Kvant Math Problem 807
Consider a regular polytope in two or three dimensions, a point $M$ inside or near it, and the feet of perpendiculars from $M$ to its sides or faces.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 13m59s
Source on kvant.digital
Problem
- From an arbitrary point $M$ inside an equilateral triangle, perpendiculars $MK_1$, $MK_2$, $MK_3$ are dropped to its sides. Prove that the sum of the vectors $\overrightarrow{MK_1}+\overrightarrow{MK_2}+\overrightarrow{MK_3}$ is equal to $\dfrac32,\overrightarrow{MO}$, where $O$ is the center of the triangle.
- From an arbitrary point $M$, perpendiculars $MK_1$, $\ldots$, $MK_n$ are dropped to all sides of a regular $n$-gon (or their extensions). Prove that $$\overrightarrow{MK_1}+\ldots+\overrightarrow{MK_n}=\dfrac n2,\overrightarrow{MO},$$, where $O$ is the center of the $n$-gon.
- From an arbitrary point $M$ inside a regular tetrahedron, perpendiculars $MK_1$, $MK_2$, $MK_3$, $MK_4$ are dropped to its faces. Prove that $$\overrightarrow{MK_1}+\overrightarrow{MK_2}+\overrightarrow{MK_3}+\overrightarrow{MK_4}=\dfrac43,\overrightarrow{MO},$$, where $O$ is the center of the tetrahedron.
V. V. Prasolov
Exploration
Consider a regular polytope in two or three dimensions, a point $M$ inside or near it, and the feet of perpendiculars from $M$ to its sides or faces. The goal is to express these perpendicular vectors in a form that allows summation and then to exploit the symmetry of the figure to evaluate the sum. For each side or face, denote by $u_i$ the unit outward normal vector pointing from the center of the polytope toward that side or face, and let $r$ be the distance from the center to the side or face. The foot $K_i$ of the perpendicular from $M$ to the side or face lies along the direction $u_i$ at a distance such that $u_i\cdot K_i = r$. Hence
$$K_i=M+(r-u_i\cdot M)u_i,$$
and therefore
$$\overrightarrow{MK_i}=(r-u_i\cdot M)u_i.$$
Summing these vectors yields
$$\sum_i \overrightarrow{MK_i} = r\sum_i u_i-\sum_i (u_i\cdot M)u_i.$$
In the plane, for a regular $n$-gon, the sum of the outward normals $\sum_i u_i$ vanishes. In three dimensions, for a regular tetrahedron, the sum of outward normals also vanishes. The remaining sum is
$$\sum_i (u_i\cdot M)u_i,$$
which can be viewed as the action of a linear operator $T$ on $M$, where
$$T(x)=\sum_i (u_i\cdot x)u_i.$$
The symmetry of the figure ensures that $T$ is proportional to the identity. Its scalar multiple is determined by computing the trace of $T$, giving the desired coefficient for the sum of perpendicular vectors.
Problem Understanding
For a given regular $n$-gon or tetrahedron and a point $M$, we aim to prove that the sum of the vectors from $M$ to the feet of perpendiculars on all sides or faces is a fixed multiple of the vector from $M$ to the center $O$. For an equilateral triangle, the coefficient should be $3/2$, for a regular $n$-gon it should be $n/2$, and for a tetrahedron $4/3$. This is a Type B problem: a proof problem, not a computation or construction. The main challenge lies in handling the symmetry to justify that the sum of rank-one operators $u_i u_i^T$ acts as a scalar multiple of the identity.
Proof Architecture
Place the origin at the center $O$ of the figure. Let $u_i$ denote the unit outward normal to the $i$-th side (or face) and $r$ the distance from $O$ to that side (or face). The foot $K_i$ of the perpendicular from $M$ to the side or face satisfies
$$\overrightarrow{MK_i}=(r-u_i\cdot M)u_i.$$
Summing over all sides or faces gives
$$\sum_i \overrightarrow{MK_i} = r\sum_i u_i-\sum_i (u_i\cdot M)u_i.$$
By symmetry, $\sum_i u_i=0$, reducing the problem to showing that $\sum_i (u_i\cdot M)u_i$ equals a scalar multiple of $M$.
In two dimensions, define the operator
$$T(x)=\sum_i (u_i\cdot x)u_i.$$
Because a rotation of $2\pi/n$ permutes the normals, $T$ commutes with a nontrivial rotation. A real symmetric $2\times2$ matrix commuting with a nontrivial rotation must be a scalar multiple of the identity. Its trace is $n$, giving $T=(n/2)I$.
In three dimensions, for a tetrahedron, let
$$T(x)=\sum_i (u_i\cdot x)u_i.$$
The linear operator $T$ is invariant under the tetrahedron's rotation group. To justify that $T$ is a scalar multiple of the identity, it suffices to prove that the rotation action on $\mathbb R^3$ is irreducible. Any nonzero invariant subspace $W$ cannot have dimension $2$, because then its orthogonal complement would be a $1$-dimensional invariant subspace. Thus it is enough to rule out nonzero invariant lines.
Let $L=\mathbb R v$ be a nonzero invariant line. Every rotation of the tetrahedron sends $v$ to a vector in $L$, hence to $\pm v$. Consider a rotation $\rho$ through $120^\circ$ about an axis passing through a vertex and the center of the opposite face. The matrix of $\rho$ has eigenvalues
$$1,\qquad e^{2\pi i/3},\qquad e^{-2\pi i/3}.$$
The only real eigenvalue is $1$, so every real vector satisfying $\rho(v)=\pm v$ must satisfy $\rho(v)=v$ and must lie on the rotation axis.
There are four such $120^\circ$ rotation axes, one for each vertex. A line invariant under the whole rotation group would have to be fixed by every $120^\circ$ rotation, hence would have to lie on all four axes simultaneously. Distinct axes intersect only at the origin, so no nonzero invariant line exists. Consequently there is no nontrivial invariant subspace, and the action is irreducible.
Schur's lemma now applies and yields $T=\lambda I$. Computing the trace gives $T=(4/3)I$. Substituting into the sum formula completes the proof.
Solution
Place the origin at the center $O$ of the regular figure. Consider a regular $n$-gon. Let $u_1,\dots,u_n$ be the unit outward normals to its sides, and let $r$ be the distance from $O$ to each side. The supporting line of the $i$-th side satisfies
$$u_i\cdot x=r.$$
The foot $K_i$ of the perpendicular from $M$ to this line satisfies
$$K_i=M+t_i u_i.$$
Substituting into the line equation gives
$$u_i\cdot(M+t_i u_i)=r.$$
Since $|u_i|=1$,
$$t_i=r-u_i\cdot M.$$
Hence
$$\overrightarrow{MK_i}=(r-u_i\cdot M)u_i.$$
Summing over all sides,
$$\sum_{i=1}^n \overrightarrow{MK_i} = r\sum_{i=1}^n u_i-\sum_{i=1}^n (u_i\cdot M)u_i.$$
For a regular $n$-gon,
$$\sum_{i=1}^n u_i=0,$$
so
$$\sum_i \overrightarrow{MK_i} = -\sum_i (u_i\cdot M)u_i.$$
Define
$$T(x)=\sum_i (u_i\cdot x)u_i.$$
By symmetry, $T$ commutes with the rotation through $2\pi/n$. A symmetric $2\times2$ matrix commuting with a nontrivial rotation is a scalar multiple of the identity. Thus
$$T=\lambda I.$$
Its trace is
$$\operatorname{tr}(T) = \sum_i \operatorname{tr}(u_i u_i^T) = \sum_i |u_i|^2 = n.$$
Since $\operatorname{tr}(\lambda I)=2\lambda$,
$$2\lambda=n, \qquad \lambda=\frac n2.$$
Therefore
$$T=\frac n2,I.$$
Consequently
$$\su$$