and (c), so we can add 575 to 672.


     It gives 672 + 575 = 1247 solutions, in all.



So, inequality (2) has 1247 solutions in the set of ordered pairs {(x,y) | x,y ∈ Z |, |x| ≤ 24, |y| ≤ 24}.


But the problem asks for DISTINCT NUMBERS x and y.  In the found set of solutions, 
there is only one pair with coinciding x and y: it is the pair (1,1).


By excluding this pair, the ANSWER to the problem 's question is 1247-1 = 1246.

Problem 3

Let all the books are 15 cm wide.


Then the total number of books is   = 16,

and we can place  ONLY  3  such books at each shelf.



So, having 5 shelves, we can place only  3*5 = 15 such books, and we need then

the 6-th shelf for the 16-th book.

1)  In the 1-st shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



2)  In the 2-nd shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



3)  In the 3-rd shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



4)  In the 4-th shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



5)  In the 5-th shelf,  I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



6)  In the 6-th shelf I can fill at least 40 cm of 56 cm.


    Indeed, if less than 40 cm is filled, then I can add any book (since it is no 
    thicker than 16 cm).



So, I can fill at least 40 cm of 56 cm in each of 6 shelves.



Taken together,  6 times 40 cm comprise  2 m 40 cm,

which means that ALL the books will be placed in 6 shelves.