Fox 322

This has not been solved yet. What'd you think?

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Fox 300 - Solutions

Bleaug:

Newzad:

Giannno:

http://www.8foxes.com/

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!!

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Fox 301

http://www.8foxes.com/

Fox 298 - Solutions

Solutions to Fox 298.

by Giannno:
by Newzad:

http://www.8foxes.com/

Fox 298

Proving this one immediately solves Foxie 296 and 297.
That's why this is the general case.

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.

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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.

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Fox 280

http://www.8foxes.com/

Fox 279

http://www.8foxes.com/