## Monday, November 16, 2009

### Fox 186

1. Thank you 8foxes

2. This comment has been removed by a blog administrator.

3. The bottom side of the bold area is 1/sin(tetha)
where tetha between asin(1/3)=19.47 and 90

and the bold area A= 1/sin(tetha)-B
B is zero where asin(2/3)=41.8 < tetha < 90

so if tetha=46.1622
A=1/sin(46.1622)=1.3863

if tetha=asin(2/3)=41.8
A=1.5 > 1.4649

so 1.4649 may not be the answer

4. I found A = 1,712 for theta = 29,4, by doing A = [3 - (upper triangle) - (lower trapezium)]. Sorry, I'm new here, and don't know the best way to show details.

5. theta = 29,4 gives A=1.712

If you send us your solution we can publish it here. Thank you.

8foxes.com

6. Yes you are right Oscar.
I have to work on it again, thank you:))

7. For the name of Humanity, I deleted my comments. They don't have the rigth that made people make mistakes

8. I noticed that i made a mistake while writing eq. before

max of the equation below gives the answer
and it is nearly 1.712 as Oscar said

1/sin(theta)-(2tan(theta)+2/tan(theta)-3sin(theta)tan(theta)-3cos(theta))*(2-3sin(theta))*1/2

9. Here goes the solution
There are 3 situations of interest:

a. 0 ≤ θ ≤ sin–1 (1/3), or 0 ≤ θ ≤ 19,47°  the intersection will be just a triangle, starting with area S = 0, going up to S = (√8)/2 = √2 = 1,414
b. sin–1 (2/3) ≤ θ ≤ π/2, or 41,81° ≤ θ ≤ 90°  the intersection will be a rhombus and will end up as a square, with area going from S = 3/2 = 1,5 down to S = 1
c. sin–1 (1/3) ≤ θ ≤ sin–1 (2/3)  the intersection will be the small rectangle ABCD, minus the triangle BDE, minus the trapezium OCFG.
We have SBDE = EB*BD/2
BD = 1 – DA = 1 – (2 – 3*sinθ) /cosθ
EB = ED*cosθ = (BD/sinθ)*cosθ
then, SBDE = [(BD/sinθ)*cosθ] * BD/2 = ½*[1 – (2 – 3*sin θ) /cosθ]2 * cosθ/sinθ
= ½*(cosθ – 2 + 3*sinθ)2 /(cosθ * sinθ)
and SOCFG = ½*(CF + OG) * 1
OG = 3/(2 – 3*sinθ)
CF = 3 – FE – EB = 3 – OG – EB
then, SOCFG = ½*(3 – EB) = ½*[3 – (BD/sinθ)*cosθ] = ½*[3 – (cosθ /sinθ) – (2 – 3*sinθ) /sinθ]
= 1/sinθ – cosθ/(2*sinθ)
Therefore, S = 3 – SBDE – SOCFG
= 3 – ½*(cosθ – 2 + 3*sinθ)2 /(cosθ * sinθ) – 1/sinθ – cosθ/(2*sinθ), or
S = 1/sinθ – 2/(sinθ*cosθ) + (12 – 9*sinθ) /(2*cosθ), for sin–1 (1/3) ≤ θ ≤ sin–1 (2/3)
Searching for the maximum of this function, we get
Smax = 1,7120 for θmax = 29,40°

10. Updated solution

http://i48.tinypic.com/2mhwqbp.gif

11. PS
on updated solution, 3. case cannot be drawn becasue 3sin(34.16)-1+cos(34.16) > 1
(forgot to say)