# Constructing Four Isosceles Trapezoids

1. Nov 14, 2014

### Bashyboy

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. Nov 14, 2014

### haruspex

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.

3. Nov 15, 2014

### Bashyboy

Ah, I am so sorry. I am so accustomed to saying/writing "isosceles triangles," rather than "isosceles trapezoids."

4. Nov 15, 2014

### Bashyboy

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
5. Nov 15, 2014

### Joffan

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.

6. Nov 15, 2014

### Bashyboy

So, that would be the only reason: it gives negative dimensions. Okay, thanks!

7. Nov 16, 2014

### Bashyboy

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
8. Nov 16, 2014

### haruspex

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.

9. Nov 16, 2014

### Bashyboy

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?

10. Nov 16, 2014

### haruspex

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.

11. Nov 16, 2014

### Bashyboy

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.

12. Nov 16, 2014

### Bashyboy

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 $\frac{y-x}{2} = 2$.

13. Nov 17, 2014

### haruspex

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.