**Two solutions to Fox 262**

**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.**

__Bleaug:__

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?

Bleaug's

ReplyDelete"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)"

answers Yu's question.

-binary

Bleaug's statement:

ReplyDelete"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)"

answers Yu's question.

-binary

I tend to agree: 260 and 262 geometric demonstrations still hold for any parabola of equation y=ax(b-x). In both problems, point V(2b/3, 2ab^2/3) appears and plays the same role. Generalized-262 area = 2.(Generalized-260 area) = 8ab^3/27

ReplyDeleteBleaug