# Thread: An equation from Diophantus I can't seem to solve properly

1. ## An equation from Diophantus I can't seem to solve properly

This is driving me nutty. I'm solving an equation from Diophantus, and I keep ending up with what appears to be wrong, and yet I can't see where. I must be messing up something pretty basic here. Help highly appreciated! Ok, the problem is:

Given a right triangle with area 7 and perimeter 12. Find it's sides.

Let x and y be the sides next to the right angle.

So, the area is $\displaystyle \frac{xy}{2} = 7$ or $\displaystyle xy = 14$ and perimeter is $\displaystyle x + y + \sqrt{x^2 + y^2} = 12$

First, I try to get rid of the radical in the second equation by isolating it on one side and squaring both sides of the equation:
$\displaystyle \sqrt{x^2 + y^2} = 12 - x - y$
$\displaystyle (\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$\displaystyle x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$\displaystyle 24x - 2xy + 24y - 144 = 0$

I turn the first equation into $\displaystyle y = \frac{14}{x}$ and substitute into the other equation:

$\displaystyle 24x - 2x(\frac{14}{x}) + 24(\frac{14}{x}) - 144 = 0$
$\displaystyle 24x + \frac{336}{x} - 172 = 0$

Multiplying everything by x to get rid of the 336/x, I get:

$\displaystyle 24x^2 - 172x + 336 = 0$

However, the book (The Story of $\displaystyle \sqrt{-1}$) says it's $\displaystyle 336x^2 - 172x + 24 = 0$

I must have made an error but I don't see it. I know I need to go to sleep, but I'll probably be having trouble sleeping with that bothering me.

2. Originally Posted by Grep
This is driving me nutty. I'm solving an equation from Diophantus, and I keep ending up with what appears to be wrong, and yet I can't see where. I must be messing up something pretty basic here. Help highly appreciated! Ok, the problem is:

Given a right triangle with area 7 and perimeter 12. Find it's sides.

Let x and y be the sides next to the right angle.

So, the area is $\displaystyle \frac{xy}{2} = 7$ or $\displaystyle xy = 14$ and perimeter is $\displaystyle x + y + \sqrt{x^2 + y^2} = 12$

First, I try to get rid of the radical in the second equation by isolating it on one side and squaring both sides of the equation:
$\displaystyle \sqrt{x^2 + y^2} = 12 - x - y$
$\displaystyle (\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$\displaystyle x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$\displaystyle 24x - 2xy + 24y - 144 = 0$

I turn the first equation into $\displaystyle y = \frac{14}{x}$ and substitute into the other equation:

$\displaystyle 24x - 2x(\frac{14}{x}) + 24(\frac{14}{x}) - 144 = 0$
$\displaystyle 24x + \frac{336}{x} - 172 = 0$

Multiplying everything by x to get rid of the 336/x, I get:

$\displaystyle 24x^2 - 172x + 336 = 0$

However, the book (The Story of $\displaystyle \sqrt{-1}$) says it's $\displaystyle 336x^2 - 172x + 24 = 0$

I must have made an error but I don't see it. I know I need to go to sleep, but I'll probably be having trouble sleeping with that bothering me.

There must be a mistake in the text of the problem:

The right triangle with the shortest perimeter is an isosceles right triangle. With the given data such a right triangle would have legs of $\displaystyle \sqrt{14}$ and a hypotenuse of $\displaystyle \sqrt{28}$.

That means the shortest perimeter has a length of

$\displaystyle p = 2\cdot \sqrt{14} + \sqrt{28} \approx12.775$

3. Originally Posted by earboth
There must be a mistake in the text of the problem:

The right triangle with the shortest perimeter is an isosceles right triangle. With the given data such a right triangle would have legs of $\displaystyle \sqrt{14}$ and a hypotenuse of $\displaystyle \sqrt{28}$.

That means the shortest perimeter has a length of

$\displaystyle p = 2\cdot \sqrt{14} + \sqrt{28} \approx12.775$

An imaginary tale: the story of [the ... - Google Books

It's originally from Diophantus' Arithmetica, book 6, number 22. Wish I had a copy of it, so I could check. Do note that it's supposed to give an imaginary solution. In other words, there's really no such possible triangle.

4. You're correct; book seems wrong. Discriminant is same in both cases anyway,
since the A and the C (336:24 , 24:336) have been switched: b^2 - 4ac remains same.

Try same problem, but change area to 6.

5. Thanks to you both. I feel much better and perhaps have the peace of mind to actually get to sleep. Thought I was losing my touch or something!

That's got to be the worst error I've seen so far in a math book. I just started it, this does not bode well. I would love to get my hands on the original text of Arithmetica (translated into English or French) so I can see what that problem is supposed to be (or if he got it wrong as well - you never know). Ah well...

6. Earlier, fresher of mind, I took another close look at this problem. Clearly my math was correct. However, so is the book. What I missed was as follows:

What Diophantus did was start this way:
$\displaystyle area*2 = 14 = (\frac{1}{x})(14x)$

So we have two perpendicular sides of $\displaystyle \frac{1}{x}$ and $\displaystyle 14x$ respectively. And from that he figures everything out, arriving at his equation.

So the x in his equation is not either of the perpendicular sides, as it was in my equation. I can figure out the actual sides by converting his x to the sides using the $\displaystyle \frac{1}{x}$ and $\displaystyle 14x$. When I correct for this, I get the right values and all equations agree.

So that explains that. His x and my x were two different things. Which, it turns out, is an important fact!

7. ## Re: An equation from Diophantus I can't seem to solve properly

Originally Posted by Grep
This is driving me nutty. I'm solving an equation from Diophantus, and I keep ending up with what appears to be wrong, and yet I can't see where. I must be messing up something pretty basic here. Help highly appreciated! Ok, the problem is:

Given a right triangle with area 7 and perimeter 12. Find it's sides.

Let x and y be the sides next to the right angle.

So, the area is $\displaystyle \frac{xy}{2} = 7$ or $\displaystyle xy = 14$ and perimeter is $\displaystyle x + y + \sqrt{x^2 + y^2} = 12$

First, I try to get rid of the radical in the second equation by isolating it on one side and squaring both sides of the equation:
$\displaystyle \sqrt{x^2 + y^2} = 12 - x - y$
$\displaystyle (\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$\displaystyle x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$\displaystyle 24x - 2xy + 24y - 144 = 0$

I turn the first equation into $\displaystyle y = \frac{14}{x}$ and substitute into the other equation:

$\displaystyle 24x - 2x(\frac{14}{x}) + 24(\frac{14}{x}) - 144 = 0$
$\displaystyle 24x + \frac{336}{x} - 172 = 0$

Multiplying everything by x to get rid of the 336/x, I get:

$\displaystyle 24x^2 - 172x + 336 = 0$

However, the book (The Story of $\displaystyle \sqrt{-1}$) says it's $\displaystyle 336x^2 - 172x + 24 = 0$

I must have made an error but I don't see it. I know I need to go to sleep, but I'll probably be having trouble sleeping with that bothering me.

I am reading that very book and I have arrived at the appropriate equation. Paul Nahin (the author) simplifies Diophantus' reduced equation $\displaystyle 1/x+14x+(\sqrt{1/x^2 + 196x^2})=12$.

First bring 1/x and 14x to the other side of the equation to simplify the radical. so
$\displaystyle \sqrt{1/x^2 + 196x^2} = 12 - 1/x - 14x$
$\displaystyle (\sqrt{1/x^2 + 196^2})^2 = (12 - 1/x-14x)^2$
$\displaystyle 1/x^2 + 196x^2 = 172 - 24/x - 336x +1/x^2 +196x^2$
$\displaystyle 0=172x-24-366x^2$

Which is the solution Pual Nahin noted

Your answer is correct if you define x to be 1/x (which is what Diophantus did for the area P1P2=14=14, he defined P1=1/x and P2= 14x)