Let f(x)=2x^2 and g(x)=36-x^2. Let R be the closed region between them. Draw a rectangle ABCD such that A and B are points on f(x), and C and D are points on g(x), and the rectangle is not parallel to the axis, or prove that to be impossible.

Fortura7i

Fortura7i

Open question

2022-08-21

Rectangle inscribed on two parabolas but not parallel to the axis
Let f ( x ) = 2 x 2 and g ( x ) = 36 x 2 . Let R be the closed region between them. Draw a rectangle ABCD such that A and B are points on f(x), and C and D are points on g(x), and the rectangle is not parallel to the axis, or prove that to be impossible.

Answer & Explanation

betoosolis7i

betoosolis7i

Beginner2022-08-22Added 12 answers

Step 1
Brute force:
First, we relax the condition to finding any parallelogram with 2 vertices on each equation.
Let A = ( a , 2 a 2 ) , B = ( a + b , 2 a 2 + 4 a b + 2 b 2 ) , D = ( d , 36 d 2 )
which makes C = ( d + b , 36 ( d + b ) 2 = 36 d 2 + 4 a b + 2 b 2 ). This gives b = 0 (rejected), d = 2 a 3 b 2 .
Now, we to make the parallelogram into a rectangle, we require A B A D ( b , 4 a b + 2 b 2 ) ( d a , 36 d 2 2 a 2 ) = 0. This simplifies to 3 / 2 b ( 2 a + b ) ( 8 a 2 + 8 a b + 3 b 2 47 ) = 0.
Step 2
Note that b = 0 is rejected, and 2 a + b means AB is parallel to the x-axis, hence rejected. So, we're looking for solutions to 8 a 2 + 8 a b + 3 b 2 47 = 0, which is an ellipse.
As it turns out, a = 3 , b = 4 23 3 is one such solution, which leads to the rectangle.
A = ( 3 , 18 ) , B ( 1.77 , 6.26 ) , C ( 5.38 , 7.01 ) , D ( 4.15 , 18.75 )
Of course, if you don't want A on the intersection point, take a = 3 + ϵ

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