## A 10-year old triangle

DanishDynamite
Posts: 2608
Joined: Mon Jun 07, 2004 4:58 pm
Location: Copenhagen

### A 10-year old triangle

Here's a fairly easy one.

Suppose you have a right triangle where the length of each side is a whole number. Suppose the length of one of the non-hypotenus sides is 1994. What is the length of the hypotenuse?

(Yes, ceptimus, I still need to get started on your Queen's walk :) )
millirem
Posts: 8
Joined: Sun Aug 29, 2004 3:51 am
The hypothenuse is 994010; the other side is 994008.
Check: 994008^2 + 1994^2 = 994010^2
DanishDynamite
Posts: 2608
Joined: Mon Jun 07, 2004 4:58 pm
Location: Copenhagen
Correct, millirem. :)

millirem
Posts: 8
Joined: Sun Aug 29, 2004 3:51 am
DanishDynamite wrote:Correct, millirem. :)

I used the same formulas as those mentioned by Abdul Alhazred's post.
b = 2mn = 1994,
from which m=997 and n=1.
millirem
Posts: 8
Joined: Sun Aug 29, 2004 3:51 am
Abdul Alhazred wrote: Is there any integer greater than 2 that cannot be the side of an integral pythagorean triangle?
No, there isn't; given a number a >2, we can always construct at least one Pythagorean triangle with it:

If a is even
b = (a/2)^2 - 1
c = (a/2)^2 + 1

If a is odd
b = (a^2 - 1)/2
c = (a^2 + 1)/2
millirem
Posts: 8
Joined: Sun Aug 29, 2004 3:51 am

### Re: Pythagorean triangles with one side equal to 120

Abdul Alhazred wrote:Pythagorean triangles with one side equal to 120:

If 120 is the hypotenuse, then we have a 24 times 3, 4, 5:
72, 96, 120

Otherwise there exist pythagorean triangles where one side equals 120 and the other (non-hypotenuse) side equals:
22
27
35
50
64
90
119
126
160
182
209
225
288
350
442
715
896
1197
1798

I have left out three of them.
This problem is a bit more tedious, since we have to consider that the sides may have a common factor k; the formulas then become:
a = k(m^2 - n^2)
b = k(2mn)
c = k(m^2 + n^2)

k=1
2mn=120 mn=60= 60*1=30*2=20*3=15*4=12*5=10*6
m=60,n=1 a=3599
m=30,n=2 a=896
m=20,n=3 a=391
m=15,n=4 a=209
m=12,n=5 a=119
m=10,n=6 a=64
(Note: the sides have no common factor iff m and n are relatively prime.)

k=2
2mn=60 mn=30 =30*1 = 15*2 = 10*3 = 6*5
m=30,n=1 a=1798
m=15,n=2 a=442
m=10,n=3 a=182
m=6,n=5 a=22

k=3
2mn=40 mn=20 = 20*1 = 10*2 = 5*4
m=20,n=1 a=1197
m=10,n=2 a=288
m=5,n=4 a=27

k=4
2mn=30 mn=15 = 15*1 = 5*3
m=15,n=1 a=896
m=5,n=3 a=64

k=5
2mn=24 mn=12 = 12*1 = 6*2 = 4*3
m=12,n=1 a=715
m=6,n=2 a=160
m=4,n=3 a=35

k=6
2mn=20 mn=10 = 10*1 = 5*2
m=10,n=1 a=594
m=5,n=2 a=126

k=10
2mn=12 mn=6 = 6*1 = 3*2
m=6,n=1 a=350
m=3,n=2 a=50

The missing sides are 3599, 391 and 594.
millirem
Posts: 8
Joined: Sun Aug 29, 2004 3:51 am
New problem:

There was a big mistake in my previous post. Find this mistake!