Lesson More complicated but still elementary logic problems

More complicated but still elementary logic problems

Problem 1

Let assume for a minute that the finite set M of points with such properties does exist.


Then you can select the pair of points (A,B) such that the distance d(A,B)  between the points is maximal among all pairs of points from M.


So, you selected such a pair.

According to properties of M, there is ANOTHER, distinct pair of points (C,D) such that the segments AB and CD  are parallel.


Then the segments AB and CD lie in one plane, so the quadrilateral ABCD is a plane quadrilateral and all its vertices lie in the same plane. 


Thus the quadrilateral  ABCD  (with the verices listed in the correct order) is either trapezoid or parallelogram.


In any case, at least one of the two the diagonals of such a qudrilateral is longer than its side AB.


It is just a contradiction, since we assumed that d(A,B) was the longest distance.


The contradiction PROVES the statement.

Problem 2

Ewan's train starts from the beginning station at 8:30 am and returns back at 8:54 am.


So, on the way, Ewan's train will meet the trains that start at or after 8:30 am, namely, at 8:30 am, 8:34 am, 8:38am, 8:42 am, 8:46 am, and 8:50 am 

    in the opposite direction.  The number of such trains is 6.


But, in addition to it, you should count those trains that started before 8:30 in the opposite direction and arrived 

to the starting station after 8:30 yet.  (Accounting for these trains and not missing them is the major point of the solution to this problem !)


These trains started from the same station at 8:06 am, 8:10 am, 8:14 am, 8:18 am and 8:24 am,  and their number is 5.


Thus there are 6 + 5 = 11 trains in all we should account for, according to the condition.


Again, the total is  6 + 5 = 11 trains.  Notice that I do not count trains that starts and arrive to the selected station at at 8:30 am, 

exactly as the condition requires.

Problem 3

1.  From the condition, it is clear that the amounts each person spent for presents, each (amount) is a perfect square of cents.



2.  If x and y  are spendings for some (any) of the three couples, then

    x^2 - y^2 = 75,    according to the condition,  or

    (x+y)(x-y) = 75.   (and we remember that the greater value is the wife's spending !)


    For integer positive  x and y  it gives these and only these opportunities


        x + y = 75
        x - y =  1     with the solution  x= 38,  y= 37  
    OR
        x + y = 25
        x - y =  3     with the solution  x= 14,  y= 11
    OR  
        x + y = 15
        x - y =  5     with the solution  x= 10,  y= 5.


    Of these solutions,  the only pair  (38,37)  has the difference of 1,  so it gives a clue to me to conclude that

                         Ann is the wife to Bill Brown.



3.  But in order for to make this conclusion ABSOLUTELY CORRECT, I must EXCLUDE that the other couples fall in the same pair/solution (38,37).

    Fortunately, the condition gives me rationality to make this conclusion.

    Indeed, it says that "Betty bought one less present than Joe Jones", which means that Betty is of the pair (10,5), while Joe Jones is of the pair (14,11).


    So, I really can conclude that Ann is the wife to Bill Brown.



4.  Finally, from the condition, it is also clear that Betty IS NOT  the wife to Joe Jones.



5.  It leaves only one opportunity  for Mary to be Mary Jones.