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NXTgen

Post subject: Triangle question Posted: Sun Sep 02, 2012 11:05 pm 

Joined: Sun Dec 13, 2009 5:48 pm

So if you are good at math (I'm decent) then I'd like some help. I have a theoretical triangle, but the problem is: it is a right triangle, and all three sides are the same length, two of which are given for the sides forming the right angle. The longest side is equal to the other sides. I would like to know if it is solvable. Here's a picture for reference: Attachment:
1344735232635.jpg [ 297.44 KiB  Viewed 1115 times ]
It is really annoying me and I MUST find the answer. And yes, this is the image I found the problem from...
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Jared

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:11 pm 

Joined: Mon Aug 18, 2008 10:16 pm Location: Somewhere Else

If it was on a sphere, it would work...


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Brandon Enright

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:12 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

In a right triangle a^2 + b^2 = c^2
If a == b == c then a == b == c == 0
Assuming that diagram is correct then x = sqrt(722) == 26.870057685088805927232085759984263493
Is there more to this problem that would help indicate the "trick"? Otherwise your description contradicts the rest of the information provided.
The other option is noneuclidean geometry (possibly spherical, hyperbolic, etc).
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will_57

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:12 pm 

Joined: Sun Mar 08, 2009 9:21 am Location: Massachusetts, USA

No, this is not solvable. If it is a right triangle, the sides cannot all be the same length, assuming we aren't talking about triangles on the surface of a sphere.
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Brandon Enright

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:36 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

I'm not so great at noneuclidean geometry but I think for this triangle to be on the surface of a sphere, the sphere would need a radius of 38 / Pi.
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NXTgen

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:43 pm 

Joined: Sun Dec 13, 2009 5:48 pm

Thanks guys for the responses, I found this and I guess my description of the problem didn't help as much as I had thought. I had saved the picture a while ago and I tried and tried but couldn't find a solution.. Thanks I'll try what bmenrigh said, using noneuclidean geometry.. If a solution is found I'd like to know about it!
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Brandon Enright

Post subject: Re: Triangle question Posted: Sun Sep 02, 2012 11:54 pm 

Joined: Thu Dec 31, 2009 8:54 pm Location: Bay Area, California

NXTgen wrote: Thanks guys for the responses, I found this and I guess my description of the problem didn't help as much as I had thought. I had saved the picture a while ago and I tried and tried but couldn't find a solution.. Thanks I'll try what bmenrigh said, using noneuclidean geometry.. If a solution is found I'd like to know about it! Well if the triangle is on the surface of a sphere see: http://www.smith.edu/physics/felder/cur ... angle.htmland https://en.wikipedia.org/wiki/Elliptic_geometryIf you assume you have a 19x19x19 triangle taking up 1/8th the surface area of a sphere then the sphere has a radius of 38 / Pi.
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KelvinS

Post subject: Re: Triangle question Posted: Mon Sep 03, 2012 2:50 am 

Joined: Mon Mar 30, 2009 5:13 pm

Yes, it must be 1/8th of the surface of a sphere of circumference 4*19 = 76, so that all 3 sides have a length of 19, and all 3 angles are 90 degrees. Simple. Just think about the curved triangular surface of a "corner" piece on a 2x2x2 puzzle ball. EDIT. The triangle could also be on an ellipsoid or other curved surface, but that would be a lot more tricky to define and figure out. So there are actually multiple solutions to this problem, which means that the question itself is actually not solvable because it doesn't provide enough information on the shape of the surface, and the person who wrote this question is not as clever as he/she thinks.
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