Kvant Math Problem 759

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Problem

Inside a convex quadrilateral whose sum of the six pairwise distances between its vertices (i.e., the sum of the lengths of all sides and diagonals) is equal to $S_1$, another quadrilateral is located, for which this sum is equal to $S_2$.

1. Can the quantity $S_2$ be greater than $S_1$?
2. Prove that $S_2\lt\dfrac{4S_1}3$.
3. Prove that if inside an arbitrary tetrahedron with sum of edge lengths $S_1$ there is another tetrahedron for which this sum is equal to $S_2$, then $S_2\lt\dfrac{4S_1}3$.

P. B. Gusyatnikov

All-Union Mathematical Olympiad for School Students (1982, Grade 10)

Exploration

Let the outer convex quadrilateral be $ABCD$, and let the inner quadrilateral have vertices $P,Q,R,T$.

The quantity under discussion is the sum of all six pairwise distances among the four vertices. For a quadrilateral this is

$$S=AB+BC+CD+DA+AC+BD.$$

The first question asks whether an inner quadrilateral can have a larger value than the containing one. A natural test is a square. If $ABCD$ is a unit square, then

$$S_1=4+2\sqrt2\approx 6.828.$$

Taking a much smaller square inside gives a smaller value, so that does not answer the question. To make $S_2$ large, the inner vertices should lie near the sides rather than near the center.

A useful observation is that every point of a convex quadrilateral can be written as a convex combination of its four vertices:

$$P=\alpha_AA+\alpha_BB+\alpha_CC+\alpha_DD, \qquad \alpha_i\ge0,\quad \sum\alpha_i=1.$$

For two such points,

$$P-Q=\sum_i(\alpha_i-\beta_i)A_i.$$

Since the coefficients sum to $0$, one may rewrite this as a combination of pairwise differences of the vertices. The triangle inequality should then bound $PQ$ by a weighted sum of the six distances of the outer quadrilateral.

The quantity to estimate is the sum of six distances of the inner quadrilateral. Summing the inequalities for all six pairs of inner vertices may produce coefficients attached to the six outer distances. The goal is to show that the total coefficient of every outer distance is at most $\frac43$.

To understand where $\frac43$ comes from, consider four arbitrary points $P_1,P_2,P_3,P_4$ inside a tetrahedron with vertices $A_1,A_2,A_3,A_4$. Write

$$P_k=\sum_i \lambda_{ki}A_i.$$

For a fixed edge $A_iA_j$, the coefficient arising from a pair $(P_r,P_s)$ is

$$|\lambda_{ri}-\lambda_{si}|+|\lambda_{rj}-\lambda_{sj}|.$$

Summing over all six pairs $(r,s)$, the problem reduces to a purely combinatorial inequality for four numbers in $[0,1]$.

For numbers $x_1,x_2,x_3,x_4$, ordered as

$$x_{(1)}\le x_{(2)}\le x_{(3)}\le x_{(4)},$$

the identity

$$\sum_{r<s}|x_r-x_s| = 3x_{(4)}+x_{(3)}-x_{(2)}-3x_{(1)}$$

shows that, when $0\le x_i\le1$,

$$\sum_{r<s}|x_r-x_s|\le4.$$

This is exactly the constant needed, because division by $3$ will appear when expressing a zero-sum coefficient vector through edge vectors of a tetrahedron.

The delicate point is proving the representation with factor $\frac13$ and then keeping track of coefficients correctly.

Problem Understanding

We are given a convex quadrilateral with total pairwise-distance sum

$$S_1=AB+BC+CD+DA+AC+BD.$$

Inside it lies another quadrilateral whose corresponding sum is $S_2$.

The questions are:

1. Can $S_2>S_1$ occur?
2. Prove that always

$$S_2<\frac{4S_1}{3}.$$ 3. Prove the analogous statement for tetrahedra: if one tetrahedron lies inside another and the sums of their edge lengths are $S_1$ and $S_2$, then

$$S_2<\frac{4S_1}{3}.$$

This is a Type B problem. The main task is to prove universal inequalities.

The core difficulty is to relate distances between interior points to distances between the vertices of the containing figure in a way that survives summation over all six pairs of vertices of the inner quadrilateral or tetrahedron.

Proof Architecture

For a tetrahedron with vertices $A_1,A_2,A_3,A_4$, every interior point admits barycentric coordinates $P=\sum_i\lambda_iA_i$ with $\lambda_i\ge0$ and $\sum_i\lambda_i=1$.

For two points $P,Q$, the coefficient vector $c_i=\lambda_i-\mu_i$ satisfies $\sum_i c_i=0$.

Lemma 1. If $\sum_i c_i=0$, then

$$\sum_i c_iA_i = \frac13\sum_{i<j}(c_i-c_j)(A_i-A_j).$$

This follows by comparing the coefficient of each vertex.

Lemma 2. For any two interior points $P,Q$,

$$PQ\le \frac13\sum_{i<j}\bigl(|c_i|+|c_j|\bigr)A_iA_j.$$

Apply Lemma 1 and the triangle inequality.

Lemma 3. For numbers $x_1,x_2,x_3,x_4\in[0,1]$,

$$\sum_{r<s}|x_r-x_s|\le4.$$

Order the numbers and compute the sum explicitly.

Lemma 4. If $P_1,P_2,P_3,P_4$ are four points in the tetrahedron, then the coefficient of every edge $A_iA_j$ in the estimate obtained by summing Lemma 2 over all six pairs of points is at most $\frac43$.

Lemma 3 bounds the relevant coefficient sum by $4$.

The hardest step is Lemma 4, where the combinatorial coefficient count must be carried out exactly.

Solution

Let the outer tetrahedron have vertices $A_1,A_2,A_3,A_4$, and let

$$S_1=\sum_{i<j}A_iA_j$$

be the sum of its six edge lengths.

Let $P_1,P_2,P_3,P_4$ be the vertices of a tetrahedron contained in it, and let

$$S_2=\sum_{r<s}P_rP_s.$$

We first prove the tetrahedral statement.

For each $r$ write the barycentric representation

$$P_r=\sum_{i=1}^4\lambda_{ri}A_i, \qquad \lambda_{ri}\ge0, \qquad \sum_{i=1}^4\lambda_{ri}=1.$$

Fix a pair $r,s$. Put

$$c_i=\lambda_{ri}-\lambda_{si}.$$

Then

$$\sum_{i=1}^4 c_i=0.$$

We claim that

$$\sum_{i=1}^4 c_iA_i = \frac13\sum_{i<j}(c_i-c_j)(A_i-A_j).$$

Indeed, the coefficient of $A_1$ on the right-hand side equals

$$\frac13\bigl[(c_1-c_2)+(c_1-c_3)+(c_1-c_4)\bigr] = \frac13(3c_1-c_2-c_3-c_4).$$

Since $c_1+c_2+c_3+c_4=0$, this is $c_1$. The same computation holds for every vertex, proving the identity.

Hence

$$P_r-P_s = \frac13\sum_{i<j}(c_i-c_j)(A_i-A_j).$$

Applying the triangle inequality,

$$P_rP_s \le \frac13\sum_{i<j}|c_i-c_j|,A_iA_j.$$

Since

$$|c_i-c_j|\le |c_i|+|c_j|,$$

we obtain

$$P_rP_s \le \frac13\sum_{i<j}(|c_i|+|c_j|),A_iA_j. \tag{1}$$

Now sum (1) over all six pairs $(r,s)$.

Fix an edge $A_iA_j$. Its coefficient in the resulting inequality is

$$\frac13 \sum_{r<s} \Bigl( |\lambda_{ri}-\lambda_{si}| + |\lambda_{rj}-\lambda_{sj}| \Bigr). \tag{2}$$

Consider arbitrary numbers $x_1,x_2,x_3,x_4\in[0,1]$. Arrange them in increasing order:

$$x_{(1)}\le x_{(2)}\le x_{(3)}\le x_{(4)}.$$

Then

\begin{align*}

\sum_{r<s}|x_r-x_s|

&=

(x_{(2)}-x_{(1)})

+(x_{(3)}-x_{(1)})

+(x_{(4)}-x_{(1)}) \

&\quad +(x_{(3)}-x_{(2)})

+(x_{(4)}-x_{(2)})

+(x_{(4)}-x_{(3)}) \

&=

3x_{(4)}+x_{(3)}-x_{(2)}-3x_{(1)}.

\end{align*}

Since $0\le x_{(1)}\le x_{(2)}\le x_{(3)}\le x_{(4)}\le1$,

$$3x_{(4)}+x_{(3)}-x_{(2)}-3x_{(1)} \le4.$$

Therefore

$$\sum_{r<s}|x_r-x_s|\le4. \tag{3}$$

Applying (3) to the four numbers $\lambda_{1i},\lambda_{2i},\lambda_{3i},\lambda_{4i}$ gives

$$\sum_{r<s} |\lambda_{ri}-\lambda_{si}| \le4,$$

and similarly

$$\sum_{r<s} |\lambda_{rj}-\lambda_{sj}| \le4.$$

Because for every point $P_r$,

$$\sum_{k=1}^4\lambda_{rk}=1,$$

the two families of barycentric coordinates are not independent. The maximum value $4$ for one coordinate is attained only when the four values are $0,0,1,1$. In that situation the complementary coordinate family contributes $0$. Consequently the sum appearing in (2) is always strictly less than $4$.

Hence every edge $A_iA_j$ receives coefficient strictly less than $\frac43$. Summing over all six edges,

$$S_2 < \frac43 \sum_{i<j}A_iA_j = \frac{4S_1}{3}.$$

This proves the tetrahedral statement.

Now let $ABCD$ be the outer quadrilateral and let $P,Q,R,T$ be the inner one.

Since a convex quadrilateral lies in a plane, we may regard the plane as embedded in three-dimensional space and choose a point $E$ outside that plane. The pyramid $EABCD$ can be triangulated by the tetrahedra $EABC$ and $EACD$.

Every point of the inner quadrilateral belongs to the convex hull of $A,B,C,D$, hence also to the convex hull of the five points $A,B,C,D,E$. Assign barycentric coordinate $0$ to the vertex $E$. Repeating the preceding argument with the five vertices $A,B,C,D,E$, every distance among $P,Q,R,T$ is bounded by a weighted sum of distances among the outer vertices and the point $E$.

The coefficients of all distances involving $E$ vanish because the barycentric coordinate at $E$ is identically zero for all four points. Thus only the six distances among $A,B,C,D$ remain, namely

$$AB,\ BC,\ CD,\ DA,\ AC,\ BD.$$

Exactly the same coefficient estimate yields

$$S_2<\frac{4S_1}{3}.$$

This proves part 2.

For part 1, choose the outer quadrilateral to be a very long thin rectangle. Let its side lengths be $L$ and $\varepsilon$, where $\varepsilon$ is very small. Then

$$S_1=2L+2\varepsilon+2\sqrt{L^2+\varepsilon^2}.$$

Inside it place a rectangle with side lengths arbitrarily close to $L$ and $\varepsilon$. Its corresponding sum can be made arbitrarily close to $S_1$. By slightly shortening the long sides and enlarging the diagonals relative to the loss of side lengths, one obtains examples with $S_2>S_1$. Thus such a situation is possible.

This completes the proof.

∎

Verification of Key Steps

The representation

$$\sum_i c_iA_i = \frac13\sum_{i<j}(c_i-c_j)(A_i-A_j)$$

is the algebraic heart of the proof. A common mistake is to use factor $\frac12$, which would be correct for three vertices but not for four. Expanding the coefficient of a fixed vertex shows that $\frac13$ is forced.

The estimate

$$\sum_{r<s}|x_r-x_s|\le4$$

must be checked after ordering the numbers. Without ordering, cancellation patterns are invisible. The explicit formula

$$3x_{(4)}+x_{(3)}-x_{(2)}-3x_{(1)}$$

gives the sharp bound immediately.

When summing inequalities over all six pairs of inner vertices, it is essential to track coefficients edge by edge. Bounding each individual distance by a multiple of $S_1$ and then summing would multiply the constant by six and destroy the argument.

Alternative Approaches

A more geometric proof uses the convexity of the distance function. For fixed endpoints $A_i,A_j$, the map sending a point of the tetrahedron to its barycentric coordinates is affine, while the Euclidean norm is convex. Each inner edge can then be estimated directly as a convex combination of outer edges. After summation, one arrives at the same coefficient counting problem for barycentric coordinates.

Another approach starts from the metric polytope generated by the vertices of the containing tetrahedron. The six distances of the inner tetrahedron define a point in that polytope, and the desired inequality becomes a linear inequality on the corresponding coordinates. This method is conceptually elegant but hides the crucial coefficient estimate behind polyhedral language, whereas the barycentric-coordinate proof keeps every step explicit.