20050910, 00:00  #1 
Jun 2003
3^{2}×5^{2}×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 
20050910, 00:51  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·599 Posts 
Quote:
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 20050910 at 00:55 

20050910, 01:04  #3  
Jun 2003
The Texas Hill Country
2×541 Posts 
Quote:


20050910, 01:24  #4 
Jun 2005
Near Beetlegeuse
110000100_{2} 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. 
20050910, 01:35  #5  
Jun 2003
3047_{8} Posts 
Quote:
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 

20050910, 01:37  #6  
Jun 2003
3^{2}·5^{2}·7 Posts 
Quote:


20050910, 01:39  #7 
Jun 2003
3^{2}×5^{2}×7 Posts 
Cheesehead an intresting solution. You are right. But there is a simpler solution, can you find it?
Citrix 
20050910, 06:54  #8 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·599 Posts 
Hey, what's simpler than drawing a couple of intersecting lines? :)

20050910, 06:56  #9 
Jun 2003
3^{2}×5^{2}×7 Posts 
Algebra 
20050910, 11:24  #10  
Jun 2003
The Texas Hill Country
2×541 Posts 
Quote:
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. 

20050910, 18:14  #11 
Jun 2003
11000100111_{2} 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 ac=n(1/c1/a) ac=n*((ac)/ac) Either a=c or n=ac in either case there are no 2 distinct solutions. Citrix 
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