Fox 15 - Solution

Well, we haven't received any comments. So here is the animated solution for Fox 15 by Bleaug.

"... Not an easy one though. I have tried to provide an animated SVG solution because packing all info into one single drawing or comment text would become clumsy. The proof is based on well-known (?) geometric properties of parabolas..."

To see the animation below, you will need any of the following browsers:
Chrome, Firefox, Opera, Safari, or IE Explorer 9.0.

http://bleaug.free.fr/8foxes/8foxes187.svg

Use mouse clicks or left-right arrows on your keyboard to browse thru the pictures.

http://www.8foxes.com/

Fox 15

This ancient fox was the general case of Fox 187, which was successfully-solved. And now, Bleaug claims that this has been solved, too. Let's see if we'll be able to confirm his claim.

http://www.8foxes.com/

Year-End Note

It is that time of the year again.

We heard a call from South,

saying, "Come - whoever you are!"
"Poor or rich"
"Full or null"
"Up or down"
"Live or dull"
"Come, just hear my cry!"

It was not the ears heard the call,
But the heart trembled with it.
How can one remain unconscious to this call?

So my friend,
never reject a good call.
Shine with your mercy
upon everyone.

Break your daily routine,

Do things out of the box.

Search the good in bad,
And never give up thinking.

--- Polar Fox

Fox 262 - Solutions

Two solutions to Fox 262

Bleaug:
OAB is minimal for any M fixed (and making AB slope vary) when M is the midpoint of AB. This is intuitively depicted in the following picture where brown and green triangles have the same area (assuming first degree differentiation approximation) which means that OAB area is extremal. This extremum is a minimum because OAB area is a continuous function and it reaches infinity when A or B reaches infinity.

This means that the point M we are looking for is such that M is midpoint of AB and AB tangent in M. This brings us to problem 260 where we already encountered point V(4/3, 8/9):
- tangent in V is parallel to UP and vector(UP) = vector(VB) because of parabola property (see Fox 260)
- vector(UP) is symmetric of vector(VO) along y axis because of parabola symmetry
- thus OVB is isoceles and V is the midpoint of AB

Thus V achieves minimal OAB area.
Area = 4/3 . 16/9 = 64/27.

By Yu:

Equation of tangent at x=a, y=-(2a-2)x+a^2.
Area of triangle = (a^4)/(4(a-1)).
When a=4/3, minimum area = 64/27.
Question: a=4/3 in Fox 260 and Fox 262. Is this a coincidence?

http://www.8foxes.com/

Fox 272 - Solution

Yu claims the following:
Without the parabolic track a particle (initially at rest) at the vertex will be in free fall and drops at maximum acceleration of g. With the parabolic track the motion is restricted by the track and the particle is no longer in free fall; it slides down the track with increasing speed. Will it leave the track at some point?

All parabolic tracks are similar (See Fox 270). Consider the track y=-x^2 which ends at the point (x,-x^2).

When a particle slides from rest at the vertex (0,0) to the point (x,-x^2), it acquires speed v=sqrt(2gx) in the direction given by tanθ=dy/dx=-2x, where θ is the angle with the x-axis.
.: cosθ = 1/sqrt(1+4x^2), sinθ=-2/sqrt(1+4x^2).

At point (x,-x^2) the particle is in free fall,
v_H=sqrt(2gx)/sqrt(1+4x^2),
v_y=-2sqrt(2gx^2)/sqrt(1+4x^2).

After a further time Δt, Δx=sqrt(2gx)Δt/sqrt(1+4x^2),
Δy=-2sqrt(2gx^2)Δt/sqrt(1+4x^2) - (1/2)g(Δt)^2
= -2xΔx-(1+(1/4x^2))(Δx)^2,
and the particle is at y=-x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.

If the track did not end at (x,-x^2) but continued to infinity,
y=-(x+Δx)^2
= -x^2-2xΔx-(Δx)^2 > -x^2 - 2xΔx -(1+(1/4x^2))(Δx)^2.
.: The particle stays on the parabolic track.

Fox 275

http://www.8foxes.com/

Fox 274

http://www.8foxes.com/

Fox 271 - Solutions

Two solutions for Fox 271.
Geometric Solution by Bleaug:

Couldn't resist a geometric approach to this one despite the obvious analytical solution...
We are looking for the osculating circle to the parabola in O. This circle is such that for M approaching O, normal in M crosses normal in O (y axis) at osculating circle's center.
Take the [y=x^2] parabola focal point F(0, 0.25). For any point M, the normal crosses the y axis in P such that FP=FM (because all blue angles are equal). Thus, the tangent in M crosses the y axis in Q such that FM=FQ=FP.
When M approaches O, so does Q so FP=FO. So osculating circle to parabola in O is such that OP=0.5 (diameter=1)

Analytic Geometry by Yu (similar submitted by Joe, Bob Ryden and Newzad):
Let r be the radius of the biggest circle that touches the parabola.
x^2 + (y-r)^2 = r^2, y=x^2. Point(s) of contact:
(0,0), ±(√(2r-1), 2r-1).
Same gradient at point(s) of contact: .: dy/dx = -(x/(y-r)) = 2x, i.e., at x=0 or y=r-(1/2).
.: 2r-1=r-(1/2), r=1/2. Hence there is only one point of contact (0,0), and the biggest circle has a diameter of 1.

http://www.8foxes.com/

Fox 272

http://www.8foxes.com/

The claim assumes that there is a positive gravity in the given direction.

So, it should hold in Pluto as well.

Himalayan fox: Here is a claim ready to be blown away with a single counterexample. How strong is your theory Polar?
Polar fox: Well, I can smell it.
Himalayan fox: It demands more than that. Are you living in a paper castle ready to be drifted away by the next wind? Or are you a real believer of something that even the strongest quake can not shake you a bit - you stand firmly on the ground?
Polar fox: No problem on my stand man! I am not shaken at all, but... the thing is... the ice is melting!

Fox 271

http://www.8foxes.com/

Fox 270

http://www.8foxes.com/

Fox 260 - Solutions

Almost Pure Geometric Solution (An Essential Property of a Parabola)
by Bleaug

Let's take as a parabola "well known" property that for any two points U, V of abscissa u, v, the tangent to point T of abscissa (u+v)/2 is parallel to vector UV. (OK! this demonstration would need some cartesian algebra but simple, promise!) From this we derive that for any point M between U and V the area of triangle UMV et less or equal to area of triangle ATV which maximizes triangle height (e.g. assume the opposite and compare area of triangle obtained from M+dM)

Then let's assume U and V achieve the maximum trapezoid area between A (x=0) and B(x=2), then necessarily Au=uv and uv=vB which implies Au=uv=vB=AB/3=2/3. Because of symmetry, maximum area is equivalent to area of rectangle AvVW = 8/9 * 4/3 = 32/27.

Geometric Translation
by Yu

Translate y=2x-x^2 to the left by 1 unit to obtain y=1-x^2.

Without going into details, the area of the trapezium is greater than the area of the quadrilateral. Area of trapezium, A = (1/2) (2x+2)(1-x^2) = (x+1)(1-x^2)
Max A = 32/27 when x = 1/3.

For more details see Fox 260.
http://www.8foxes.com/

Fox 262

http://www.8foxes.com/
See 204 for a similar one.

Fox 260

An obvious extension...

http://www.8foxes.com/

Fox 12

Yu sent a confirmed solution.http://www.8foxes.com/

Fox 11

An old fox resurrects since a new solution has been received. http://www.8foxes.com/

Fox 197

Those apes keep juglin'

http://www.8foxes.com/

Fox 193

Inspired from a question submitted by Ajit.

Fox 188

Fox 187