Sunday, December 19, 2010

Year-End Note

It is that time of the year again.

We heard a call from South,

saying, "Come - whoever you are!"
"Poor or rich"
"Full or null"
"Up or down"
"Live or dull"
"Come, just hear my cry!"

It was not the ears heard the call,
But the heart trembled with it.
How can one remain unconscious to this call?

So my friend,
never reject a good call.
Shine with your mercy
upon everyone.

Break your daily routine,

Do things out of the box.

Search the good in bad,
And never give up thinking.

--- Polar Fox

Sunday, December 12, 2010

Fox 323

Polar Fox:
Hey Red, these points seem to have identity issues. Are they black or white?

Red Fox: An obvious illusion! They are white for sure.

Polar Fox: So what I perceive is not the exact reality?

Red Fox: This is just a simple game of colors playing on your retina. It is perfectly explainable and part of the reality.

Polar Fox: Yes, most definitely. But, can this be an artful clue of an even bigger reality that there is truth beyond what you see?

Red Fox: Here we go again!

Polar Fox: A brush mark from a picture? A letter from a book? A note from a symphony?

Red Fox: And who is the artist, juggling with colors to deceive our perception?

Polar Fox: Deceive? I am not so sure. But looks like someone who knows us inside and out. And I think he wants to talk.

Red Fox: Why me? Why me?

Polar Fox: Yep, and you have bigger ears than mine.

Friday, December 10, 2010

Fox 316 and 318 - Solutions

Both solutions to 316 and 318 are by Bleaug. Also César Lozada has commented out a very similar solution to 316. Obviously, the maximum area is achieved when P is the mid-point, which is also the case for 321. But for Fox 322, it is not that obvious. That's why it still remains as unsolved. Thank you all for the good work.

Fox 316 Solution:
"I couldn't get a pure geometric demonstration of these two problems. At best, with some lazy observations you are able to show that for both #316 and #318, maximum area is achieved for P midpoint of BC:
a) angle(QPR) in P remains constant when P varies
b) PR/PB is constant
c) PQ/PC is constantHence area(PQR) is proportional to PB.PC which is maximized when P is the midpoint of BC.
The rest is obtained by calculus."

Fox 318 Solution:

Monday, December 6, 2010

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 !

Wednesday, November 24, 2010

Fox 317

Based on our earlier analysis, the sample picture below will never happen.
By "equal-separation" we mean 120-120-120 split.
Otherwise, some may consider 12:00:00 as "equally-separated".
Peace out!

Wednesday, November 17, 2010

Fox 315

Polar Fox: I have a problem with this one!

Red Fox: What's wrong with it?

Polar Fox: Nothing moves continuously?

Red Fox: Come again?!

Polar Fox: The universe is discrete! THERE IS NO CONTINUUM.

Red Fox: Then how do the things move, flow, or slide?

Polar Fox: Nothing moves! Absolutely nothing moves!!

Red Fox: But the time ticks away, no? Clock arms DO advance.

Polar Fox: No, they don't! But they die in one instance and resurrect in the next one. In between no measurable time passes. Matter oscillates between existence and non-existence continuously.

Red Fox: Continuously? Isn't that ironic?

Polar Fox: Between any two existence, there is nothing but emptiness.

Red Fox: I am having the feeling that your intelligence fall into non-existing state just now.

Polar Fox: Hold on. I think I am about to jump back into the reality. Wait a sec.

Red Fox: Take your time dude. Just take your time!

Tuesday, November 16, 2010

Fox 312 - Solutions

Two solutions for Fox 312 are identified below.

As in Fox 262 (see its solution here) A area (= OQR area below) extremal/constant implies P midpoint of QR. Because:
A constant <=> dA = 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).

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-2A = 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 = 2A -> xy = A/2 (equilateral hyperbola)
and y = f(x) = A/2x.

Answer is D) for A =1.

Sunday, November 14, 2010

Fox 314

Bleaug adds new meaning to "laziness". Hmmm, do your lazy tricks lazy scientists to solve this one, huh? Although not confirmed, the similarity of the terms with Fox 3 is striking. Enjoy...

Saturday, November 6, 2010

Fox 313

Bleaug submitted the following fox, saying: "a problem which can be expressed in geometric terms and that is elegantly solved using pure algebraic arguments, still giving deep insight."
General case of this problem can be found in literature (with at least 14 proofs :)
But let's try this smaller version here.

Thursday, October 28, 2010

Fox 311

A natural extension of Fox 3.
The claim may inspire purely-geometric solutions, but looks hard.

Wednesday, October 27, 2010

Sunday, October 24, 2010

Thursday, October 14, 2010

Sunday, October 3, 2010

Tuesday, September 21, 2010

Fox 306 - Solution

To see the animation below, you will need any of the following browsers:
Chrome, Firefox, Opera, Safari, or IE Explorer 9.0.

Bleaug converts Fox 306 to Fox 302 by "folding" twice. I have seen translation, rotation, similarity, etc., but haven't seen anything like this before. Enjoy and please do respect to human intelligence!

Thursday, September 16, 2010

Fox 307

This is related to Proposition 4 in Book of Lemmas by Archimedes.

Tuesday, September 7, 2010

Fox 302 - Solution

To see the animation below, you will need any of the following browsers:
Chrome, Firefox, Opera, Safari, or IE Explorer 9.0.

Bleaug teases the human intelligence, saying:
"I know that analytic solution to this Fox 302 is obvious (although cumbersome). Here is an alternative geometric demonstration. Use Home/PageUp/PageDown/End keys/ or mouse wheel to go though demo steps."
Do enjoy!

Friday, August 6, 2010

Fox 305

Yet again,
try to see the goodness,
see the beauty,
surrounding you.
Forget about the numbers, summations, subscripts.
Leave behind the accounts, stocks, papers, statistics.
Just leave yourselves to the arms of an ocean,
full of love and compassion.
Drift away with the blowing wind...
-- Dervish Fox

Monday, August 2, 2010

Fox 54 - Discussion

Bob Ryden provided a solution below. This has been the first attempt to solve this fox. His solution has not been confirmed yet, but it is published here to start the discussion. Wanna say somethin', comment it out!
Let the radius of the spheres = 1.
Inside is a tetrahedron whose vertices are the centers
of the spheres. Its edge length = 2
and its volume = (1/8) √3

On each face of the tetrahedron, build a triangular prism.
Volume of each prism = (1/2) 2 (√3) 1
Total volume of four prisms = 4√3

On each edge of the tetrahedron, build a cylindrical sector.
The angle of the sector = 360 – 90 – 90 – dihedral angle of the tetrahedron
= 360 – 90 – 90 – arccos (1/3) = approx. 109.47°
Length of each cylindrical sector = 2, radius = 1
Total volume of the six sectors
= 6 π (1^2) 2 (109.47 / 360) ≈ 11.46
Finally, there are pieces of the four original spheres that are not covered by any of the above. The four pieces together make one complete sphere, V = (4/3)π.

Total volume is the sum
tetrahedron + 4 prisms + 6 cylindrical sectors + sphere
= approx. 23.52

Friday, July 30, 2010

Fox 304

Note, the statement should hold in 3-D as well. Don't you agree?
And YES, "plane" is misspelled !
No, no annoying philosophical stand here.
Just pure simple layyziness.

Thursday, July 22, 2010

Fox 303

Fox Weber: Why are you doing this? What about the human civilization. What happened to the progress they made on internal combustion engine?
Karl Fox: And how about the resources. They were unlimited, right? Those bourgeois deserve worse than this.
Fox Weber: But there are thousands of people making a living on this: honest, hardworking people.
Karl Fox: Oh, you must be talking about the “small people”.
Fox Weber: You’re a pity old man, lost the cause, and now doodling fancy things. See we lost another potential sponsor, thanks to you Karli!
Karl Fox: Don't worry about that. I am working on Chávez as a sponsor. A barrel of oil is still paying pretty good you know.
Fox Weber: Dude, you're hopeless!

Tuesday, July 20, 2010

Fox 300

We have been delaying this for a while, but Ajit and Vihaan have already solved it. So there is no point of keeping it a secret :)
Note that this may be a general case for Foxes: 296, 297, and 298. It is also related to Fox 301.
It would be great if there are purely geometric solutions!!

Saturday, July 17, 2010

Fox 160 - Solutions

Three distinct solutions were found for Fox 160.
The solution commented out by Julian deserves to be presented here. We often had questions inspired from the "things" we observe around us. Julian's solution is the other way. It uses a very casual observation from real-life for the solution. That's perspective lines converge linearly. See the solution below:
by Julian: This one has a simple, intuitive proof. Imagine a long, hollow square-prism. Something like a cardboard tube with a square cross-section. Imagine lines connecting the diagonally-opposite corners of the opposite ends. Due to the symmetry of the tube these four lines will meet at a single point half-way along it.

Now imagine looking through the tube from one end to the other. You see two squares: a large square at the near end, and a smaller square at the far end. By subtly changing the direction the tube is pointing, the far square may move off centre so that the construction looks like the problem diagram. This will not alter the fact that the four lines meet at a single point. QED!

Analytic Geometry by Bob Ryden:

Observe that proving DH implies CG as well (due to rotation). Neat!

Similarity of Triangles by Giannno:

Wednesday, July 14, 2010