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.
Originally posted by phgaoa. 5 (side lengths 1,1,1,2). You can't get shorter, as [1,1,1,1] is not allowed because parallel sides must have unequal length.
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 ...[text shortened]... as different only if they are not congrunt. In particular [a,b,c,d] and [a,d,c,b] are the same.
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.
c. 14, in the form [1,5,5,3], [1,5,4,4] or a mirror image: If we have one right angle between a parallel side P and a non-parallel side Q, we must have another between the same non-parallel side R and the other parallel. That forms a 'U' shape, only the sides of the U can't be of equal height. By Pythagoras, the length of the remaining side S is given by S = Q^2 + (P-R)^2. The smallest solution to this in positive integers is S = 5, with one of:
Q = 3, P = 5, R = 1
Q = 3, P = 1, R = 5
Q = 4, P = 4, R = 1
Q = 4, P = 1, R = 4
d. No side can be longer than 4 by the triangle inequality; the parallel sides cannot be the same length; and mirror images are considered equivalent. Hence an equivalence class of trapezints with perimeter 9 can be indicated by two unordered pairs of integers between 1 and 4 whose sum is 9: the first pair represents the parallel sides, which must be unequal, and the second pair the other sides. These equivalence classes are as follows:
1,2 3,3
1,2 2,4
1,3 2,3
1,3 1,4
2,3 2,2
2,3 1,3
2,4 1,2
3,4 1,1
Originally posted by AcolyteI'm no math wizz, but I don't think you can construct a trapezint with lengths 1,1,2,2. What would the interior angles of your 1,1,2,2, trapezint be?
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.
I think the actual correct answer is 7 (1,2,3,1).
Originally posted by The PlumberMy bad - that trapezint would go a bit flat, wouldn't it 😳 There's a restriction I failed to spot: a non-parallel side has to be shorter than |P-Q| + R, where P and Q are the parallel sides and R is the other side. Both 1,2,2,2 and 1,2,3,1 work though - well spotted.
I'm no math wizz, but I don't think you can construct a trapezint with lengths 1,1,2,2. What would the interior angles of your 1,1,2,2, trapezint be?
I think the actual correct answer is 7 (1,2,3,1).
Hmm, some of those classes of length 9 look dodgy too. Here's an edited list:
1,2 3,3
1,3 2,3
2,3 2,2
2,4 1,2
3,4 1,1
Originally posted by doublezOriginally posted by phgao
3,4,1,1 and others where a >c don't exist, it states that a is the shortest side
Let a be the length of the shorter of the parallel sides
If you'll notice, I put the trapezint classes as a pair of parallel sides, followed by a pair of non-parallel sides. So the class 3,4 1,1 consists of the trapezint [3,1,4,1].