Fox 339 - Solutions

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

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Fox 330 - Solutions

Below are three solutions to Fox 330.

By Six:




By Bleaug:

By Henk Reuling:

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Fox 320 - Solutions

This fox has been discussed extensively. See here...

1. Calculus by Bleaug

He says: "I spotted this problem in a Paul Halmos book "Problems for mathematicians, young and old, 1991", actually in a French translation. He provided the solution below (reformulated by myself) as being proposed by Hugh Montgomery in 1985 in some Math Conference."






2. Checkerboard Solution by Rochberg and Stein:

1. Call sub rectangles as TILEs.


2. Start from the lower left corner of the overall rectangle (let's call this point as the origin)Draw horizontal and vertical lines separated by 1/2=0.5 units starting from the origin.


3. This will create 0.5 x 0.5 squares (and possibly rectangles around 2 edges -top and/or RHS)


4. Color the first square (that has the origin as one of its corners) as black the next one as white, and so on to generate a checkerboard-look.


5. Each TILE will have equal areas of black and white (Why?)


6. Therefore the overall rectangle will have equal areas of black and white.


7. So overall rectangle has at least one integer side.
   

H means the horizontal side is integer, and V means that the vertical side is integer. a1 is the square at the origin. There could have been a smarter combination, but this simple one illustrates the process clearly. This looks like the closest geometric solution we can get -at this time.

When there's hardly no day nor hardly no night
There's things half in shadow and halfway in light

3. Induction by Robinson

1. Assume that each H-tile has a width of 1, and each V-tile has a height of 1. Note that any rectangle can be converted this way without distorting the original problem. (This may increase the number of tiles significantly though)


2. Chose any H-tile, say T(0). (If there is no H-tile, then the result is immediate)


3. If there are H-tiles whose lower border shares a segment with T(0)'s upper border, choose one and call it T(1).


4. Otherwise only V-tiles share this border. In this case, we can expand T(0) upward 1 unit. This does not increase the number of H-tiles. Also, the cut V-tiles still have height 1. (They are still V-tiles)


5. Continue expanding T(0) until either the top of the rectangle is reached or a choice of an adjacent H-tile T(1) is possible.


6. Then repeat the same process from T(1). (Continue upward similarly from T(1) to get T(2), and so on...)


7. This will result in a chain of T(0), T(1), T(2), ... , T(m).


8. Starting from T(0) again, work downward similarly to obtain a bigger chain:
   T(-n), T(1-n), ... , T(0), T(1), ... , T(m-1), T(m) of H-tiles stretching from bottom to top.


9. Remove these tiles and slide the rest together to get a rectangle with fewer H-tiles.


10. Induction applied to this smaller rectangle yields the result for the original rectangle.
    

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Fox 319 - Solutions

Below are 2 distinct solutions for Fox 319...

A solution based on symmetry by Bleaug:


Trigonometry and calculus by Six:

x + y + z = 360 and sinx*siny*sinz

-> f(x,y) = sin(x)*sin(y)*sin(360-x-y)
-> f(x,y) = sin(x)*sin(y)*-sin(x+y)

Now to find the critical points of f(x,y), we just need to find the partial derivative with respect to x and y and solve for 0.

f_x(x,y) =

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)

solve for 0.

sin(x)sin(y)(-cos(x+y)-cos(x)sin(y)sin(x+y)=0

-> tan(x) = -tan(x+y)

Since the function is symmetric, we should get the same partial derivative for y.

-> tan(y) = -tan(y+x)

-> tan(x)=tan(y)

-> x = y or they are opposites. However, if they are opposites, the original function just becomes 0. Thus, x = y.

Now substitute in x for y in the original equation and find its critical points.

Eventually, you will get sin(3x)=0

x = 120 degrees
y = 120 degrees
z = 120 degrees

Answer (D)

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Fox 309 - Solutions

Two solutions to Fox 309...

Bleaug:


Bob Ryden:

Fox 308 - Solution

Bleaug says "a logical follow-up indeed", and solves:

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Fox 307 - Solutions

Two solutions to Fox 307...

by Bleaug:


by Bob Ryden:

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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!

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!

Fox 300 - Solutions

Bleaug:

Newzad:

Giannno:

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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 182 - Solution

Here are two solutions to Fox 182 by Giannno.

Analytic Geometry:

Pythagoras and quadratics:

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Fox 298 - Solutions

Solutions to Fox 298.

by Giannno:
by Newzad:

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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 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 295

Let the one who lives, live with an evidence,

Let the one who dies, perish with an evidence.

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Fox 293 - Solutions

According to the order of submission...
by Bob Ryden: (Ajit commented a very similar one)


by Bleaug:

Two tangents to a given circle define a symmetric 'kite' figure. By construction, the two kites defined by OUV and OTU are isometric because they have identical short legs. Since angles in T, U, V are right angles, blue and green angles are supplemental. Hence, angle a is angle(OUV). From Thales theorem we get cos(a) = 1/3

by Yu:
Several steps remain implicit in the figure. More words may be needed for a formal proof. But the solution holds.


by Binary Descartes:

Common tangents from the same point have equal lengths.
See the two identical deltoids (kites) sharing the same side.
Then cos(a) = e/3e = 1/3.

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Fox 124 - Solution

Bob Ryden provided this solution to Fox 124.

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Fox 285 - Solution

Bleaug constructs:
Here is a geometric construction (i.e. compass and ruler) of an inscribed triangle with required property for any angle from 0 to 40°:
1- take any point b on AB and build point c such as angle(Abc)=2α
2- build a inside ABC such as abc is equilateral
3- build O such that angle(Oca)=angle(Oba)=α
4- build P intersection of AO and BC
5- build PQR homothetic to Obc

PQR is such that angle(BPQ)=α, angle (AQR)=2α, angle(CRP)=3α

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Fox 281 - Solution

This solution to 281 was submitted by Bob Ryden. It illustrates Joe's solution as well.

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