Sunday, October 24, 2010

Fox 3

We formatted this old one upon César's solution:


  1. From the figure geometry, we've:a=xsin(θ-60) or a=(x/2)[sin(θ)-√3cos(θ)]. Similarly,b==(x/2)[sin(θ)+√3cos(θ)]. Thus, a^2+b^2+ab=(x^2/4)[3] using the expressions above or x =2√[(a^2+ab+b^2)/3]


  3. Trace a perpendicular to parallelles through the middle vertex, named P. Two new triangles are produced. In first of this, cos(p1)=a/x. In second, cos(p2)=b/x, where p1 and p2 are angles in P and p1+p2=120º, then
    Use sin(u)=sqrt(1-cos²(u)) and get x.

  4. Having x=f(a,b) proves that, only one type of equilateral can be drawn on 3 parallel lines, which is pretty obvious. Fox 290 implies the same:

    If this was a multiple-choice question, below are few lazy observations:

    * The formula must be "symmetric" a and b due to rotation (a and b are replacable). For example, x=sqrt(a^2 + ab) can be rejected right away.

    * The unit of x must reduce to the unit of length. For example, x=a^2 + ab + b^2 can NOT be an answer, because it reduces to length^2. It looks like and area rather than a length.

    * a=b => x=2a

    * a=0 => x=2b/sqrt(3)

    * Finally, x >= max{a,b} always holds

  5. I would add a 5th lazy (sort of) observation:

    * x^2 is a quadratic form in a and b: in vector plane of base (I,J) A=a/x.J and B=b/x.J can be obtained from I by composition of linear apps: rotation theta, n x rotations pi/3, and projection on y axis. Hence A and B are linear in sin(theta) and cos(theta). A and B modules are independent so this system is inversible. We get the result by computing sin^2 + cos^2.

    Now from observation 5 we get:
    x^2 = p.a^2 + q.ab + r.b^2

    From observation 1: p=r
    From observation 4: p=4/3
    From observation 3: q=p=4/3
    x^2= 4/3(a^2 + ab + b^2)

    Lazy doesn't mean useless...