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

  1. #1
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    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.

    P.S. Answers are complex, FYI.
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  2. #2
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    Quote Originally Posted by Grep View Post
    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.

    P.S. Answers are complex, FYI.
    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
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  3. #3
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    Quote Originally Posted by earboth View Post
    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
    Thanks a lot for your reply.

    Here's a link to that page on Google Books:
    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.
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  4. #4
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    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.
    Last edited by Wilmer; November 22nd 2010 at 04:18 AM. Reason: added "same" remark
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  5. #5
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    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...
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  6. #6
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    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.

    So that explains that. His x and my x were two different things. Which, it turns out, is an important fact!
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    Re: An equation from Diophantus I can't seem to solve properly

    Quote Originally Posted by Grep View Post
    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.

    P.S. Answers are complex, FYI.
    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.

    First bring 1/x and 14x to the other side of the equation to simplify the radical. so
    \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

    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)
    Last edited by neffk; March 18th 2013 at 02:31 PM.
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