Author |
Topic: non-similar triangles (Read 4745 times) |
|
Christine
Full Member
Posts: 159
|
 |
non-similar triangles
« on: May 24th, 2013, 10:23am » |
 Quote  Modify
|
The triangle (5,7, 8) has one of its angle measuring 60 degrees.
Law of cosines:
5^2 + 8^2 - (2 * 5 * 8)cos(60) = 49 = 7^2
The triangle has 60 degrees angle between the sides of length 5 and 8.
How many non-similar triangles w/ integer sides and an angle of 60 degrees are there?
// turned off smileys so 8) shows as intended --towr
|
| « Last Edit: May 24th, 2013, 11:37am by towr » |
 IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender: 
Posts: 13730
|
|
Re: non-similar triangles
« Reply #1 on: May 24th, 2013, 12:27pm » |
Quote Modify
|
So, basically how many solutions can we find to a2+b2 -ab = c2, where a,b,c are coprime.
I'll assume there are solutions of the form:
a = r m2 + s mn + t n2
b = u m2 + v mn + w n2
c = x m2 + y mn + z n2
Which gives us
(r m2 + s mn + t n2)2 +(u m2 + v mn + w n2)2 - ((r m2 + s mn + t n2)*(u m2 + v mn + w n2)) - ( x m2 + y mn + z n2)2 = 0
From which we get the system:
2 r s - r v - s u+2 u v-2 x y = 0
2 s t - s w - t v+2 v w-2 y z = 0
2 r t - r w + s2- s v- t u+2 u w+ v2-2 x z- y2 = 0
r2 - r u + u2 - x2 = 0
t2 - t w + w2 - z2 = 0
Some searching gives us lots of possible solutions, among which
a = m2 + 2 mn
b = 2 mn + n2
c = m2 + mn + n2
So for distinct m,n we get an infinite number of integer triangles with a 60 degree angle.
And I think if m and n are coprime a, b and c are also; but I'll leave that for another time/person to check.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: non-similar triangles
« Reply #2 on: May 24th, 2013, 12:39pm » |
Quote Modify
|
Hmm, well, the last bits not true unfortunately, so there remains some work to be done.
Maybe just take m,n prime?
[edit]That also fails (e.g 5,11 -> 135,231,201 which are divisible by 3) [/edit]
[edit2]n=m+1 seems to work[/edit2]
|
| « Last Edit: May 24th, 2013, 1:02pm by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: non-similar triangles
« Reply #3 on: May 24th, 2013, 1:53pm » |
Quote Modify
|
Is it in general the case that if there is some integer triangle with an angle A, that there are infinitely many non-similar integer triangles with an angle A?
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: non-similar triangles
« Reply #4 on: May 24th, 2013, 3:12pm » |
Quote Modify
|
I computed the solutions up to 1000 and found some regularity that leads to....
Consider the 1/2 equilateral triangle with sides (d, 2d, sqrt(3)d). It has a 60 degree angle between d and 2d. It is a right triangle.
To complete the triangle and get integer sides, let's find a 2nd right triangle with sides (sqrt(3)d, e, f) with f the hypothenuse.
We want
3d2 + e2 = f2
That is easy when d is odd. We can choose
e+1 = f
and we need
(e+1)2 - e2 = 3d2
<=> 2e+1 = 3d2
<=> e = (3d2-1)/2
When we put together the triangles, we get a triangle with sides:
a = 2d, b = d+e, c = f
Since d must be odd, let's write d = 2k+1. We can express a, b, c in terms of k:
a = 4k + 2
b = 6k2 + 8k + 2
c = 6k2 + 6k + 2
Since all are even, we can divide them by 2
a' = 2k + 1
b' = 3k2 + 4k + 1
c' = 3k2 + 3k + 1
You can verify these satisfy a'2 - a'b' + b'2 = c'2 (or ab - a2 = b'2 - c'2)
And a',b',c' are co-prime because a'-2b'+2c' = 1
So there are an infinity of solutions
|
| « Last Edit: May 26th, 2013, 2:32pm by Grimbal » |
IP Logged |
|
|
|
Christine
Full Member
Posts: 159
|
|
Re: non-similar triangles
« Reply #5 on: May 24th, 2013, 7:11pm » |
Quote Modify
|
towr,
Thanks for turning off the smileys. I tried but no success.
How did you do it? Just in case it happens again in the future
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: non-similar triangles
« Reply #6 on: May 25th, 2013, 12:04am » |
Quote Modify
|
When you make a post or edit, there are two checkboxes under the textarea, notify of replies and disable smilies. So you need to check the second one.
Another option is to always put a space after an 8 and before the ), because then it doesn't form that smiley.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: non-similar triangles
« Reply #7 on: May 25th, 2013, 2:23am » |
Quote Modify
|
I searched by computed for solutions to
a2 + b2 - 2ab cos(alpha) = c2
for small rational values of cos(alpha). There seems to always be plenty of solutions. Even for values >1 or <-1.
|
|
IP Logged |
|
|
|
Christine
Full Member
Posts: 159
|
|
Re: non-similar triangles
« Reply #8 on: May 25th, 2013, 9:33am » |
Quote Modify
|
c^2 = a^2 + b^2 - a*b
the 60 degrees angle is between a and b.
the sides a and b are integers
if b = 1
c^2 = a^2 + 1 - a
<=> (2c + 2a - 1)(2c - 2a + 1) = 3
Let x and y be two rational numbers such that x*y = 3
suppose
(2c + 2a - 1) = x
(2c - 2a + 1) = y
then
c = (x^2 + 3)/4x and a = (x-1)(x+3)/4x
therefore, triangles with sides
a = (x-1)(x+3), b = 4x and c = (x^2 + 3)
where x >= 2
will have integer sides and a 60 degrees angle
Can you show that these triangles will be similar for x=2 or x=5
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: non-similar triangles
« Reply #9 on: May 25th, 2013, 11:00am » |
Quote Modify
|
on May 24th, 2013, 1:53pm, towr wrote:| Is it in general the case that if there is some integer triangle with an angle A, that there are infinitely many non-similar integer triangles with an angle A? |
|
You get a right triangle with cos(alpha)=p/q by just taking p as a side and q as the hypothenuse. The third side is sqrt(q2-p2).
So you can do the same construction as I did above with the half of an equilateral triangle. You get:
a = 2nq
b = n2(q2-p2) + 2np - 1
c = n2(q2-p2) + 1
You can verify that (a2+b2-c2)/(2ab) = p/q
And the triangles are dissimilar because b/a increases with n.
|
| « Last Edit: May 25th, 2013, 11:01am by Grimbal » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print