Thursday, December 17, 2009

Fox 212


  1. let be I,J,K the projections of P on AB,AC,BC
    AIP and AJP are right triangles,center D is the midpoint of AP
    hence A(AID)=1/2A(AIP)
    the same way with the other right triangles


  3. Hola César,

    Your comment above was delayed automatically, because Blogger labelled it as spam.

    Gracias por tu trabajo.

  4. Thank you.

    And congratulations: Your Spanish is better than my English.

  5. César, I have looked at your solution. It looks great and very impressive!

    One thing we need to make clear here. Occasionally we ask very hard questions, but we can make the following generalization:

    If a fox' claim is very concise such as A=B, or a+b=c+d, or S=1/2 A (as in this one), then there better be a creative solution which involves a few assisting lines, or seeing an essential feature. That's one of the reasons we use the motto "Please try to see the good side :)"

    For example, in this one: connect AP, and try to see that ADP is co-linear and D is their mid-point (because of the right angle in a circle.) If you do the same thing for BP and CP, then you can easily see that for every triangle in the shaded area, there is an equal triangular area next to it. So S=1/2A holds.

    An the claim is true for any acute triangle. Let's illustrate this as Fox 339, coming sometime tomorrow.

    Buenas noches...

  6. UNTAMED, proof:
    Let’s look at one circle, around the center E, which passes through points, lets say M in BC and N in AB. The opposite angles in M and N of the rectangle BMPN are right, so it is the inscribed rectangle. Also, the angle AMP (or angle PNB) is right so BP is diameter of the circle.
    Look at the two triangles MPE and MPB. They have the same basis MP and the heights to points E and B in proportions 1:2. Accordingly, area(BMP) = 2 x area(MPE). Similarly, area(APN) = 2 x area(EPN), and also in all of the three triangles. Summing all we get: area(ABC) = 2 x shaded area .

  7. Rastko, thank you for a solution.