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What is missing in the original solution is the following observation: If $F([a,c])$ F([a,t^\ast])$does not lie a hyperplane, then we can find$\epsilon>0$such that the extremum of$F$|P\circ F|$ on any one-dimensional projection $P$ orthogonal to $F(c)-F(a)$ F(t^\ast)-F(a)$does not occur on$(c-\epsilon,c]$. (t^\ast-\epsilon,t^\ast]$. Similarly for $[c,a]$.[t^\ast,a]$. With this, you just pick$n-2$points on$(c-\epsilon,c+\epsilon)$. (t^\ast-\epsilon,t^\ast+\epsilon)$. We may assume that the space spanned by $F'$ at each of these points is $n-1$-dimensional and does not contain the vector $F(c)-F(a)$ F(t^\ast)-F(a)$(if it necessarily does, then adding one additional point makes the collection linearly dependent). Now throw in$F(c)-F(a)$F(t^\ast)-F(a)$ to obtain a hyperplane $L$ and take an orthogonal projection to obtain an extreme point on each side of $c$. t^\ast$. Its derivative is zero, and it doesn't lie in the original collection by the remark. 1 What is missing in the original solution is the following observation: If$F([a,c])$does not lie a hyperplane, then we can find$\epsilon>0$such that the extremum of$F$on any one-dimensional projection orthogonal to$F(c)-F(a)$does not occur on$(c-\epsilon,c]$. Similarly for$[c,a]$. With this, you just pick$n-2$points on$(c-\epsilon,c+\epsilon)$. We may assume that the space spanned by$F'$at each of these points is$n-1$-dimensional and does not contain the vector$F(c)-F(a)$(if it necessarily does, then adding one additional point makes the collection linearly dependent). Now throw in$F(c)-F(a)$to obtain a hyperplane$L$and take an orthogonal projection to obtain an extreme point on each side of$c\$. Its derivative is zero, and it doesn't lie in the original collection by the remark.