## Thursday, December 10, 2009

### Fox 206 We have some work on this, but not the final solution.

1. I have a proof but.... forget it:)

2. Don't be shy share with us. This is really hard!

3. sqrt(15)/4, of course i am not sure:)

4. You can ask the biggest area too

5. sqrt(15)/4 looks reasonable but, I actually don't have the answer. Max area will be a different fox, may be later...

6. I am sure that the answer is 0.949489745, i will post the solution later if i find a shorter one, now i have too sleep:)

7. it is interesting that max area-min area=3

8. Newzad's two last posts are right!

Let A,B,C the vertixes on inner, middle and outer circles. Put the circles centered on origen O and place A at coordinates [1,0]. Let b=<XOB and c=<XOC. Of course, c depends on b, because of the rectangle angle, so the triangle depends only on angle b.

We have the A, B, C coordinates:
A[1,0]; B[2cos(b),2sin(b)];C[3cos(c),3sin(c)]

and must satisfy:
vector BA orthogonal vector BC and their scalar product BA.BC is 0. This lead to equation:

u1 = 2cos(c-b)-cos(c)=(4-2cos(b))/3

The area of ABC is half the modulus of vectorial product BA x BC. From here we obtain a second equation:

u2 = 2sin(c-b)-sin(c)=(2Area-2sin(b))/3

Let's calculate u1^2 + u2^2. It is 5-4cos(b)!!. We are lucky!!!. All c's are gone away!!

Now whe have:

(2Area-2sin(b))^2/9+(4-2cos(b))^2/9 = 5-cos(b)

and, from here, you can obtain Area as function of b:

Area=sin(b)+sqrt(54-(2cos(b)+5)^2)/2

Now that my luck is over, I used Maple to find these values (sorry!):
minimal = 0.9494897434
maximal = 3.949489742

9. 