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

• Nov 21st 2010, 11:26 PM
Grep
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 $\frac{xy}{2} = 7$ or $xy = 14$ and perimeter is $x + y + \sqrt{x^2 + y^2} = 12$

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:
$\sqrt{x^2 + y^2} = 12 - x - y$
$(\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$24x - 2xy + 24y - 144 = 0$

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

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

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

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

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

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. (Headbang)

• Nov 22nd 2010, 12:20 AM
earboth
Quote:

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

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:
$\sqrt{x^2 + y^2} = 12 - x - y$
$(\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$24x - 2xy + 24y - 144 = 0$

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

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

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

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

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

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. (Headbang)

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 $\sqrt{14}$ and a hypotenuse of $\sqrt{28}$.

That means the shortest perimeter has a length of

$p = 2\cdot \sqrt{14} + \sqrt{28} \approx12.775$
• Nov 22nd 2010, 12:32 AM
Grep
Quote:

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 $\sqrt{14}$ and a hypotenuse of $\sqrt{28}$.

That means the shortest perimeter has a length of

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

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.
• Nov 22nd 2010, 12:33 AM
Wilmer
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.
• Nov 22nd 2010, 12:48 AM
Grep
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! (Happy)

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...
• Nov 22nd 2010, 05:52 PM
Grep
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:
$area*2 = 14 = (\frac{1}{x})(14x)$

So we have two perpendicular sides of $\frac{1}{x}$ and $14x$ 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 $\frac{1}{x}$ and $14x$. When I correct for this, I get the right values and all equations agree. (Whew)

So that explains that. His x and my x were two different things. Which, it turns out, is an important fact!
• Mar 18th 2013, 12:57 PM
neffk
Re: An equation from Diophantus I can't seem to solve properly
Quote:

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

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:
$\sqrt{x^2 + y^2} = 12 - x - y$
$(\sqrt{x^2 + y^2})^2 = (12 - x - y)^2$
$x^2 + y^2 = 144 - 12x - 12y - 12x +x^2 + xy - 12y + xy + y^2$
$24x - 2xy + 24y - 144 = 0$

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

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

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

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

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

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. (Headbang)

I am reading that very book and I have arrived at the appropriate equation. Paul Nahin (the author) simplifies Diophantus' reduced equation $1/x+14x+(\sqrt{1/x^2 + 196x^2})=12$.
$\sqrt{1/x^2 + 196x^2} = 12 - 1/x - 14x$
$(\sqrt{1/x^2 + 196^2})^2 = (12 - 1/x-14x)^2$
$1/x^2 + 196x^2 = 172 - 24/x - 336x +1/x^2 +196x^2$
$0=172x-24-366x^2$