## Problem 039

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nZifnab
Posts: 4
Joined: Thu Sep 25, 2008 10:46 pm

### Problem 039

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
funktio
Posts: 14
Joined: Mon May 19, 2008 6:55 pm
Location: Helsinki, Finland

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

nZifnab wrote:what does "integral length sides" mean?
They must be integers.
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nZifnab
Posts: 4
Joined: Thu Sep 25, 2008 10:46 pm

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

funktio wrote:
nZifnab wrote:what does "integral length sides" mean?
They must be integers.
oh...dur.

Thanks :p
macfreek
Posts: 1
Joined: Thu Nov 13, 2008 4:17 pm

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

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.
rayfil
Posts: 1407
Joined: Sun Mar 26, 2006 5:30 am
Contact:

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

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.
When you assume something, you risk being wrong half the time.
eay
Posts: 1
Joined: Thu Oct 22, 2009 5:07 am

### Re: Problem 039

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?
Last edited by eay on Thu Oct 22, 2009 3:36 pm, edited 5 times in total.
elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Problem 039

Erm...you should edit that out. Solutions + open forum = baaaaaaaaad....
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?
JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 11:35 pm

### Re: Problem 039

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?
rayfil
Posts: 1407
Joined: Sun Mar 26, 2006 5:30 am
Contact:

### Re: Problem 039

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
When you assume something, you risk being wrong half the time.
davidlively
Posts: 3
Joined: Sun Aug 19, 2018 9:49 pm

### Re: Problem 039

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."
Last edited by davidlively on Sun Aug 19, 2018 10:44 pm, edited 1 time in total.
v6ph1
Posts: 131
Joined: Mon Aug 25, 2014 7:14 pm

### Re: Problem 039

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.

davidlively
Posts: 3
Joined: Sun Aug 19, 2018 9:49 pm

### Re: Problem 039

Thanks for the information, and that'll be helpful on other problems.

However, they're *not* geometrically identical. One points north, the other points east. (Depending on your perspective, I guess!)

(Qualifier: I'm a graphics programmer, so these things come up. A lot.)

Thanks again.
griffrawk
Posts: 4
Joined: Mon May 31, 2021 12:03 pm
Location: Wirral, UK

### Re: Problem 039

Going back in time a bit here re. #39 which I solved quite some time ago. I'm currently stuck with #75 which currently I'm treating as a version of #39. So I revisited my code for #39 and I found something I found with relevance to why I may be stuck, and wondering how I managed to get the right solution for #39 when I did.

Euclid's formula states that "..given an arbitrary pair of integers m and n with m > n > 0... the integers a = k(m^2 − n^2) , b = k(2 m n), c = k(m^2 + n^2) form a Pythagorean triple. The triple generated by Euclid's formula is primitive if and only if m and n are coprime and not both odd."

The example given in the text for #39 states that there are 3 solutions when p = a + b + c = 120.

Now, this is true for p = 120 when not checking m, n are coprime (therefore not primitive), but the correct solution for the whole problem is only found when you do check m, n are coprime. In that case only 2 solutions for p = 120 appear.

Is the wording of the problem intentional? I appreciate there are other ways of solving the problem where this won't come up.
hk
Posts: 11318
Joined: Sun Mar 26, 2006 10:34 am
Location: Haren, Netherlands

### Re: Problem 039

It's not stated that the solutions must be primitive.
In fact none of the given solutions is primitive:
{20,48,52}=4*(5,12,13)
{24,45,51}=3*(8,15,17)
{30,40,50}=10*(3,4,5)
griffrawk
Posts: 4
Joined: Mon May 31, 2021 12:03 pm
Location: Wirral, UK

### Re: Problem 039

Ah.. sorry I probably shouldn't have stressed the primitive (given that k is involved). I misworded that.

However I still get the correct solution when checking m, n are coprime, but only 2 solutions for p=120. Yet the the problem suggests 3 solutions for p=120 which I only get when I don't check m, n are coprime.

Hmm.. The missing solution I get for p=120 when I don't check coprime is {20,48,52}. But I get that for k=1, not 4 as you show. Ok I need to check my code again and find out why I don't get {20,48,52}=4*(5,12,13) when I am checking coprime.

In addition, and probably also due to this, I never get a result for p=30, which I haven't been able to track down.

It's a miracle this ever worked the first time!

Thanks hk.
hk
Posts: 11318
Joined: Sun Mar 26, 2006 10:34 am
Location: Haren, Netherlands

### Re: Problem 039

There is another condition: m and n should have different parity.
The coprime triple that generates (5,12,13) is m=3 and n=2
The other set (m,n) is 5,1: 25+1=26. 25-1=24 and 2*5*1=10 which gives 10,24,26. But because m and n are both odd it's not a valid coprime triple.
griffrawk
Posts: 4
Joined: Mon May 31, 2021 12:03 pm
Location: Wirral, UK

### Re: Problem 039

Agreed. I think I'm generating m, n incorrectly. I thought I was, but probably didn't check thoroughly enough. I suspect I'm skipping valid pairs.
griffrawk
Posts: 4
Joined: Mon May 31, 2021 12:03 pm
Location: Wirral, UK

### Re: Problem 039

Yes, skipping valid m,n pairs. All fixed now and onto #75 with that fix.

Thanks for the pointers, hk. Much appreciated.