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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)

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I think that it isn't possible to draw such square....Can you provide drawing ? – pedja Feb 19 '12 at 18:39
that is the problem i m also facing..but i have got this problem in my geometry problem sheet and the answer is non-zero – Prateek sharma Feb 19 '12 at 18:42
Hm, 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. – msh210 Feb 19 '12 at 18:48
not in R4 ...we are talking about R2 – Prateek sharma Feb 19 '12 at 18:50
It 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? – D. Thomine Feb 19 '12 at 19:32

Time for some reverse engineering. Suppose we start with a square; translate it so that it is centred at the origin. Then for some $s$ and $\theta$, the vertices are $(s \cos(\theta), s \sin(\theta))$, $(s \sin(\theta), -s \cos(\theta))$, $(-s \cos(\theta), -s \sin(\theta))$ and $(-s \sin(\theta), s \cos(\theta))$, and the area is $2 s^2$. The cubic that interpolates these points is $$f(x) = \frac{4}{s^2 \sin(4 \theta)} x^3 - \frac{3 + \cos(4 \theta)}{\sin(4 \theta)} x$$ Note that for $b \ge \sqrt{8}$, $b = (3 + \cos(4 \theta))/\sin(4 \theta)$ has real solutions, namely $$\sin(4 \theta) = \frac{3b \pm \sqrt{b^2-8}}{b^2 - 1}$$ Note also that these solutions have $\sin(4 \theta) > 0$. We can then take $s=2/\sqrt{\sin(4\theta)}$ to make the cubic $f(x) = x^3 - b x$, and the area is $$2s^2 = \frac{8}{\sin(4\theta)} = 3 b \mp \sqrt{b^2 - 8}$$ The given problem has $b = 235$ and area $705 \mp \sqrt{55217}$.

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If you take D. Thomine's hint, and let $X=x+9$ and $Y=y-1477$ to centre the cubic at $(0,0)$ as $Y=X^3-235X$. So if $(X_0,Y_0)$ are vertices of the square centred on the origin then so too are $(Y_0,-X_0)$, $(-X_0,-Y_0)$ $(-Y_0,X_0)$.

So you need to solve $(X^3-235X)^3-235(X^3-235X)=-X$. There are nine solutions including $X=0$; the others come in two sets of four for the two possible squares.

One square has a corner near $(0.065234579,-15.32984851)$ and just rotate by one, two or three right angles to find the others. The area is just over $470$.

The other square has a corner near $(15.29691915,-15.36215329)$ and again rotate to find the others. The area is just under $940$.

To find the corners of the original squares on the original cubics, just add $(-9, 1477)$ to the coordinates of the four corners of the squares, but you do not ned to do this to find the areas.

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Note: This answers an earlier version of the question.

It is impossible to draw a square of positive area in $\mathbb R^2$ all of whose points lie on the graph of $y=x^3+27x^2+8x+91$. Proof: Any side of the square will have to lie on that graph, so that an interval will have to lie on the graph, but an interval has zero second derivative whereas $\frac{d^2}{dx^2}x^3+27x^2+8x+91=6x+54$ which is not zero on an interval.

Thus, if you're seeking the area of a square drawn so its sides are all on that curve, the answer is $0$.

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only the end points lie on the graph and not necessarily the edges – Prateek sharma Feb 19 '12 at 19:00
@Prateeksharma, then you should indicate as much in the question. – msh210 Feb 19 '12 at 19:00
@msh210 Isn't that obvious though? I mean, of course a square cannot be drawn on a non-linear curve so that all points of the square lie on the curve. – Quinn Culver Feb 20 '12 at 2:35
@QuinnCulver, only because of the proof in my answer (or similar). – msh210 Feb 28 '12 at 19:52
@msh210 It's obvious because a square isn't a function. But notice that the OP has edited the question to require only that the vertices lie on the curve. – Quinn Culver Feb 29 '12 at 3:20

Suppose the points $V_i = (x_i, f(x_i))$, $i=1,2,3,4$, form the vertices of a square. Since there is only one interval in which $f' < 0$, if $x_1 < x_2 < x_3 < x_4$ the vertices in counterclockwise order will be $V_1, V_3, V_4, V_2$. For these to form a square we need $V_4 - V_2 = V_3 - V_1$, $\|V_3 - V_1\|^2 = \|V_2 - V_1\|^2$, and $(V_3 - V_1) \cdot (V_2 - V_1) = 0$. Solving this system of four equations in four unknowns using Maple, besides the trivial solution $x_1=x_2=x_3=x_4$ I get the following real solutions:

$$[-24.36215898, 1461.700388], [-24.29691914, 1492.362157],[6.29691914, 1461.637842], [6.362158975, 1492.299610],\ \text{ area } 705+ \sqrt{55217}$$

$$[-24.32984851, 1476.934762], [-9.065234, 1492.329713], [-8.934766, 1461.670288], [6.32984851, 1477.065233],\, \text{ area } 705- \sqrt{55217}$$

The decimal numbers are approximations to complicated expressions involving radicals, but the areas are exact. Offhand I don't see why the area should come out so neatly.

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I get areas half yours, suggesting one of us has miscalculated – Henry Feb 19 '12 at 21:18
Strange. I don't know where that factor of $2$ crept in. I'll edit my answer. – Robert Israel Feb 20 '12 at 1:13

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