[jerbil]

For what integral numbers a, b and c, where 0 < a < b < c < 90, is the following equation satisfied, and why?

cos²aº + cos²bº + cos²cº = 1.

This post has been edited by jerbil: 07 November 2009 - 05:00 AM

[ljb]

Spoiler for My opinion

I would think your "why" question means there aren't any solutions

This post has been edited by ljb: 09 November 2009 - 01:39 PM

[ljb]

Not quick enough to edit

Spoiler for

I checked them all and couldn't find any that worked. If it was <= 90 on the limit, it would work.

[psychic_mind]

I'm really having no luck solving this.

Just something to note: If you use a computer to find solutions it may miss some or create false solutions if you simply check if the sum equals one. This is due to rounding errors given that the computer can only store values to a limited accuracy.

[mmiguel1]

Spoiler for Using computer

36, 60, 72
Here's my python script: with indenting signified by ~'s
#!/usr/bin/python
import math
for a in range(0,88):
~for b in range(a+1,89):
~~for c in range(b+1,90):
~~~z = math.cos(a*math.pi/180)**2;
~~~z += math.cos(b*math.pi/180)**2;
~~~z += math.cos(c*math.pi/180)**2;
~~~if math.fabs(z-1) < 0.000001:
~~~~print a, b, c;

[mmiguel1]

Sorry, I meant to edit, not repost

Spoiler for Using computer

36, 60, 72
Here's my python script: with indenting signified by ~'s
#!/usr/bin/python
import math
for a in range(0,88):
~for b in range(a+1,89):
~~for c in range(b+1,90):
~~~z = math.cos(a*math.pi/180)**2;
~~~z += math.cos(b*math.pi/180)**2;
~~~z += math.cos(c*math.pi/180)**2;
~~~if math.fabs(z-1) < 0.000001:
~~~~print a, b, c;

--------------------------
Note that though this was found without checking equality,
I was able to verify equality symbolically on my calculator (not using floating point calculations with round off errors)

This post has been edited by mmiguel1: 10 November 2009 - 08:17 AM

[mmiguel1]

Spoiler for Another verification

cos^2(x) = (1/2)(1+cos(2x))
Using this formula,
we get (1/2)(3 + cos(2a) + cos(2b) + cos(2c) ) = 1
or

cos(2a) + cos(2b) + cos(2c) = -1
letting a,b,c = 36, 60, 72
We get
Cos(72) + Cos(120) + Cos(144) = -1
This is the same as
Cos(2 pi/5) + Cos(2 pi/3) + Cos(4 pi/5) = -1

From a table,
Cos(2 pi/5) = (1/4)( Sqrt[5] - 1 )
Cos(2 pi/3) = -1/2
Cos(4 pi/5) = (1/4)(-1 - Sqrt[5] )
Summing these together gives -1
Meaning that the equation is satisfied and a,b,c = 36, 60, 72 satisfies the original relation.

[jerbil]

Nicely done, solvers. I append my own ideas, which also explains why I entitled the problem the way I did.

 Explanation of Problem.doc (94.5K)
Number of downloads: 33

[mmiguel1]

Very nice solution.
Other close calls:
20, 71, 84
31, 61, 80
39, 54, 77
39, 55, 75
42, 53, 73

These are all off by an error of the order 10^(-5)