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Pythagoras Unconventional (Posted on 2011-02-10) Difficulty: 3 of 5

No Solution Yet Submitted by brianjn    
Rating: 4.0000 (1 votes)

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Mathematica search | Comment 3 of 10 |

to simplify things we have three integer values, name them x,y,z
that leaves 3 more irrational rod lengths and only one more digit that can be used.  Thus we can name the list of rods as
x,y,z,sqrt(w),sqrt(x),sqrt(y)
with x,y,w not perfect squares.  Of course this does not always give them in order.

I used the following Mathematica code to count the number of right triangles formed with each of the valid sets {x,y,z,w} of digits 1-9.

dgs=Table[i,{i,1,9}];

subs=Subsets[dgs,{4}];

slng=Length[subs];

pyths=Table[{},{i,0,20}];

ProgressIndicator[Dynamic[si/slng]]

For[si=1,siTslng,++si,

lst1=subs[[si]];

perms=Permutations[lst1];

For[k=1,kT24,++k,

lst=perms[[k]];

x=lst[[1]];

y=lst[[2]];

z=lst[[3]];

w=lst[[4]];

If[!IntegerQ[Sqrt[x]] && !IntegerQ[Sqrt[y]] && !IntegerQ[Sqrt[w]],

rods={x,y,z,Sqrt[w],Sqrt[x],Sqrt[y]};

count=0;

tsubs=Subsets[rods,{3}];

tlng=Length[tsubs];

For[ti=1,tiTtlng,++ti,

tri=tsubs[[ti]];

a=tri[[1]];

b=tri[[2]];

c=tri[[3]];

If[a^2+b^2c^2 || a^2+c^2b^2 || b^2+c^2a^2,

++count;

];

];

rods=Sort[rods];

If[!MemberQ[pyths[[count+1]],rods],

pyths[[count+1]]=Append[pyths[[count+1]],rods];

];

];

];

];

It gives 3 rod length sets that give exactly 4 right triangles and they are as follows:
{1,2,5,Sqrt(2),Sqrt(3),Sqrt(5)}

{1,2,3,Sqrt(2),Sqrt(3),Sqrt(6)}

{1,2,3,Sqrt(2),Sqrt(3),Sqrt(7)}

my next post will give the full listing.  As a side note Ady TZIDON, the list you found is the only one with 5 right triangles and is also the list with the most right triangles.


  Posted by Daniel on 2011-02-10 14:12:27
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