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Old 2005-09-10, 00:00   #1
Citrix
 
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Default 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.

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Old 2005-09-10, 00:51   #2
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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
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Old 2005-09-10, 01:04   #3
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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.
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Old 2005-09-10, 01:24   #4
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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.
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Old 2005-09-10, 01:35   #5
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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.

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Old 2005-09-10, 01:37   #6
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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.
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Old 2005-09-10, 01:39   #7
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Cheesehead an intresting solution. You are right. But there is a simpler solution, can you find it?

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Old 2005-09-10, 06:54   #8
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Hey, what's simpler than drawing a couple of intersecting lines? :-)
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Old 2005-09-10, 06:56   #9
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Algebra
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Old 2005-09-10, 11:24   #10
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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.
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Old 2005-09-10, 18:14   #11
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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.

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