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Constructing Four Isosceles Trapezoids

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose we have four isosceles triangles with the same area, which must some whole number less than 29. Denote the the lower base and upper base of the i-th triangle with ##y_i## and ##x_i##, respectively. Furthermore, suppose that the angle between the side of length ##y_i## and the adjacent sides is ##45^\circ##. What will the lengths ##x_i## and ##y_i## be?


    2. Relevant equations


    3. The attempt at a solution
    Because the angle is ##45^\circ##, the leg of the triangle and the height of the isosceles trapezoid must be equal; that is, ##h = \frac{y_i - x_i}{2}##. Therefore, the area is given by

    ##\displaystyle A = \frac{y_i + x_i}{2} h##

    ##\displaystyle A = \frac{y_i + x_i}{2} \frac{y_i - x_i}{2}##

    In general, the two factors will not be whole numbers; for instance, ##\displaystyle \frac{y_i + x_i}{2}## will result in a whole number only if ##y_i + x_i## is an even whole number, which can occur when ##x_i## and ##y_i## have the same parity.

    This is where I am unsure of how to proceed. I was told that I should look at all the integers in ##\{1,2,3,...,29\}## which could be factored into four different ways, where each term in the factorization has opposite parity. This doesn't really make sense two me. Does ##\displaystyle A = \frac{y_i + x_i}{2} \frac{y_i - x_i}{2}## basically say that the area can be factored into two even integers, and therefore we would want to look for four factorizations with even terms?
     
  2. jcsd
  3. Nov 14, 2014 #2

    haruspex

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    I finally twigged that you meant four isosceles trapezoids. And it should also be given that the xi and yi are integers, yes?
    I see no reason why the parities should not be the same. Just look for values of A that permit four different factorisations.
     
  4. Nov 15, 2014 #3
    Ah, I am so sorry. I am so accustomed to saying/writing "isosceles triangles," rather than "isosceles trapezoids."
     
  5. Nov 15, 2014 #4
    So, I see that 24 has four factorizations. These would be

    ##24 \cdot 1##

    ##12 \cdot 2##

    ##8 \cdot 3##

    ##6 \cdot 4##

    So, for instance, with regard to the first trapezoid, I could write

    ##\displaystyle 24 \cdot 1= \frac{y_1 + x_1}{2} \frac{y_1 - x_1}{2}##

    What would the justification for writing ##\displaystyle 24 = \frac{y_1 + x_1}{2}## and ##1 = \frac{y_1 - x_1}{2}## ? Would it solely be that, if I were to associate 1 with ##\frac{y_1 + x_1}{2}## and 24 with ##\frac{y_1 - x_1}{2}##, this would lead to one of the variables being a negative number, which we do not want? Or is there some other reason?

    EDIT: I don't know why this is not rendering properly. I ran each piece of code through the LaTex previewer, and it worked perfectly.
     
    Last edited: Nov 15, 2014
  6. Nov 15, 2014 #5
    It doesn't produce a feasible isosceles trapezoid, and it only meets the area condition if you regard part of the enclosed space of the self-intersecting polygon as negative.
     
  7. Nov 15, 2014 #6
    So, that would be the only reason: it gives negative dimensions. Okay, thanks!
     
  8. Nov 16, 2014 #7
    The more that I think about this, the more I think to myself, "is this observation necessary to state?" Does anyone else feel this way?
     
    Last edited: Nov 16, 2014
  9. Nov 16, 2014 #8

    haruspex

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    Your argument was a bit backwards. When you wrote ## A = \frac{y_i + x_i}{2} \frac{y_i - x_i}{2}## you appear to have assumed that represents the actual factorisation, rather than ## A = \frac{y_i + x_i}{1} \frac{y_i - x_i}{4}## etc. If we start with ##4A = {y}^2 - {x}^2##, all being integers, it's easy to show that x and y have the same parity. It follows that the factorisation as you wrote it necessarily produces integer factors.
     
  10. Nov 16, 2014 #9
    I do not understand what made you write ##A = \frac{y_i + x_i}{1} \frac{y_i - x_i}{4}## and then ##4A = y^2 - x^2##. What do you mean by "actual factorization?" Actual factorization of what?
     
  11. Nov 16, 2014 #10

    haruspex

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    In the OP, you wrote correctly that ##A = \frac{y_i + x_i}{2} \frac{y_i - x_i}{2}##, then discussed 'the two factors'. But all you had shown is ##A = \frac{(y_i + x_i)(y_i - x_i)}{4}##. So at this point 'the two factors' is not well defined. You assumed a particular factorisation and deduced that x and y had the same parity. That's backwards. You should have proved they had the same parity and deduced the factorisation.
     
  12. Nov 16, 2014 #11
    I don't see how it is not well-defined. ##\frac{(y_i + x_i)(y_i-x_i)}{4}## is the exact same thing as ##\frac{(y_i + x_i)}{2}\frac{(y_i-x_i)}{2}##. So, I can look at each individual factor in the product, and determine that they have the same parity. If I want ##\frac{(y_i + x_i)}{2}## to evaluate to some positive integer, then I need the number to be some multiple of ##2##; similarly, the numerator of the quotient ##\frac{(y_i-x_i)}{2}## must be even, if I wish it to be a positive integer.
     
  13. Nov 16, 2014 #12
    Here are two things with which I am having trouble. Why does knowing about parity matter? Second, I still do not understand how I can just factor 24 into, say, 2 and 12, and then write ##\frac{y+x}{2} = 12## and ##\frac{y-x}{2} = 2##.
     
  14. Nov 17, 2014 #13

    haruspex

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    Numerically, yes, but in the OP it seemed you went beyond that and required ##\frac{(y_i + x_i)}{2} \times \frac{(y_i-x_i)}{2}## to be a factorisation of A, and, hence, that ##\frac{(y_i + x_i)}{2}## and ##\frac{(y_i-x_i)}{2}## are both integers. Do I misunderstand? I'm saying that you should first have shown that those two terms are integers, and from that deduced that the two terms constitute a factorisation.
    Can you be more specific?
    You supposed a pair (x, y), y > x, to be a solution to the question. You showed (except as I've explained) that this implies A can be factored into integers ##\frac{(y + x)}{2}## and ##\frac{(y-x)}{2}##. Therefore every solution corresponds to such a factorisation, so if you go through all the factorisations you can be sure to find all solutions.
     
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