Key words: Geometry, Unusual geometry, Math, Physics, Chemistry, High school, Geometry Olympiads, Free Geometry, Euclidean Geometry, Calculus, Geometric Construction. Oh yes, going-nowhere discussions, haikus, and poems too.

## Tuesday, November 30, 2010

### Fox 320

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## Monday, November 29, 2010

### Fox 319

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## Thursday, November 25, 2010

### Fox 318

Dervish Fox: I see some beauty between this one and Fox 316.

Red Fox: Would you please help me see that beauty?

Dervish Fox: The term "sin(B)+sin(C)-sin(A)" has evolved into "sin(B+C-A)" in this one. And the rest is identical.

Red Fox: And you found that beautiful?

Dervish Fox: Both answers are very different and very similar at the same time. Don't you feel anything about that?

Red Fox: I am a rational guy. I depend on my intelligence only. I can easily prove both and there is nothing magical or irrational about that. Everything's explainable. Pure and simple!

Dervish Fox: And is that all you can be? Is that the only direction you can grow? Is that who you are, or are you more than that? Why walk when you can fly? Why mutter when you can sing? Why scribble when you can paint?

Red Fox: I think I am starting to feel little bit of annoyance.

Dervish Fox: Not bad for a start my friend, not bad at all !

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## Wednesday, November 24, 2010

### Fox 317

Otherwise, some may consider 12:00:00 as "equally-separated".

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## Tuesday, November 23, 2010

### Fox 316

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## Wednesday, November 17, 2010

### Fox 315

*continuously*?

*Continuously*? Isn't that ironic?

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## Tuesday, November 16, 2010

### Fox 312 - Solutions

**Bleaug**:

As in Fox 262 (see its solution here)

**A**area (= OQR area below) extremal/constant implies P midpoint of QR. Because:

**A**constant <=> d

**A**= 0 <=> area(PRR') = area(PQQ') <=> area(P'RR') = area(P'QQ') => PR = PQ when dP tends towards 0.

Hence A = 1 = 2xy or y = 1/2x i.e. answer (D).

**Migue**:

If area is A the answer is f(x)=y=**A**/2x.

f(x) is the envelope of all straight lines that close in with axes X-Y constant area equals to **A**.

Equation of one of this lines is: x/a + y/b = 1 (a,0) in X & (0,b) in Y.

Notice that a & b are two variable parameters.Another equation is: ab/2 = A (area of triangle that straight line draws with axes).

(1) Let f(x,y,a,b)= bx+ay-ab = 0

(2) Let g(a,b)=ab-2**A** = 0

(3) f'a · g'b - f'b · g'a = 0 (Jacobian of derivatives respect to a and b parameters).

f'a is the derivative of f respect a, ...

y/b - x/a = 0 (3)

If we resolve this system of equations (1),(2) & (3) we obtain:

a = 2x & b = 2y -> ab = 4xy = 2**A** -> xy = **A**/2 (equilateral hyperbola)

and y = f(x) = **A**/2x.

Answer is D) for **A** =1.

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## Sunday, November 14, 2010

### Fox 314

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## Thursday, November 11, 2010

### Fox 310

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## Saturday, November 6, 2010

### Fox 313

*a problem which can be expressed in geometric terms and that is elegantly solved using pure algebraic arguments, still giving deep insight.*"

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## Friday, November 5, 2010

### Fox 311 - Solution

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