A square is drawn on the curve $y = x^3 + 27\cdot x^2 + 8\cdot x + 91$. What would be the area of the square. All the points of the square should lie on the curve. All four points lie on the curve (not the sides)
Square on the curve
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geometry
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0I think that it isn't possible to draw such square....Can you provide drawing ? – 2012-02-19
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0that is the problem i m also facing..but i have got this problem in my geometry problem sheet and the answer is non-zero – 2012-02-19
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0Hm, you can fit a square in the graph in $\mathbb R^4$ of $y=x^3+27x^2+8x+91$, but presumably that's not what the problem-sheet author seeks. – 2012-02-19
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0not in R4 ...we are talking about R2 – 2012-02-19
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1It is the same thing as looking for a square whose vertices lie on the graph $Y=X^3-235X$. I would look for square whose center is at $X=0$, $Y=0$. There seems to be a solution, but the computations are quite cumbersome; what answer is provided? – 2012-02-19
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0I would encourage the PO to write a more accurate description since it is not possible that "all four points line on the curve". – 2012-02-20