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 2005-09-10, 00:00 #1 Citrix     Jun 2003 32×52×7 Posts right triangles? Find 2 right triangles such that their areas and hypotonuse is the same or show that no such distinct triangles exist. An example is needed in the first case or a proof for the second case. Citrix
2005-09-10, 00:51   #2

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·599 Posts

Quote:
 Originally Posted by Citrix Find 2 right triangles
... with differing sets of nonhypotenutal(spelling?) side lengths, I presume.

Let the nonhypotenutal sides be x and y. Then xy = 2A, where A is the triangle area and is constant. Graphing that, it's a hyperbola. We need consider only the branch where x > 0 and y > 0. Note that the branch is symmetrical around the line x = y.

In order that the hypotenuse be constant H, we need x^2 +y^2 = H^2. Graphing that gives us a circle centered at (0,0) with radius H.

So the solutions x, y are the 1 or 2 points of intersection of the circle and the hyperbola branch. If there are 2 distinct intersections (x1,y1) and (x2,y2), then x1 = y2 and x2 = y1, so the triangles are equivalent.

Last fiddled with by cheesehead on 2005-09-10 at 00:55

2005-09-10, 01:04   #3
Wacky

Jun 2003
The Texas Hill Country

2×541 Posts

Quote:
 Originally Posted by Citrix Find 2 right triangles such that their areas and hypotonuse is the same or show that no such distinct triangles exist. An example is needed in the first case or a proof for the second case. Citrix
Do your own homework! This isn't a puzzle, it is a standard problem from Plane Geometry.

 2005-09-10, 01:24 #4 Numbers     Jun 2005 Near Beetlegeuse 1100001002 Posts For each integer a, there exists a unique pair of integers x, y such that x + 1 = y x + y = a^2 a^2 + x^2 = y^2 Proof (a^{2}+1) / 2 = y a^2 - y = x therefore x + 1 = y and x + y = a^2 Now, when y = x + k y^2 = x^2 + k(x + y) and since we already know that x + y = a^2 y^2 = x^2 + a^2 Therefore, for each x there is a unique a and y, and no two such distinct triangles exist.
2005-09-10, 01:35   #5
Citrix

Jun 2003

30478 Posts

Quote:
 Originally Posted by Wacky Do your own homework! This isn't a puzzle, it is a standard problem from Plane Geometry.

Wacky, I think I have proven it. What I am wondering is why no one else has come up with the proof so far? I wasn't sure if this problem was a solved or an unsolved problem that is why I posted it here

What is this problem called? Could you provide me some links on this problem.

Citrix

2005-09-10, 01:37   #6
Citrix

Jun 2003

32·52·7 Posts

Quote:
 Originally Posted by Numbers For each integer a, there exists a unique pair of integers x, y such that Therefore, for each x there is a unique a and y, and no two such distinct triangles exist.
I don't agree with your solution, lets discuss it when more people answer to the puzzle, so we don't have to hide the solution.

 2005-09-10, 01:39 #7 Citrix     Jun 2003 32×52×7 Posts Cheesehead an intresting solution. You are right. But there is a simpler solution, can you find it? Citrix
 2005-09-10, 06:54 #8 cheesehead     "Richard B. Woods" Aug 2002 Wisconsin USA 22·3·599 Posts Hey, what's simpler than drawing a couple of intersecting lines? :-)
 2005-09-10, 06:56 #9 Citrix     Jun 2003 32×52×7 Posts Algebra
2005-09-10, 11:24   #10
Wacky

Jun 2003
The Texas Hill Country

2×541 Posts

Quote:
 Originally Posted by Citrix Algebra
For an algebraic solution, Cheesehead's solution is simple enough.
Why don't you post your solution that you consider simpler?

Then think about how you would go about proving it in Geometry class rather than in Analytic Geometry class.

As for "links" to the problem, there weren't any when I took Geometry.

That could have been because there wasn't any Internet then, or for a number of years thereafter. And hyperlinking is an even more recent addition.

 2005-09-10, 18:14 #11 Citrix     Jun 2003 110001001112 Posts let the 2 conditons be a^2+b^2=c^2+d^2 and ab=cd=n (1 and 2) adding 1 and 2 , a+b=c+d substituting b=n/a and c=n/c into 3 and rearranging a-c=n(1/c-1/a) a-c=n*((a-c)/ac) Either a=c or n=ac in either case there are no 2 distinct solutions. Citrix

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