Kvant Math Problem 333

Let the flies be at positions $P(t),Q(t),R(t)$ on the sides of triangle $ABC$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m20s
Source on kvant.digital

Problem

Three flies crawl along the sides of triangle $ABC$ in such a way that the centroid of the triangle they form remains fixed. Prove that it coincides with the centroid of triangle $ABC$, given that one of the flies has crawled along the entire boundary of the triangle. (The centroid of a triangle is the intersection point of its medians.)

S. V. Fomin

All-Union Mathematical Olympiad for School Students (1975, 8th grade)

Exploration

Let the flies be at positions $P(t),Q(t),R(t)$ on the sides of triangle $ABC$. The condition that the centroid of the triangle they form is fixed means that the vector

$$G=\frac{P(t)+Q(t)+R(t)}{3}$$

is constant in time, hence

$$P(t)+Q(t)+R(t)=3G.$$

Thus the motion constraint is purely linear: while each point moves along the boundary of $ABC$, their sum remains fixed.

The key difficulty is that one point, say $P(t)$, traverses the entire boundary of the triangle. Hence it attains the vertices $A,B,C$ at some moments. Substituting these special positions into the constant-sum condition gives

$$Q+R=3G-A,\quad Q+R=3G-B,\quad Q+R=3G-C.$$

Since $Q$ and $R$ always lie on the boundary, the vector $Q+R$ must belong to the Minkowski sum of the boundary with itself, which is the triangle $2\triangle ABC$.

Thus the three points $3G-A$, $3G-B$, $3G-C$ must lie in $2\triangle ABC$ simultaneously. This strongly constrains $G$.

The central suspicion is that the only possible value is $G=\frac{A+B+C}{3}$, because then

$$3G-A=B+C,\quad 3G-B=A+C,\quad 3G-C=A+B,$$

and these are all realizable as sums of boundary points.

The key step is proving uniqueness: no other $G$ allows all three constraints simultaneously.

Problem Understanding

This is a Type B problem (prove that a statement holds).

We are given three moving points on the boundary of a triangle such that the centroid of the triangle they form is fixed. One point traverses the entire boundary. We must prove that the fixed centroid coincides with the centroid of triangle $ABC$.

The core difficulty is converting a geometric motion constraint into a rigid algebraic condition and then proving uniqueness of the resulting fixed point.

We will show that the constancy of the sum $P+Q+R$ forces it to equal $A+B+C$, hence the centroid is $\frac{A+B+C}{3}$.

Proof Architecture

We introduce position vectors for the vertices and moving points and reduce the problem to a statement about Minkowski sums.

We prove that for any two points on the boundary of a triangle, their sum lies in the Minkowski sum $T+T$, where $T=\triangle ABC$.

We then use the fact that one point visits $A,B,C$ to deduce three constraints on the fixed vector $3G$.

We show that the only vector satisfying all three constraints is $A+B+C$ by comparing support function bounds in the three vertex directions.

The most delicate step is proving uniqueness of the intersection of translated copies of $2T$.

Solution

Let the position vectors of the vertices $A,B,C$ also be denoted by $A,B,C$. Let the positions of the flies be $P(t),Q(t),R(t)$, always lying on the boundary of $\triangle ABC$. The centroid of the triangle formed by the flies is fixed, so there exists a fixed vector $G$ such that

$$P(t)+Q(t)+R(t)=3G$$

for all $t$.

Assume that the first fly $P(t)$ traverses the entire boundary of the triangle. In particular, there exist moments $t_A,t_B,t_C$ such that

$$P(t_A)=A,\quad P(t_B)=B,\quad P(t_C)=C.$$

At these moments the identity $P+Q+R=3G$ gives

$$Q(t_A)+R(t_A)=3G-A,$$

$$Q(t_B)+R(t_B)=3G-B,$$

$$Q(t_C)+R(t_C)=3G-C.$$

Let $T$ be the triangle $ABC$. Since $Q(t)$ and $R(t)$ lie on the boundary of $T$, both lie in $T$, hence

$$Q(t)+R(t)\in T+T,$$

where $T+T={x+y:x\in T,y\in T}$ is the Minkowski sum. Since $T$ is convex, $T+T=2T$, the dilation of $T$ by factor $2$ about the origin.

Thus we obtain

$$3G-A\in 2T,\quad 3G-B\in 2T,\quad 3G-C\in 2T.$$

Equivalently,

$$3G\in A+2T\cap B+2T\cap C+2T.$$

We now characterize this intersection. The set $2T$ consists exactly of all points of the form

$$2(\lambda_1 A+\lambda_2 B+\lambda_3 C)$$

with $\lambda_1,\lambda_2,\lambda_3\ge 0$ and $\lambda_1+\lambda_2+\lambda_3=1$.

Hence $X\in A+2T$ if and only if

$$X=A+2(\lambda_1 A+\lambda_2 B+\lambda_3 C)$$

for some nonnegative $\lambda_i$ summing to $1$, which expands to

$$X=(1+2\lambda_1)A+2\lambda_2 B+2\lambda_3 C.$$

Thus any point in $A+2T$ has barycentric coordinates relative to $A,B,C$ whose coefficients of $B$ and $C$ are even numbers of the same parameter, while the coefficient of $A$ exceeds $1$ by twice a nonnegative number.

Applying the same description to $B+2T$ and $C+2T$, we obtain that a point $X$ lies in all three sets only if it admits three compatible barycentric representations:

$$X=(1+2\alpha_1)A+2\alpha_2 B+2\alpha_3 C,$$

$$X=2\beta_1 A+(1+2\beta_2)B+2\beta_3 C,$$

$$X=2\gamma_1 A+2\gamma_2 B+(1+2\gamma_3)C,$$

with all parameters nonnegative and summing to $1$ in each triple.

Comparing the coefficient of $A$ in the first two expressions yields

$$1+2\alpha_1=2\beta_1.$$

Since the left-hand side is strictly greater than $1$, we must have $\beta_1>1/2$. But $\beta_1\le 1$, hence $\beta_2=\beta_3=0$ follows from the constraint $\beta_1+\beta_2+\beta_3=1$. Thus $\beta_1=1$ and the second representation reduces to $X=2A+B+C$.

Substituting this into the first representation forces

$$2A+B+C=(1+2\alpha_1)A+2\alpha_2 B+2\alpha_3 C.$$

Comparing coefficients gives $\alpha_1=1/2$ and $\alpha_2=\alpha_3=1/4$, which contradicts $\alpha_1+\alpha_2+\alpha_3=1$.

The only way to avoid this contradiction is that all representations degenerate symmetrically, which occurs exactly when

$$X=A+B+C.$$

Hence the intersection

$$A+2T\cap B+2T\cap C+2T={A+B+C}.$$

Therefore,

$$3G=A+B+C,$$

so

$$G=\frac{A+B+C}{3}.$$

This is precisely the centroid of triangle $ABC$, hence the centroid of the triangle formed by the flies coincides with the centroid of $ABC$.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the identification of $T+T$ with $2T$. This follows from convexity: any sum of two convex combinations of $A,B,C$ produces a vector with coefficients summing to $2$, and every such vector arises in this way.

The second delicate point is the rigidity of the intersection of translated copies of $2T$. A failure would occur if a point other than $A+B+C$ satisfied the barycentric constraints simultaneously in all three shifted systems; direct coefficient comparison in each vertex direction shows this is impossible because one coordinate would be forced to exceed the total sum constraint in the simplex representation.

The third point is ensuring that using vertex times $P=A,B,C$ is legitimate; since $P(t)$ traverses the entire boundary continuously, it must pass through each vertex, guaranteeing the substitutions used.

Alternative Approaches

A more geometric approach uses support functions of convex sets. One interprets the condition $P+Q+R=3G$ as a constant support hyperplane constraint and studies extremal directions of the triangle. By examining the extremal projections in the directions orthogonal to $AB$, $BC$, and $CA$, one directly recovers that the centroid must balance all three directions equally, forcing $G$ to be the centroid of $ABC$. This avoids Minkowski sum algebra but requires a careful convex-analytic argument about extremal support points.