Kvant Math Problem 134

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Problem

What set of points is traced out by the centroids of triangles whose three vertices lie respectively on the three sides $AB$, $BC$, and $AC$ of a given triangle $ABC$?

L. G. Makarov

Exploration

Let the variable triangle have vertices $P\in AB$, $Q\in BC$, $R\in AC$. We seek the locus of the centroid

$$G=\frac{P+Q+R}{3}.$$

The condition that each point lies on a specified side suggests using affine or barycentric coordinates.

Write

$$P=(1-u)A+uB,\qquad Q=(1-v)B+vC,\qquad R=(1-w)A+wC,$$

with $0\le u,v,w\le1$.

Then

$$3G=P+Q+R.$$

Collecting coefficients of $A,B,C$ gives

$$3G=(2-u-w)A+(u+1-v)B+(v+w)C.$$

Hence the barycentric coordinates of $G$ with respect to $ABC$ are

$$\alpha=\frac{2-u-w}{3},\qquad \beta=\frac{u+1-v}{3},\qquad \gamma=\frac{v+w}{3}.$$

Since $u,v,w\in[0,1]$, these coordinates satisfy

$$\alpha+\beta+\gamma=1.$$

The question is which triples $(\alpha,\beta,\gamma)$ arise.

From the formulas,

$$\beta+\gamma=\frac{u+1-v+v+w}{3} =\frac{1+u+w}{3}.$$

Because $0\le u+w\le2$,

$$\frac13\le\beta+\gamma\le1.$$

Since $\alpha=1-(\beta+\gamma)$,

$$0\le\alpha\le\frac23.$$

By symmetry,

$$0\le\beta\le\frac23,\qquad 0\le\gamma\le\frac23.$$

Thus every centroid lies in the region of $ABC$ cut off by the three lines

$$\alpha=\frac23,\qquad \beta=\frac23,\qquad \gamma=\frac23.$$

In barycentric geometry, $\alpha=\frac23$ is a line parallel to $BC$, passing through the points on $AB$ and $AC$ that divide those sides in the ratio $1:2$ from $A$. The three such lines form the medial-type central hexagon.

The remaining issue is surjectivity: does every point of that hexagon occur as a centroid? This is the only delicate step. We need to show that any barycentric triple satisfying

$$\alpha+\beta+\gamma=1,\qquad 0\le\alpha,\beta,\gamma\le\frac23$$

can be represented by suitable $u,v,w\in[0,1]$.

Using

$$u+w=2-3\alpha,\qquad v+w=3\gamma,$$

one may solve for $u,v$ in terms of $w$. The constraints become interval conditions for $w$. The crucial point is proving that these intervals always overlap.

Problem Understanding

We are given a fixed triangle $ABC$. A variable triangle $PQR$ is formed by choosing $P$ on side $AB$, $Q$ on side $BC$, and $R$ on side $AC$. For every such triangle we consider its centroid. The task is to determine the complete locus of all possible centroids.

This is a Type A problem. We must determine exactly which points occur as centroids and prove both inclusion directions.

The core difficulty is showing that every point of the candidate region is actually attainable. Deriving necessary conditions is straightforward; proving sufficiency requires a careful construction.

The expected answer is the central hexagon bounded by the three lines parallel to the sides of $ABC$ through the trisection points nearest the corresponding vertices. The reason is that the barycentric coordinates of the centroid cannot exceed $\frac23$, and these three inequalities describe precisely that hexagon.

Proof Architecture

Lemma 1. If $P\in AB$, $Q\in BC$, $R\in AC$, then the barycentric coordinates $(\alpha,\beta,\gamma)$ of the centroid $G$ satisfy $0\le\alpha,\beta,\gamma\le\frac23$.

Sketch. Parametrize $P,Q,R$ on their sides and compute the barycentric coordinates of $G$ directly.

Lemma 2. The set of points of $ABC$ whose barycentric coordinates satisfy $0\le\alpha,\beta,\gamma\le\frac23$ is a hexagon bounded by the lines $\alpha=\frac23$, $\beta=\frac23$, $\gamma=\frac23$.

Sketch. Each equation defines a line parallel to the opposite side, and the three lines cut from $ABC$ the corner triangles adjacent to the vertices.

Lemma 3. Every barycentric triple $(\alpha,\beta,\gamma)$ with $\alpha+\beta+\gamma=1$ and $0\le\alpha,\beta,\gamma\le\frac23$ arises from some choice of $P,Q,R$.

Sketch. Express $u+w=2-3\alpha$ and $v+w=3\gamma$, then choose $w$ in the common interval determined by the constraints $0\le u,v,w\le1$.

The hardest direction is Lemma 3. The most likely point of failure is proving that the interval for $w$ is always nonempty.

Solution

Let

$$P=(1-u)A+uB,\qquad Q=(1-v)B+vC,\qquad R=(1-w)A+wC,$$

where

$$0\le u,v,w\le1.$$

Let $G$ be the centroid of triangle $PQR$. Since

$$G=\frac{P+Q+R}{3},$$

we obtain

$$3G=(2-u-w)A+(u+1-v)B+(v+w)C.$$

Hence the barycentric coordinates of $G$ with respect to triangle $ABC$ are

$$\alpha=\frac{2-u-w}{3},\qquad \beta=\frac{u+1-v}{3},\qquad \gamma=\frac{v+w}{3}.$$

Because $0\le u,v,w\le1$, we have

$$0\le2-u-w\le2,$$

and therefore

$$0\le\alpha\le\frac23.$$

Similarly,

$$0\le u+1-v\le2,$$

which gives

$$0\le\beta\le\frac23,$$

and

$$0\le v+w\le2,$$

which gives

$$0\le\gamma\le\frac23.$$

Thus every centroid belongs to the set

$$\alpha+\beta+\gamma=1,\qquad 0\le\alpha,\beta,\gamma\le\frac23.$$

This proves the necessity.

Now let a point $G$ of triangle $ABC$ be given with barycentric coordinates satisfying

$$\alpha+\beta+\gamma=1, \qquad 0\le\alpha,\beta,\gamma\le\frac23.$$

We shall construct $P,Q,R$ whose centroid is $G$.

From the formulas above we need

$$u+w=2-3\alpha, \qquad v+w=3\gamma.$$

Set

$$s=2-3\alpha,\qquad t=3\gamma.$$

Since $0\le\alpha,\gamma\le\frac23$,

$$0\le s\le2,\qquad 0\le t\le2.$$

Choose $w$ satisfying

$$\max(0,s-1,t-1)\le w\le \min(1,s,t).$$

Such a choice is possible because

$$s\le2,\qquad t\le2$$

imply

$$s-1\le1,\qquad t-1\le1,$$

and also

$$s-1\le s,\qquad t-1\le t,\qquad 0\le s,\qquad 0\le t.$$

Hence the left endpoint does not exceed the right endpoint.

Define

$$u=s-w,\qquad v=t-w.$$

From the choice of $w$,

$$0\le u\le1,\qquad 0\le v\le1,\qquad 0\le w\le1.$$

Therefore points $P,Q,R$ on the prescribed sides exist.

For these points,

$$u+w=s=2-3\alpha, \qquad v+w=t=3\gamma.$$

Using $\alpha+\beta+\gamma=1$,

$$u+1-v =(2-3\alpha-w)+1-(3\gamma-w) =3-3(\alpha+\gamma) =3\beta.$$

Substituting into the expressions for the barycentric coordinates of the centroid yields precisely

$$\alpha,\qquad\beta,\qquad\gamma.$$

Thus the centroid of $PQR$ is the prescribed point $G$.

We have proved that the locus consists exactly of all points of $ABC$ whose barycentric coordinates satisfy

$$0\le\alpha,\beta,\gamma\le\frac23.$$

The line $\alpha=\frac23$ is parallel to $BC$ and passes through the points of $AB$ and $AC$ dividing those sides in the ratio $1:2$ from $A$. Analogous statements hold for $\beta=\frac23$ and $\gamma=\frac23$. These three lines bound a central hexagon.

Hence the required locus is that hexagon.

$$\boxed{\text{The locus is the central hexagon cut from }ABC\text{ by the three lines parallel to }BC,CA,AB\text{ through the trisection points nearest }A,B,C.}$$

Verification of Key Steps

The first delicate step is the derivation of the barycentric coordinates. Starting from

$$3G=P+Q+R,$$

the coefficient of $A$ equals

$$(1-u)+(1-w)=2-u-w,$$

the coefficient of $B$ equals

$$u+(1-v)=u+1-v,$$

and the coefficient of $C$ equals

$$v+w.$$

Their sum is

$$(2-u-w)+(u+1-v)+(v+w)=3,$$

so division by $3$ indeed gives barycentric coordinates.

The second delicate step is proving the interval for $w$ is nonempty. The lower endpoint is

$$L=\max(0,s-1,t-1),$$

and the upper endpoint is

$$U=\min(1,s,t).$$

Each of the three numbers defining $L$ does not exceed each of the three numbers defining $U$:

$$0\le1,\quad 0\le s,\quad 0\le t,$$

$$s-1\le1,\quad s-1\le s,\quad s-1\le t$$

because $s\le2$ and $t\ge0$,

and similarly

$$t-1\le1,\quad t-1\le t,\quad t-1\le s.$$

Hence $L\le U$.

The third delicate step is recovering $\beta$. After enforcing

$$u+w=2-3\alpha,\qquad v+w=3\gamma,$$

we compute

$$u+1-v =(u+w)+1-(v+w) =(2-3\alpha)+1-3\gamma =3\beta.$$

Thus all three barycentric coordinates match exactly.

Alternative Approaches

An affine-geometric proof avoids explicit interval calculations. Since the centroid depends affinely on each vertex, the set of all centroids is the image of the cube of parameters $(u,v,w)\in[0,1]^3$ under an affine map into the plane. The image of a cube under an affine map is a convex polygon. Computing the images of the eight vertices of the cube shows that the image is precisely the central hexagon bounded by the lines $\alpha=\frac23$, $\beta=\frac23$, and $\gamma=\frac23$.

Another approach uses vectors. Taking $A,B,C$ as position vectors, write

$$G=\frac23A+\frac13B +\frac13(u(B-A)+v(C-B)+w(C-A)).$$

The variable part is the Minkowski sum of three segments. The sum of three planar segments is a centrally symmetric hexagon. Evaluating its six vertices yields the same hexagon inside triangle $ABC$. The barycentric method is preferable because it identifies the bounding lines immediately and proves both necessity and sufficiency in a uniform way.