Fox 345

A 2-circle covering problem would have been easier. But Let's start with this one.

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Fox 344

The answer is not confirmed.http://www.8foxes.com/

Two flow together
one rises one falls
one after another
one shines one glooms

Fox 312 - Solutions

Two solutions for Fox 312 are identified below.

Bleaug:
As in Fox 262 (see its solution here) A area (= OQR area below) extremal/constant implies P midpoint of QR. Because:
A constant <=> dA = 0 <=> area(PRR') = area(PQQ') <=> area(P'RR') = area(P'QQ') => PR = PQ when dP tends towards 0.

Hence A = 1 = 2xy or y = 1/2x i.e. answer (D).

Migue:
If area is A the answer is f(x)=y=A/2x.
f(x) is the envelope of all straight lines that close in with axes X-Y constant area equals to A.
Equation of one of this lines is: x/a + y/b = 1 (a,0) in X & (0,b) in Y.
Notice that a & b are two variable parameters.Another equation is: ab/2 = A (area of triangle that straight line draws with axes).
(1) Let f(x,y,a,b)= bx+ay-ab = 0
(2) Let g(a,b)=ab-2A = 0
(3) f'a · g'b - f'b · g'a = 0 (Jacobian of derivatives respect to a and b parameters).

f'a is the derivative of f respect a, ...

y/b - x/a = 0 (3)

If we resolve this system of equations (1),(2) & (3) we obtain:
a = 2x & b = 2y -> ab = 4xy = 2A -> xy = A/2 (equilateral hyperbola)
and y = f(x) = A/2x.

Answer is D) for A =1.

Fox 312

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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?

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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.
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Fox 262

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See 204 for a similar one.

Fox 260

An obvious extension...

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Fox 253 Solution

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Bleaug goes on:

Energy acquired in free fall from rest position is 1/2.mV2=mg(h0-1/2)

Vertical velocity after bounce is V reduced by a factor equal to cos(2a), where a is the angle of tangent to 1/x at B: Vy = V.cos(2a) and tan(a)=-1/4

Highest point reached after bounce is such that: 1/2.mVy2=mg(h-1/2), so h = 1/2 + cos(2a)2.(h0-1/2)

From simple trigonometry we get cos(2a) = 1-tan(a)2/(1+tan(a)2) = 15/17

So h = 1/2 + (15/17)2 . (h0-1/2) = 1/2 + (225/289). 289/2 = 113.

Fox 253

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The answer is now confirmed! Will post later...