Kvant Math Problem 1400
A shortest closed route that visits all four faces can be replaced by a polygonal route whose vertices lie on the faces.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m52s
Source on kvant.digital
Problem
Inside a regular tetrahedron with edge length $a$, a fly is flying. What is the shortest closed path the fly must travel in order to visit all the faces of the tetrahedron?
I. F. Sharygin
Exploration
A shortest closed route that visits all four faces can be replaced by a polygonal route whose vertices lie on the faces. If a face is visited several times, removing redundant visits shortens the route. Thus the extremal object should be a quadrilateral with one vertex on each face.
The regular tetrahedron has a large symmetry group. Any candidate for the shortest quadrilateral should be compared with its images under all symmetries of the tetrahedron. Since the perimeter is a convex function of the vertex positions, averaging all symmetric copies cannot increase the perimeter. The averaged configuration must be fixed by every symmetry. The only point of a given face fixed by all symmetries preserving that face is the center of the face. This strongly suggests that the extremal quadrilateral has its vertices at the four face centers.
The delicate point is proving that symmetry really gives a global minimum rather than merely a stationary configuration. Convexity of the perimeter functional provides exactly the required justification.
The distance between the centers of two faces of a regular tetrahedron is easy to compute from the inradius. If the inradius is $r=\dfrac{a\sqrt6}{12}$, and $n_i,n_j$ are unit outward normals of two faces, then the corresponding face centers are $rn_i,rn_j$. Since $n_i\cdot n_j=-\dfrac13$,
$$|rn_i-rn_j| =r\sqrt{2-!2!\left(-\frac13\right)} =r\sqrt{\frac83} =\frac a3.$$
Hence the quadrilateral formed by the four face centers has perimeter $4a/3$.
Problem Understanding
We seek the shortest closed path inside a regular tetrahedron of edge length $a$ that visits every face.
This is a Type C problem. We must determine the minimum possible length, exhibit a path attaining it, and prove that no shorter closed path exists.
The central difficulty is establishing a global lower bound. Because the tetrahedron is completely symmetric, one expects the optimal path to use the four face centers. The task is to convert that symmetry intuition into a rigorous minimization argument.
The expected answer is
$$\boxed{\frac{4a}{3}},$$
attained by the quadrilateral whose vertices are the centers of the four faces.
Proof Architecture
Let $F_1,F_2,F_3,F_4$ be the faces, and let $x_i\in F_i$.
The first claim is that every shortest closed path visiting all faces may be represented by a quadrilateral $x_1x_2x_3x_4$ with one vertex on each face. This follows because any extra visits create unnecessary segments.
The second claim is that the perimeter
$$P(x_1,x_2,x_3,x_4) = |x_1x_2|+|x_2x_3|+|x_3x_4|+|x_4x_1|$$
is a convex function of the four variables. Each distance term is a norm of an affine function, hence convex.
The third claim is that averaging a feasible quadrilateral over all rotational symmetries of the tetrahedron does not increase its perimeter. This is a direct consequence of convexity and symmetry.
The fourth claim is that the averaged quadrilateral has vertices at the face centers. The face center is the unique point of a face fixed by all symmetries preserving that face.
The final claim is that the distance between any two face centers equals $a/3$. Computing this distance gives the perimeter $4a/3$.
The most delicate step is the symmetry-averaging argument, because it is the step that converts symmetry into a rigorous global minimum.
Solution
Let $F_1,F_2,F_3,F_4$ be the four faces of the regular tetrahedron.
A closed path visiting all four faces can be shortened, if necessary, until each face is visited at a single point and consecutive visits are connected by straight segments. Hence a shortest path may be represented by a quadrilateral
$$x_1x_2x_3x_4, \qquad x_i\in F_i.$$
Its length is
$$P(x_1,x_2,x_3,x_4) = |x_1x_2| + |x_2x_3| + |x_3x_4| + |x_4x_1|.$$
Consider the set of all quadruples $(x_1,x_2,x_3,x_4)$ with $x_i\in F_i$. This set is convex, because each face is convex.
Each summand $|x_ix_j|$ is the Euclidean norm of an affine function of the variables, hence is convex. Therefore $P$ is a convex function on the feasible set.
Now average an arbitrary feasible quadrilateral over all rotational symmetries of the regular tetrahedron. Since the tetrahedron is regular, every rotation permutes the faces. Applying a rotation to a feasible quadrilateral produces another feasible quadrilateral with the same perimeter.
Let $\overline x_i$ be the average of all images of $x_i$ under the symmetry group. Convexity of $P$ gives
$$P(\overline x_1,\overline x_2,\overline x_3,\overline x_4) \le P(x_1,x_2,x_3,x_4).$$
Thus averaging cannot increase the perimeter.
The averaged configuration is fixed by every rotational symmetry. For a given face $F_i$, the only point of $F_i$ fixed by all rotations preserving $F_i$ is its center. Consequently $\overline x_i$ is the center of $F_i$.
Hence among all feasible quadrilaterals, one of minimum perimeter is obtained by taking the four face centers.
Let $r$ be the inradius of the tetrahedron. The center of face $F_i$ is the tangency point of the insphere with that face, so it has the form $rn_i$, where $n_i$ is the unit outward normal to $F_i$.
For a regular tetrahedron,
$$n_i\cdot n_j=-\frac13 \qquad (i\ne j).$$
Therefore the distance between the centers of two faces is
$$r,|n_i-n_j| = r\sqrt{2-!2!\left(-\frac13\right)} = r\sqrt{\frac83}.$$
The inradius of a regular tetrahedron of edge length $a$ equals
$$r=\frac{a\sqrt6}{12}.$$
Substituting,
$$r\sqrt{\frac83} = \frac{a\sqrt6}{12}\cdot\frac{2\sqrt2}{\sqrt3} = \frac a3.$$
Thus every side of the quadrilateral formed by the four face centers has length $a/3$. Its perimeter is
$$4\cdot\frac a3 = \frac{4a}{3}.$$
Since this quadrilateral is a minimizer, no shorter closed path exists.
Hence the shortest closed path visiting all four faces has length
$$\boxed{\frac{4a}{3}},$$
and it is realized by the quadrilateral joining the centers of the four faces.
Verification of Key Steps
The convexity step must be checked carefully. For fixed indices $i,j$, the function
$$(x_1,x_2,x_3,x_4)\mapsto |x_i-x_j|$$
is the norm of an affine map. Norms are convex, and a sum of convex functions is convex. Thus the whole perimeter functional is convex.
The averaging step uses only two facts. First, every symmetry preserves perimeter. Second, Jensen's inequality for a convex function states that the value at the average does not exceed the average of the values. Since all symmetric copies have the same perimeter, the averaged configuration cannot have larger perimeter.
The distance computation between face centers depends on the relation
$$n_i\cdot n_j=-\frac13.$$
Then
$$|n_i-n_j|^2 = |n_i|^2+|n_j|^2-2n_i\cdot n_j = 1+1+\frac23 = \frac83.$$
Combining this with $r=\dfrac{a\sqrt6}{12}$ indeed yields $a/3$.
Alternative Approaches
A different approach uses the dual tetrahedron. The four face centers form the vertices of a regular tetrahedron dual to the original one. The problem then becomes a minimization of the perimeter of a quadrilateral with one vertex on each face. By applying repeated symmetrizations with respect to the rotational symmetries of the tetrahedron, one arrives directly at the face-center configuration.
Another route uses Lagrange multipliers. One minimizes the perimeter subject to the four planar constraints defining the faces. The stationarity equations force a completely symmetric solution, which is precisely the quadrilateral formed by the face centers. The symmetry-averaging argument is preferable because it establishes global optimality immediately and avoids a lengthy analysis of critical points.