I've got another puzzle that I'w working on that has stumped me. This one, I'm sure, has a much easier solution than the one I posted earlier.

The problem is: find six different nondegenerate triangles with integer sides for which a=16 and A=60 degrees.

In this problem the three angles are labeled A,B,C and the sides opposite them are a,b,c.

What exactly do you mean by nondegenerate?

The following formula is valid for any triangle.a² = b² + c² - 2*b*c*cos(A)For your problem, this becomes:a² = b² + c² - b*cTry fooling around with the above.

Originally posted by Mephura
What exactly do you mean by nondegenerate?

A nondegenerate triangle is a triangle with no angles equal to zero.

Originally posted by Dinosaur
Try fooling around with the above.

Oh I have.

Perhaps there is something I'm not seeing though.

Thinking out loud here:
a=16

a^2=256=(b-c)^2+b*c
therefore, 256-b*c is a square number

b-c (b-c)^2 b*c
1 1 255
2 4 252
...

This question has me stumped. It is easy to see that an equilateral triangle works. However, the other 5 seem impossible


Following on from Mephura's logic.

b*c <=256

max diff between b and c is 16

256-b*c is a square

possible values of b*c with factors
255 5*17*3
252 2*2*7*9
247 13*19
240 2*2*2*2*3*5
231 3*7*11
220 2*2*5*11
207 3*3*23
192 2*2*2*2*2*2*3
175 5*5*7
156 13*2*2*2
135 5*3*3*3
112 7*2*2*2*2
87 3*29
60 2*2*3*5
31 31
0 0

Possibly I have misunderstood the question or have missed a combination of b and c that would meet the constraints, but I am pretty sure that the equilateral triangle is the only possible solution.

Cheers!

I picked c equal values from 1 to 6, then solve with those values for b.
Then solve for the angles C & B.

http://members.rogers.com/force1/angleBandLength.jpg

I needed 2 seperate set of angles because b had 2 different answers.
http://members.rogers.com/force1/angleC.jpg

Specialist: No need to go to all that trouble. You only need find integer values which satisfy either of the following.256 - b*c = (b - c)²

256 = b² + c² - b*cI did a quick analysis and did not find anything other than b = c = 16, which results in the equilateral triangle.

I was too lazy to try all values of b from 1 to 16 and solve for 16 possible values of c.

Huh, actually 3/4 of the information on my picture is the value of the angle for the length that suppliments the triangle.

Only solutions for c = 1,2,3,4,5,6 were calculated (the question asked for six :)) then for each c value, 2 b values were produced from the quadratic equation. That makes 12 in total but some length might have to be discarded since it's negative.

There is no way around it, if you substitute 6 values for either b or c, your still left with a quadratic equation that will give another two solutions. 2x6=12

Unless you're implying I guess values to satisfy the equation...

Originally posted by contrarian
Following on from Mephura's logic.


while I appreciate the credit, it wasn't mine.
I'm thinking there is hell of alot (i don't want ot say infinite with out thinking more) of solutions.

I think you were refering to dinosuar or geodesic.

My logic on this one is probably flawed.

Edit: just caught the bit about integer sides...
definitely not infinite. That makes it a bit more restrictive. Not even a hell of alot now.
My bad.

Other than the equilateral triangle, there are no triangles which satisfy the conditions. Some algebra results in the following.b1 = [c + SquareRoot(1024 - 3c²)] / 2

b2 = [c + SquareRoot(1024 - 3c²)] / 2I used MathCad7 to try all values of c from 1 to 18. 16 results in the equilateral triangle. No other solutions were integers. For c greater than 18, b1 & b2 are imaginary.