Fox 348

We'll return back to Sampy's question later. But, for a change let's see a simple one - or at least it looks simple. We're looking for the shortest proof of the claim below. May be purely-geometric one? Or is the claim wrong? Let us know...

Fox 338

This is a little related to the recent ones, but much simpler. Here we'd appreciate a pure geometric solution if one found.

http://www.8foxes.com/

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.

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!

http://www.8foxes.com/

Fox 297 - Solutions

by Giannno:

by Bleaug:

Start from an arbitrary AOC triangle such that OC > OA.
Set length unit as (OC-OA)/2 and OA as x.

Reasoning is identical to Fox 296. See its solution.

Polar Fox adds:

By definition OB=x+1 => OBH is an isosceles triangle (1).
Also known that, m(BHO)=90 degrees (2).

Nope, aint gonna happen.

i.e. (1) and (2) can't happen at the same time.

http://www.8foxes.com/

Fox 297

Let's mess around little bit more with this:

http://www.8foxes.com/

Fox 296 - Solutions

Among similar ones, Bleaug finds a contradiction for Fox 296:Take an arbitrary AOC triangle such that OC=3OA. Build B as intersection of triangle's circumscribed circle and AOC internal bisector. B is such that ABC is isoceles, i.e. AB=BC. Symmetry in OABI implies AB=BI, hence IBC is isoceles. Orthogonal projection of B on OC coincides with H such that IH= HC=OA. OHB is rectangle, therefore
OB^2 = (2OA)^2 + BH^2 => OB > 2OA iff angle(AOC) > 0.

Strictly speaking, if angle(AOC)=0 there is a solution if we admit that a straight line is a circle with center at infinity (e.g. in projective plane).

And Giannno solves too:

http://www.8foxes.com/

Fox 296

This seems to defy the logic. One may ask why not. Looks perfectly possible.
So the claim may very well be wrong.

If you believe so, you may try to find a counter-example.

http://www.8foxes.com/

Fox 179