Fox 340

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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:

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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:

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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 124 - Solution

Bob Ryden provided this solution to Fox 124.

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Fox 257 - Solution

http://www.8foxes.com/

See Fox 257.

Fox 264

http://www.8foxes.com/
Die no soon, before all your story are told.
--- Yoda Fox

Fox 261 - Solutions

Four good solutions for Fox 261, all drawing lines in the sand...

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Fox 263

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One more before releasing a general case.

Fox 261

The claim below makes some information in Fox 252 redundant.

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This is an intriguing construction! We can easily post half a dozen more good foxes. Keep ckeckin' !!
Also there are at least 3 purely-geometric solutions...

Fox 257

Let's build upon this one...
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Fox 256 Solution

Submitted by Takyu:

Area of kite = 2 × area of triangle AFG
= area of parallelogram ABEF = area of parallelogram ABCD
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Fox 256

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Hey you, tired of theory?

Start flying a kite.

Put aside the numbers,

Play like a child!

Fox 252 - Solution-2

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See Fox 252

Fox 252 - Solution-1

http://www.8foxes.com/
See Fox 252

Fox 252

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All along the wall, between us

I see a teacher, there for us

I look at the wall, I see right through it

I lean on the wall, there for us

Fox 248 - Solution-2

http://www.8foxes.com/

Bleaug states:
Let's take a, b, c to be the sides of A, B, C squares
By construction, the sum of the two shorter sides of D is equal to c. Same result for E.
All right triangles in the picture are similar, so are D and E.
Similar triangles with equal lengths implies D and E are identical (assertion 2)
The two shorter sides of D and E are a and b. So a+b=c (assertion 1)
Triangle D is to F as G is to E, as b is to a. So F/D=E/G=(b/a)^2. Since E=D we get F.G=E^2=D^2 (assertion 3)

Fox 248 - Solution-1

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Fox 248

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Never delay a good idea.

Once you let it go,

it may never have a chance to resurrect.

--- Himalayan Fox

Fox 242

http://www.8foxes.com/

A nice counterexample was found!