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 316 and 318 - Solutions

Both solutions to 316 and 318 are by Bleaug. Also César Lozada has commented out a very similar solution to 316. Obviously, the maximum area is achieved when P is the mid-point, which is also the case for 321. But for Fox 322, it is not that obvious. That's why it still remains as unsolved. Thank you all for the good work.

Fox 316 Solution:
Bleaug:
"I couldn't get a pure geometric demonstration of these two problems. At best, with some lazy observations you are able to show that for both #316 and #318, maximum area is achieved for P midpoint of BC:
a) angle(QPR) in P remains constant when P varies
b) PR/PB is constant
c) PQ/PC is constantHence area(PQR) is proportional to PB.PC which is maximized when P is the midpoint of BC.
The rest is obtained by calculus."



Fox 318 Solution:

http://www.8foxes.com/

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 311 - Solution

This good geometric solution to Fox 311 is by Newzad.

http://www.8foxes.com/