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Math Help - Trapezints

  1. #1
    Junior Member phgao's Avatar
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    Trapezints

    Trapezints are those special trapezia whose sides have integer lengths. The parallel sides are of unequal length. Let a be the length of the shorter of the parallel sides and b,c,d the lengths of the other sides in clockwise order. We write [a,b,c,d] as the name of the trapezint. For example [1,3,5,4] is a trapezint with perimeter 13.

    a. What is the shorter perimeter that a trapezint can have? Explain why there is not a smaller one.

    b. What is the smallest perimeter of a trapezint with a non-parallel sides having different lengths? Explain why there is not a smaller one.

    c. What is the smallest perimeter of a trapezint with at least one angle a right angle? Explain why there is not a smaller one.

    d. Find all trapezints with perimeter 9.
    Note: here we regard two trapezints as different only if they are not congrunt. In particular [a,b,c,d] and [a,d,c,b] are the same.
    Last edited by MathGuru; June 9th 2005 at 09:01 AM.
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  2. #2
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    trapezint

    This is my first time to encounter trapezints, so I only depend on your posted desciptions re trapezint:
    "Trapezints are those special trapezia whose sides have integer lengths. The parallel sides are of unequal length."
    The reading is from the top parallel side, then clockwise.
    ----------
    a. What is the shorter perimeter that a trapezint can have? Explain why there is not a smaller one.

    You mean "shortest", instead of the posted "shorter"?
    I assume "shortest, so,

    Let us start with the smallest positive integer, 1. So the shorter of the parallel sides is 1.
    The next smallest positive integer is 2. Then, the other parallel side is 2.
    For the other two sides, the two legs, we use 1 for each.
    Hence this trapezint is [1,1,2,1].
    It is an isosceles trapezint.
    Its perimeter is 5. It is the shortest perimeter that a trapezint can have.

    There is none smaller than 5 perimeter because there is no positive integer that is lower than 2. This side 2 cannot be lower than 2. Otherwise, if it could be lower, then it could be 1 only, and the perimeter is 4 only. But then the two parallel sides are equal. That cannot be, because it is said that the two parallel sides must be of unequal lengths.

    -----------
    b. What is the smallest perimeter of a trapezint with a non-parallel sides having different lengths? Explain why there is not a smaller one.


    That means all 4 sides are unequal in lengths.

    The shortest distance between two parallel lines is the perpendicular distance between them.
    The right triangle with all sides as integers that has the shortest sum of the 3 sides is the 3-4-5 triangle.

    Using those, here is the trapezint for this part (b).
    The top parallel side is 1.
    The right leg is 3. It is perpendicular to the top and bottom sides. It is the shorter leg of the 3-4-5 right triangle.
    The bottom parallel side is 1+4 = 5.
    The left leg is 5. It is the hypotenuse of the 3-4-5 triangle.

    In short, the trapezint is [1,3,5,5].
    Its perimeter is 14.

    Another trapezint for this part (b) is [1,4,4,5], whose perimeter is also 14.
    This is the same as the [1,3,5,5], except that here, the leg 4 of the 3-4-5 right triangle is the perpendicular distance between the top and bottom parallel sides.

    Why there is no trapezint for this part (b) that has a perimeter lower than 14? Because there is no right triangle of integer sides that is smaller than the 3-4-5 triangle.

    These special right triangles are Pythagorean Triples: 3-4-5, 5-12-13, etc.

    ------------
    c. What is the smallest perimeter of a trapezint with at least one angle a right angle? Explain why there is not a smaller one.

    My aswer here is exactly the same as those in part (b) above: [1,3,5,5] or [1,4,4,5].

    -------------
    d. Find all trapezints with perimeter 9.
    Note: here we regard two trapezints as different only if they are not congrunt. In particular [a,b,c,d] and [a,d,c,b] are the same.

    The trapezints here are all isosceles trapezints, meaning, the non-parallel legs are equal in lengths.
    They are:

    >>>If the top parallel side is 1:
    [1,1,6,1], [1,2,4,2], [1,3,2,3]

    >>>and if the top parallel side is 2:
    [2,1,5,1], [2,2,3,2]

    >>>and if the top parallel side is 3:
    [3,1,4,1]

    Meaning, I found 6 trapezints whose perimeters are 9 each.
    Last edited by MathGuru; June 9th 2005 at 09:01 AM.
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  3. #3
    Junior Member phgao's Avatar
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    trapezia

    I think....b

    6 (side lengths 1,1,2,2). You can't get shorter as each pair of opposite sides has unequal length, so one of each pair must have length >1.

    But can you check that?
    Last edited by MathGuru; June 9th 2005 at 09:02 AM.
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  4. #4
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    trapezint

    Is [1,1,2,2] a trapezint for part b?

    How would you draw that?
    The top parallel side is 1.
    The right leg is 1.
    The bottom parallel side is 2.
    Then, would a left leg of 2 close the trapezint? How?
    Last edited by MathGuru; June 9th 2005 at 09:02 AM.
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  5. #5
    hpe
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    trapezint

    Quote Originally Posted by ticbol
    Is [1,1,2,2] a trapezint for part b?

    How would you draw that?
    The top parallel side is 1.
    The right leg is 1.
    The bottom parallel side is 2.
    Then, would a left leg of 2 close the trapezint? How?
    That would be a flat trapezint (all vertices lie on a straight line).
    Last edited by MathGuru; June 9th 2005 at 09:02 AM.
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  6. #6
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    All the vertices lie in a straight line.

    You are not joking, are you?

    How was a trapezint described again?
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  7. #7
    hpe
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    OK, here goes

    Assume as before that a is the smaller of the parallel sides and that b, c=a+e, d are the other sides in clockwise direction. Thus e > 0. We can also assume that d >= b. See the picture below

    We can think of the trapezint as consisting of a parallelogram with sides a,b,a,b (clockwise), glued to a triangle with sides b, e, d. The triangle now has integer sides. Any such triangle gives rise to an infinite family of trapezints, just take a = 1,2,3,... and c = 1+e,2+e,.... Reversely, any trapezint defines a triangle with integer sides. The perimeter of any such trapezint is b+d+e+2a, where a = 1,2,3... is a free parameter. A trapezint with shortest perimeter within a certain class (e.g. at least one right angle, or non-parallel sides of different length) corresponds to a triangle with integer sides b,e,d that has shortest perimeter b+e+d within the corresponding class (right triangle, or b not equal d)

    For this triangle to be non-degenerate, we need b > e+d, e > b+d, d > b+e. If there is equality in one of these inequalities, this is a degenerate triangle, and thus the trapezint is also degenerate (it lies on a straight line). If a reverse inequality holds, the triangle doesn't close up, so there is no trapezint. E.g., the trapezint (1,1,4,1) doesn't exist, since the triangle (1,3,1) does not exist.

    The trapezint (1,1,2,2) has b=e=1, d=2, so this is a degenerate trapezint.

    The trapezint (1,1,2,1) has b=e=d=1, so this is a nongenerate trapezint with interior angles either 60 or 120 degrees.

    For nonparallel sides to have different lengths and to be non-degenerate, we need b not equal d, thus b<d. Then b=1, d=2, e=1 does not work, but b=1, d=2, e=2 is OK, with a perimeter of 5 for the triangle. If b>1, then d>2, and the perimeter will be larger than 5, so this is the minimal triangle.
    Since a=1 or larger, this implies that (1,1,3,2) is the minimal trapezint with unequal non-parallel sides.

    For a trapezint to have a right angle, we need a right triangle with integer sides. As ticbol already explained, these are the Pythagorean triangles, (3,4,5) is the smallest one. This leads to the smallest perimeter trapezint (1,3,5,5). The triangle (3,5,4) leads to the trapezint (1,3,6,4) which also has smallest perimeter.

    Now as to trapezints with perimeter 9. This means we must find integer triangles with perimeter 3 or 5 or 7, since any such triangle delivers a trapezint with perimeter 9 (take a = 3 or a = 2 or a = 1). Remember that we may assume b <= d.

    Perimeter 3: The triangle is (1,1,1), the trapezint is (3,1,4,1).
    Perimeter 5: The triangles are (1,2,2) or (2,1,2), the trapezints are (2,1,4,2) or (2,2,3,2)
    Perimeter 7: The triangles are (1,3,3) or (2,2,3) or (2,3,2) or (3,1,3), the trapezints are (1,1,4,3) or (1,2,3,3) or (1,2,4,2) or (1,3,2,3).
    Attached Thumbnails Attached Thumbnails Trapezints-trapezint.bmp  
    Last edited by hpe; May 11th 2005 at 05:36 PM.
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  8. #8
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    That one about the smallest trapezint with unequal legs, [1,1,3,2], is very good.

    That degenerate trapezint is no good. If there is no trapezium, there is no trapezint.
    Degenerate? Whatever. A straight line does not look like a trapezium to me.

    The whole answer/explanation is good, though.
    Last edited by ticbol; May 12th 2005 at 11:01 PM.
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  9. #9
    Junior Member phgao's Avatar
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    I concur, I loved the explanation, and also that before i was wrong. The answer to b is as said; perimeter 7.

    Also with d.....Why cant there be these ones? Any explanation would be helpful.

    If the top parallel sides is 1:
    [1,2,4,2],

    if the top parallel side is 2:
    [2,1,5,1]
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  10. #10
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    Umm, now that you asked this, let me say that the [1,1,6,1] and [2,1,5,1] that I thought then were trapezints are not trapezints. Because the bottom parallel side is greater than the sum of the other 3 sides in each of those two.
    In [1,1,6,1], 6 > 1+1+1.
    In [2,1,5,1], 5 > 2+1+1.
    They will not close to form trapezia.

    The [1,2,4,2], as an isosceles trapezint is okay.
    Lay 1 and 4, the parallel sides, such that their midpoints lie on a perpendicular line to both. Then lay the 2 and 2 legs, adjusting the perpendicular distance between the 1 and 4 accordingly, until an isosceles trapezium is formed.
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  11. #11
    Junior Member phgao's Avatar
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    Thanks a lot Ticbol! I appreciate your help, and generous working out. Also to Hpe as well for the excellent explanation.

    PS I think I found a mistake, but only a minor one in hpe's working...

    When you say

    "For a trapezint to have a right angle, we need a right triangle with integer sides. These are the Pythagorean triangles, (3,4,5) is the smallest one. This leads to the smallest perimeter trapezint (1,3,5,5). The triangle (3,5,4) leads to the trapezint (1,3,6,4) which also has smallest perimeter."

    With one of your answers, ie. (1,3,6,4) , i think it doesnt work as if 1 is the smallest side, being the top side, and the others go around in clockwise order, we get the hypotenuse of the 'triangle' having a side of 3 or 4, but this cannot be, as side d of each triangle is 4 and 5 respectively, so that means one of the sides of the triangles is longer than the hypotenuse!

    I am sure 14 as the perimeter is correct, and that there is only one solution, discounting that (1,3,5,5) is also the same as (1,5,5,3).
    Last edited by phgao; May 17th 2005 at 03:04 AM.
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