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 or and perimeter is
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:
I turn the first equation into and substitute into the other equation:
Multiplying everything by x to get rid of the 336/x, I get:
However, the book (The Story of ) says it's
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.
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.
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.
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...
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:
So we have two perpendicular sides of and 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 and . 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!
I am reading that very book and I have arrived at the appropriate equation. Paul Nahin (the author) simplifies Diophantus' reduced equation .
First bring 1/x and 14x to the other side of the equation to simplify the radical. so
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)