Fox 347

Below is the general case submitted by Roland Sampy.
There seems to be at least 2 expressions for r.
We had a solution, but it looked a little clumsy. Moreover, we lost it :)

So, do help us to recover... Oh yes, do enjoy as well...

Fox 343

x, y, and z are perpendicular to the chord.

What makes a man a man
is not the fire
nor the bullets,

but pure,
simple,
childish forgiveness.

-- Dervish Fox

Fox 342

Ahmet Arduç submitted this one. Geometric solutions will be appreciated.

http://www.8foxes.com/

Fox 339 - Solutions

Two geometric solutions are below for this fox. First by bleaug, next by Bob Ryden. Enjoy...

http://www.8foxes.com/

Fox 341

http://www.8foxes.com/

Fox 339

A general case of an old one, as promised last week... http://www.8foxes.com/

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

http://www.8foxes.com/

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!!

http://www.8foxes.com/

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.

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 288

http://www.8foxes.com/

Fox 276 - Solution

Solution to 276 was already known but figure below explains more. By Jim Wilson:

http://www.8foxes.com/

The locus of point P is the arc of a circle. Because the central angle must be twice the subtended arc in measure, the central angle is 60. Thus the major arc has length 10π/6. Since the trajectory would also be over a second arc that is a reflection in AB, the total trajectory is 10π/3.

Fox 276

http://www.8foxes.com/

Fox 269 - Solution

By Yu:

0 ≤ α ≤ 60

Large Ellipse

Semi-major axis: (1/2)(√3 + cot(α/2))
Semi-minor axis: (1/2)(√3 - tan(α/2))
Area = π*major*minor=(π/2)(√3 cotα + 1).

Small Ellipse

Semi-major axis: (1/2)(tan(α/2) + √3)
Semi-minor axis: (1/2)(cot(α/2) - √3)
Area = π*major*minor = (π/2)(√3 cotα - 1).

Ratio of their areas (large : small) = (√3 cotα + 1)/(√3 cotα - 1)
= [(√3/2)cosα + (1/2)sinα] / [(√3/2)cosα - (1/2)sinα]

= sin(60+α) / sin(60-α)

60 ≤ α ≤ 90
Large Ellipse: Area = (π/2)(1 + √3 cotα)

Small Ellipse: Area = (π/2)(1 - √3 cotα)

Ratio of their areas (large : small) = (1 + √3 cotα)/(1 - √3 cotα)
= [(1/2)sinα + (√3/2)cosα] / [(1/2)sinα - (√3/2)cosα]

= sin(α+60) / sin(α-60).

An observation:
When α=60, the ratio is undefined because the area of the small ellipse is 0. The third vertex maps out a straight line.