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### Problem 039

Posted: Tue Sep 30, 2008 5:36 pm

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p<1000, is the number of solutions maximised?

Now as i see it, the sides of a right triangle must match the well-known pythagorean a^2+b^2=c^2

However the wording of the question throws me off when it says "with integral length sides"...I know integral is generally used in the realm of calculus, something I haven't really dabbled in since high school (i'm now out of college), but I fail to see how it relates to the problem at hand. Am I looking for perimeter values with special rules for their side lengths, or are any side lengths that match the pythagorean theorem fair game?

If I need a special rule for it, what does "integral length sides" mean?

Thanks for the help ### Re: Problem 39 - Integral Length Sides?

Posted: Tue Sep 30, 2008 5:54 pm
nZifnab wrote:what does "integral length sides" mean?
They must be integers.

### Re: Problem 39 - Integral Length Sides?

Posted: Tue Sep 30, 2008 5:55 pm
funktio wrote:
nZifnab wrote:what does "integral length sides" mean?
They must be integers.
oh...dur.

Thanks :p

### Re: Problem 39 - Integral Length Sides?

Posted: Thu Nov 13, 2008 4:24 pm
To the moderators: there is an off-by-one error in the problem itself :
• The summary of this problem says p ≤ 1000
• The long description of this problem says p < 1000
Of course, the solution remains the same.

### Re: Problem 39 - Integral Length Sides?

Posted: Sun Nov 16, 2008 4:38 am
macfreek wrote:To the moderators: there is an off-by-one error in the problem itself :
• The summary of this problem says p ≤ 1000
• The long description of this problem says p < 1000
Of course, the solution remains the same.
Thanks for pointing that out. Correction made.

### Re: Problem 039

Posted: Thu Oct 22, 2009 4:13 am
removed posted solution

I have an efficient solution to this problem that was not posted by anyone (I believe) in the solutions, since the thread is locked, is there anyway I can get it added?

### Re: Problem 039

Posted: Thu Oct 22, 2009 4:20 am
Erm...you should edit that out. Solutions + open forum = baaaaaaaaad....

### Re: Problem 039

Posted: Sun Sep 18, 2011 4:05 am
I'm now stump on this problem. Are we supposed to figure out what value below or equal to 1000 has the most count? In addition, does "integral length sides only" means ones that are Pythagorean triples which obviously goes with integers only?

### Re: Problem 039

Posted: Sun Sep 18, 2011 5:35 am
JMW1994 wrote:I'm now stump on this problem. Are we supposed to figure out what value below or equal to 1000 has the most count?
Yes. The example shows a "count" of 3 for the right angle triangles with integral length sides having a perimeter of 120. The required answer is the value of that perimeter, not to exceed 1000, which would have the highest count.
In addition, does "integral length sides only" means ones that are Pythagorean triples which obviously goes with integers only?
Yes

### Re: Problem 039

Posted: Sun Aug 19, 2018 8:52 pm

The problem description says the solutions (for the example) are

{20,48,52}, {24,45,51}, {30,40,50}

But, if we allow degenerate triangles, then

{60,0, 60} and {0,60,60}

are valid (or, really, just pick one of them.)

Are degenerate triangles not valid, or is the description incorrect?

For any given perimeter it only adds two (one) possibilities, but perhaps the text of the question should be clarified to specify "non-zero integers."

### Re: Problem 039

Posted: Sun Aug 19, 2018 9:32 pm
Right angle triangle should be clear enough.
Only 1 angle of a degenerated triangle can be determined exactly: 0° between the two identical sides.
The other two angles are undetermined.

In any problem description, corner cases are named extra. (Integer 0, degenerated polygons,...)
But this does not change the solution of this problem.