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Given three points in $\mathbb{R}^2$, $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$, and a $C^2$ surface through them $f(x,y)$, is there always a point in the interior of the triangle in formed by the three points where the tangent plane is parallel to the plane through $(x_i,y_i,f(x_i,y_i))$.

I've not seen this as a generalisation of the MVT, so I'm assuming it should be either blindingly obvious, or there should be some simple counter example - but I can't think of either myself.

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The hypothesis is false.

Proof

Define the plane interpolating $(x_i,y_i,f(x_i,y_i))$ as $L(x,y)$, as $\hat{f}=f-L$.

It is clear that the hypothesis is equivalent to asking whether given three zeros of $\hat{f}$, is there a point where $\nabla\hat{f}=0$ the interior of the triangle formed by the zeros.

Now consider $\hat{f}=1-r^2$. This has zeros on the unit circle, but it's only maxima is at the origin.

The points $(1,0)$ and $(\cos(\theta),\sin(\theta))$ are zeros of $\hat{f}$, but for small enough $\theta$ their convex hull does not contain the origin, thus there is no point in their interior where $\nabla\hat{f}=0$, providing a counter-example to the claim.

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