Question 148450

Once again, a drawing helps:


<center>
<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/47-15-08.jpg" alt="Photobucket - Video and Image Hosting"><br><br>
</center>

I'm not real sure how to go about this problem, but I would start by pairing up corresponding sides. So side "5" would correspond to side "4", side "5" would correspond to side "8", side "5" would correspond to side "y", etc. This pairing goes on for quite a while. So here are the pairs that I've drawn up


5/4=6/y    Solution  {{{y=24/5}}}
5/4=6/8    No Solution

5/8=6/y    Solution  {{{y=48/5}}}
5/8=6/4    No Solution

5/y=6/8    Solution  {{{y=20/3}}}
5/y=6/4    Solution  {{{y=10/3}}}

-------------

6/4=5/y    Solution  {{{y=10/3}}}
6/4=5/8    No Solution

6/8=5/y    Solution  {{{y=20/3}}}
6/8=5/4    No Solution

6/y=5/8    Solution  {{{y=48/5}}}
6/y=5/4    Solution  {{{y=24/5}}}

-------------

x/4=5/y   Infinite Solutions
x/4=5/8   Solution  {{{x=5/2}}}

x/8=5/y   Infinite Solutions
x/8=5/4   Solution  {{{x=10}}} 

x/y=5/8   Infinite Solutions
x/y=5/4   Infinite Solutions

---------------

This list is probably far from exhaustive, but it gives you an idea of the sides that the triangles can take. To see if a triangle is formed from 3 sides, you must test to see if the sum of any two given sides are greater than the length of the third side (you must test this for all three sides and for every possible case. So the problem gets a lot longer).


Also, to find every solution has stumped me. From the list, it's hard to tell for sure if I've listed every case. Sorry for the trouble.