Kvant Math Problem 529
A homothety with ratio $k<0$ reverses directions through its center.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m17s
Source on kvant.digital
Problem
- The polygon $M'$ is the image of the convex polygon $M$ under a homothety with ratio $k=-\dfrac12$. Prove that there exists a translation $T$ such that $T(M')\subset M$.
- For which homothety ratios $k\lt0$ does an analogous statement hold for a convex polyhedron $M$ in space?
N. Yu. Netsvetaev
Exploration
A homothety with ratio $k<0$ reverses directions through its center. If the center is taken as the origin, the image of a point $x$ is $kx$.
For the planar problem, $k=-\frac12$. We seek a vector $t$ such that
$$kM+t\subset M.$$
Since $k=-\frac12$, this becomes
$$-\frac12 M+t\subset M.$$
A natural language for inclusions of convex bodies is that of support functions. If $h_M(u)=\max_{x\in M}\langle u,x\rangle$, then
$$h_{aM+b}(u)=ah_M(u)+\langle u,b\rangle \qquad (a\ge0),$$
and
$$h_{-M}(u)=h_M(-u).$$
Hence
$$h_{-\frac12M+t}(u) = \frac12 h_M(-u)+\langle u,t\rangle .$$
The inclusion is equivalent to
$$\frac12 h_M(-u)+\langle u,t\rangle\le h_M(u) \quad\text{for all }u.$$
This is a family of linear inequalities in $t$:
$$\langle u,t\rangle \le h_M(u)-\frac12 h_M(-u).$$
A standard criterion says that such inequalities are solvable iff for every finite collection $u_i$ and nonnegative coefficients $\lambda_i$ with
$$\sum \lambda_i u_i=0,$$
one has
$$\sum \lambda_i!\left(h_M(u_i)-\frac12 h_M(-u_i)\right)\ge0.$$
The crucial point is to prove this condition. In the plane, if $\sum\lambda_i u_i=0$, then after normalizing we obtain a weighted set of directions on the circle whose centroid is at the origin. Since the space is two-dimensional, Carathéodory's theorem reduces the verification to triples of directions. Thus it suffices to treat
$$\lambda_1u_1+\lambda_2u_2+\lambda_3u_3=0, \qquad \lambda_i>0.$$
The three vectors surround the origin. Using subadditivity of the support function,
$$\lambda_1h(u_1) = h(\lambda_1u_1) = h(-\lambda_2u_2-\lambda_3u_3) \le \lambda_2h(-u_2)+\lambda_3h(-u_3).$$
Cyclic permutations give three inequalities. Summing them yields
$$\sum \lambda_i h(u_i) \le 2\sum \lambda_i h(-u_i).$$
Hence
$$\sum \lambda_i!\left(h(u_i)-\frac12 h(-u_i)\right)\ge0.$$
This proves existence in the plane.
For a polyhedron in space, the same argument works in dimension $3$ provided
$$|k|\le\frac13.$$
Indeed, with $k=-a$, $a>0$, the solvability condition becomes
$$\sum\lambda_i h(u_i)-a\sum\lambda_i h(-u_i)\ge0.$$
Carathéodory reduces matters to at most four vectors. If
$$\sum_{i=1}^4\lambda_i u_i=0,$$
then
$$\lambda_i h(u_i) \le \sum_{j\ne i}\lambda_j h(-u_j).$$
Summing over $i$ gives
$$\sum\lambda_i h(u_i) \le 3\sum\lambda_i h(-u_i).$$
Therefore the required inequality follows whenever $a\le\frac13$.
To see necessity, consider a tetrahedron. Let $u_i$ be the outer normals to its four faces, weighted by face areas, so that
$$\sum \lambda_i u_i=0.$$
For a tetrahedron the support function can be normalized so that
$$h(u_i)=1,\qquad h(-u_i)=0.$$
Then the solvability condition becomes
$$\sum\lambda_i-a\sum0\ge0,$$
which is violated for $a>\frac13$ after applying the dual inequality corresponding to the four normals. Thus $\frac13$ is the sharp threshold.
The critical issue is establishing the support-function criterion and extracting the sharp constant $1/3$ from the four-vector relation in $\mathbb R^3$.
Problem Understanding
We are given a convex polygon $M$ and its image $M'$ under a homothety with ratio $k=-\frac12$. We must prove that some translation of $M'$ lies entirely inside $M$.
Then we must determine all negative homothety ratios $k$ for which the analogous statement is true for every convex polyhedron in three-dimensional space.
This is a Type A problem. We must determine exactly the set of admissible ratios.
The core difficulty is converting the geometric inclusion problem into a system of inequalities and finding the largest negative scaling factor that always permits a translation. In dimension $2$ the required factor is fixed, while in dimension $3$ one must identify the sharp threshold.
The answer is that every convex polygon admits such a translation for $k=-\frac12$, and for convex polyhedra in space the statement holds exactly for
$$-\frac13\le k<0.$$
Proof Architecture
Lemma 1. For a convex body $M$ and vector $t$, the inclusion $-aM+t\subset M$ is equivalent to
$$\langle u,t\rangle\le h_M(u)-a,h_M(-u)$$
for every vector $u$.
The proof uses support functions and the characterization of inclusion by comparison of support functions.
Lemma 2. A system of inequalities
$$\langle u,t\rangle\le c(u)$$
has a solution $t$ if and only if for every finite family $\lambda_i\ge0$ with $\sum\lambda_i u_i=0$,
$$\sum\lambda_i c(u_i)\ge0.$$
This is the classical compatibility criterion for linear inequalities.
Lemma 3. In dimension $d$, if $\sum\lambda_i u_i=0$ with $\lambda_i>0$, then
$$\sum\lambda_i h(u_i)\le d\sum\lambda_i h(-u_i).$$
The proof uses Carathéodory's theorem to reduce to at most $d+1$ vectors and then applies subadditivity of the support function.
Lemma 4. Consequently, $-aM+t\subset M$ is always possible in $\mathbb R^d$ whenever
$$a\le\frac1d.$$
The proof combines Lemmas 2 and 3.
Lemma 5. The constant $\frac1d$ is sharp. For a simplex in $\mathbb R^d$, the statement fails for every $a>\frac1d$.
The proof uses the facet normals of a simplex.
The hardest direction is the proof of sharpness in dimension $3$.
Solution
Place the center of the homothety at the origin. Then the image of $M$ under a homothety with ratio $k=-a$ is $-aM$, where $a=-k>0$.
Let $h=h_M$ be the support function of $M$:
$$h(u)=\max_{x\in M}\langle u,x\rangle .$$
For any vector $t$,
$$h_{-aM+t}(u)=a,h(-u)+\langle u,t\rangle .$$
Hence
$$-aM+t\subset M$$
is equivalent to
$$a,h(-u)+\langle u,t\rangle\le h(u) \qquad\text{for all }u.$$
Thus we seek a vector $t$ satisfying
$$\langle u,t\rangle\le h(u)-a,h(-u) \qquad\text{for all }u.$$
By the compatibility criterion for systems of linear inequalities, such a vector exists if and only if for every finite family of vectors $u_i$ and coefficients $\lambda_i\ge0$ satisfying
$$\sum_i\lambda_i u_i=0,$$
one has
$$\sum_i\lambda_i\bigl(h(u_i)-a,h(-u_i)\bigr)\ge0. \tag{1}$$
We first establish a general estimate in dimension $d$.
Assume
$$\sum_i\lambda_i u_i=0, \qquad \lambda_i>0.$$
By Carathéodory's theorem, it suffices to consider at most $d+1$ vectors. Write
$$\lambda_i u_i=-\sum_{j\ne i}\lambda_j u_j .$$
Using positive homogeneity and subadditivity of the support function,
$$\lambda_i h(u_i) = h(\lambda_i u_i) = h!\left(-\sum_{j\ne i}\lambda_j u_j\right) \le \sum_{j\ne i}\lambda_j h(-u_j).$$
Summing over all $i$ gives
$$\sum_i\lambda_i h(u_i) \le d\sum_i\lambda_i h(-u_i), \tag{2}$$
because each term $\lambda_i h(-u_i)$ appears exactly $d$ times on the right.
Suppose now that
$$a\le\frac1d.$$
From (2),
$$\sum_i\lambda_i h(u_i) - a\sum_i\lambda_i h(-u_i) \ge \sum_i\lambda_i h(u_i) - \frac1d,d\sum_i\lambda_i h(-u_i) \ge0.$$
Thus condition (1) holds, and a suitable translation exists.
For polygons, $d=2$. Taking $a=\frac12$ yields the desired statement:
$$-\frac12 M+t\subset M$$
for some translation vector $t$.
We now turn to polyhedra in $\mathbb R^3$. Here $d=3$, so the argument shows that for every convex polyhedron and every
$$a\le\frac13,$$
there exists a translation $t$ such that
$$-aM+t\subset M.$$
Equivalently, the statement holds for
$$-\frac13\le k<0.$$
It remains to prove that this bound is sharp.
Let $M$ be a tetrahedron. Denote by $u_1,u_2,u_3,u_4$ the outer unit normals to its four faces and by $\lambda_i$ the corresponding face areas. The equilibrium relation for a tetrahedron is
$$\sum_{i=1}^4\lambda_i u_i=0.$$
Translate the tetrahedron so that each supporting plane corresponding to $u_i$ has equation
$$\langle u_i,x\rangle=1.$$
Then
$$h(u_i)=1.$$
Since the opposite vertex lies on the side where $\langle u_i,x\rangle<1$, one obtains
$$h(-u_i)=\frac13.$$
Substituting these values into condition (1) gives
$$\sum_i\lambda_i\left(1-\frac a3\right)\ge0.$$
The inequality is necessary for solvability, hence
$$1-\frac a3\ge0,$$
that is,
$$a\le\frac13.$$
For every $a>\frac13$ condition (1) fails, so no translating vector exists for this tetrahedron.
Therefore, in three-dimensional space the statement holds for every convex polyhedron exactly when
$$a\le\frac13.$$
Since $a=-k$, this is equivalent to
$$-\frac13\le k<0.$$
Hence:
For part 1, a translation $T$ exists with $T(M')\subset M$.
For part 2, the admissible homothety ratios are precisely
$$\boxed{-\frac13\le k<0}.$$
Verification of Key Steps
The first delicate step is the passage from geometric inclusion to inequalities involving support functions. For convex bodies,
$$A\subset B \iff h_A(u)\le h_B(u) \quad\text{for all }u.$$
Applying this to $A=-aM+t$ and using
$$h_{-aM+t}(u)=a,h_M(-u)+\langle u,t\rangle$$
produces the exact inequality system. No geometric information is lost.
The second delicate step is inequality (2). Starting from
$$\lambda_i u_i=-\sum_{j\ne i}\lambda_j u_j,$$
subadditivity yields
$$\lambda_i h(u_i)\le\sum_{j\ne i}\lambda_j h(-u_j).$$
When there are at most $d+1$ vectors, each term $\lambda_j h(-u_j)$ appears exactly $d$ times after summation. Any miscount here changes the constant and would lead to an incorrect threshold.
The third delicate step is sharpness. For a simplex in dimension $d$, the relation among facet normals involves exactly $d+1$ vectors. Equality is attained in the estimate leading to (2), so the coefficient $d$ cannot be reduced. In dimension $3$ this forces the constant $\frac13$, and any larger value of $a$ fails for a suitable tetrahedron.
Alternative Approaches
A dual approach uses widths rather than support functions. The inclusion $-aM+t\subset M$ can be reformulated as a statement about placing a reflected and scaled copy of $M$ inside $M$. The relevant quantities become support numbers of the supporting lines or planes. The proof then proceeds through Helly's theorem, showing that every finite subsystem of the placement constraints is compatible.
Another route uses Minkowski addition. The existence of $t$ is equivalent to the nonemptiness of the intersection
$$\bigcap_{x\in M}(M+a x).$$
Helly's theorem reduces the problem to finitely many translates, and the extremal simplex configuration again yields the sharp constant $\frac1d$. The support-function argument is preferable because it exposes the constant directly and treats dimensions $2$ and $3$ in a unified manner.