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While solving exam questions, I came across this problem:

Let $M = \{(x, y, z) \in \mathbb{R^3}; (x - 5)^5 + y^6 + z^{2010} = 1\}$. Show that for every unit vector, $v \in \mathbb{R^3}$, exists a single vertex $p \in M$ such that $N(p) = v$, where $N(p)$ is the outer surface normal to $M$ at $p$.

I don't know where to start. Ideas?

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Are you sure you copied the question correctly? This surface intersects every line parallel to the $x$ axis. I suspect the $x-5$ should have an even exponent. – Robert Israel Jul 17 '12 at 18:00
up vote 2 down vote accepted

It is straightforward, but tedious, to analytically show that the assertion is false. A picture illustrates the problem more clearly.

First, note that if $z=0$, then the normal will also have a zero $z$ component, so we can take $z=0$ and look for issues there. Plotting the contour of $(x-5)^5+y^6 = 1$ gives the picture below: enter image description here

Pick a normal $v$ to the surface around $(2,2.5)$. Then note that in the vicinity of $(6,1)$, it is possible to find a normal that matches $v$.

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