What is the intersection of the standard simplex with a hyperplane and a hypersphere Fix $n>2$, $0 < k \leq 1$, and $0 < m \leq 1$. Consider the following three regions in $\mathbb R^n$: 


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*$\Delta^{n-1} = \{ \vec x \in \mathbb R^n : \sum_i x_i = 1, x_i \geq 0\}.$

*$H^n_m = \{ \vec x \in \mathbb R^n: \sum_i^{n-1} x_i = m\}$

*$S^n_k = \{ \vec x \in \mathbb R^n: (1-x_1)^2 + \sum_{i>1} x_i^2 = k \}$


$\Delta^{n-1}$ is the standard $n-1$-simplex. $H^n_m$ is a hyperplane with scalar equation
$$x_1 + x_2 + x_3 + \ldots + x_{n-1} = m.$$
And $S_k$ is the hypersphere of radius $k$ centered on $(1, 0, \ldots, 0)$. 
For $n=3$, the intersection of $H^n_m$, $\Delta^{n-1}$, and $S^n_k$ contains exactly one point. 
I have two questions: 


*

*What about for $n=4$? Toying around with online equation solvers suggests it may also be true, so I'm looking for (a) a hint to show that it is true or (b) a counterexample. 

*Can we say anything in general about the intersection of $H^n_m$, $\Delta^{n-1}$, and $S^n_k$?
 A: Visualization: The hyperplane $H_m^n$ is parallel to one facet of the simplex, and a simplex is a cone; their intersection is a scaled-down copy of the facet.  The vertex of the intersection corresponding to $e_1$ is closer to $e_1$ than the rest of the vertices, so a sphere of suitable radius centred at $e_1$ will have one vertex inside and the rest outside, and so will meet the intersection at more than one point.
Computation: Let $e_i$ denote the standard basis vectors.  The set $\triangle^{n-1}\cap H_m^n$ is convex and contains the points
$$ \{ me_i + (1-m)e_n : i=1,\dotsc,n-1 \} $$
(In fact it is exactly the convex hull of these points, but we won't need that fact.)  Let $v_i = me_i + (1-m)e_n$.  We have
$$ \|v_i-e_1\| = \begin{cases}
\sqrt2(1-m) &\text{if $i=1$,} \\
\sqrt2\sqrt{(1-m)^2+m} &\text{if $i\ne 1$.}
\end{cases} $$
So if we take $k\in[0,1]$ such that
$$ \sqrt2(1-m) < k < \sqrt2\sqrt{(1-m)^2+m} $$
(and $m\in[1-\frac1{\sqrt2},1]$ so that this is possible), then by continuity the sphere will meet $\triangle^{n-1}\cap H_m^n$ at points of the form $(1-\lambda_i)v_1+\lambda_iv_i$ for each $i=2,\dotsc,n-1$, for suitable $\lambda_i\in[0,1]$.

Unpacking that last part a bit as requested:  Define $f\colon[0,1]\to\mathbb R^n$ by $f(\lambda) = \|(1-\lambda)v_1+\lambda v_i-e_1\|$.  Then
$$ f(0) = \sqrt2(1-m) < k < \sqrt2\sqrt{(1-m)^2+m} = f(1) $$
By the intermediate value theorem, there exists $\lambda_i\in[0,1]$ such that $f(\lambda_i)=k$, which means $(1-\lambda_i)v_1+\lambda_i v_i\in S_k^n$.
