Kvant Math Problem 1016

For a polygon circumscribed about a circle of radius $r$, let the sides be $s_1,\dots,s_n$, with corresponding side lengths $\ell_1,\dots,\ell_n$.

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Problem

A polygon is circumscribed around a circle with center $O$. Let $P$ be the centroid of the polygon, and $K$ the centroid of its boundary. Prove that the points $P$, $O$, and $K$ lie on a single straight line, with $PO=2PK$. (When defining the centroid $P$, the polygon is considered as a uniform lamina, and $K$ as a wireframe of uniform density.)

I. Z. Weinstein

Exploration

For a polygon circumscribed about a circle of radius $r$, let the sides be $s_1,\dots,s_n$, with corresponding side lengths $\ell_1,\dots,\ell_n$. The center of the circle is $O$.

A natural decomposition of the polygon is into the triangles formed by joining $O$ to the endpoints of each side. Since the circle is tangent to every side, the altitude from $O$ to each side equals $r$. Hence the area of the triangle based on side $s_i$ is

$$A_i=\frac12 r\ell_i.$$

The total area is

$$A=\frac12 r\sum_i \ell_i.$$

The centroid of a lamina obtained by gluing pieces together is the area-weighted average of the centroids of the pieces. Thus the centroid $P$ of the polygon can be expressed through the centroids of these triangles.

For the boundary centroid $K$, the boundary consists of the sides themselves, regarded as uniform rods. The centroid of side $s_i$ is its midpoint $M_i$, and $K$ is the length-weighted average of the points $M_i$.

The centroids of the triangles are easy to locate. If $M_i$ is the midpoint of side $s_i$, then the centroid $G_i$ of the triangle with vertex $O$ and base $s_i$ lies on $OM_i$ and divides that median in the ratio $OG_i:G_iM_i=2:1$. Since the distance from $O$ to the side is $r$, one obtains

$$\overrightarrow{OG_i}=\frac23,\overrightarrow{OM_i}.$$

This suggests that the area-weighted average defining $P$ is exactly $\frac23$ of the length-weighted average defining $K$. If vectors are taken with origin at $O$, this would give

$$\overrightarrow{OP}=\frac23,\overrightarrow{OK}.$$

Then $O,P,K$ are collinear and

$$OP=\frac23,OK, \qquad PK=\frac13,OK,$$

hence $OP=2PK$.

The step most likely to conceal an error is the passage from the centroids of the individual triangles to the centroid of the whole polygon. The weights must be checked carefully and expressed in terms of the side lengths.

Problem Understanding

We are given a polygon circumscribed about a circle with center $O$. The centroid of the polygonal lamina is $P$, and the centroid of the boundary, regarded as a uniform wire, is $K$.

The task is to prove that $P$, $O$, and $K$ lie on one line and that

$$PO=2PK.$$

This is a Type B problem. The statement is already specified, so the goal is a proof.

The core difficulty is relating two different centroids: one obtained from area and one obtained from boundary length. The crucial observation is that, because the polygon is tangential, every triangle formed by $O$ and a side has area proportional to the length of that side, with the same proportionality constant.

Proof Architecture

Let $M_i$ be the midpoint of side $s_i$ of length $\ell_i$, and let $G_i$ be the centroid of the triangle formed by $O$ and $s_i$.

Lemma 1. The area of the triangle based on side $s_i$ equals $\frac12 r\ell_i$; this follows because the altitude from $O$ to every side is the inradius $r$.

Lemma 2. The centroid $G_i$ satisfies $\overrightarrow{OG_i}=\frac23\overrightarrow{OM_i}$; this follows from the fact that the centroid divides each median in the ratio $2:1$ from the vertex.

Lemma 3. The centroid $P$ of the polygon satisfies

$$\overrightarrow{OP} = \frac{\sum_i A_i,\overrightarrow{OG_i}} {\sum_i A_i};$$

this is the standard additivity property of centroids for a union of laminae.

Lemma 4. The centroid $K$ of the boundary satisfies

$$\overrightarrow{OK} = \frac{\sum_i \ell_i,\overrightarrow{OM_i}} {\sum_i \ell_i};$$

this is the corresponding additivity property for a wire made from line segments.

The hardest step is combining Lemmas 1 through 4 correctly and keeping track of the weights.

Solution

Let the sides of the polygon be $s_1,\dots,s_n$, with lengths $\ell_1,\dots,\ell_n$. Let $M_i$ denote the midpoint of $s_i$.

Joining the center $O$ to all vertices decomposes the polygon into $n$ triangles. Let $T_i$ be the triangle whose base is $s_i$ and whose third vertex is $O$.

Since the polygon is circumscribed about a circle centered at $O$, the distance from $O$ to every side equals the radius $r$ of the circle. Hence the area of $T_i$ is

$$A_i=\frac12 r\ell_i.$$

Let $G_i$ be the centroid of $T_i$.

The median of $T_i$ from the vertex $O$ joins $O$ to the midpoint $M_i$ of the opposite side. The centroid divides this median in the ratio $2:1$ measured from the vertex. Therefore

$$\overrightarrow{OG_i} = \frac23,\overrightarrow{OM_i}.$$

The centroid $P$ of the polygonal lamina is the area-weighted average of the centroids of the triangles $T_i$. Taking $O$ as the origin of vectors,

$$\overrightarrow{OP} = \frac{\sum_i A_i,\overrightarrow{OG_i}} {\sum_i A_i}.$$

Substituting $A_i=\frac12 r\ell_i$ and $\overrightarrow{OG_i}=\frac23\overrightarrow{OM_i}$ gives

$$\overrightarrow{OP} = \frac{\sum_i \left(\frac12 r\ell_i\right) \left(\frac23\overrightarrow{OM_i}\right)} {\sum_i \frac12 r\ell_i}.$$

The common factor $\frac12 r$ cancels:

$$\overrightarrow{OP} = \frac23 \frac{\sum_i \ell_i,\overrightarrow{OM_i}} {\sum_i \ell_i}.$$

Now consider the boundary of the polygon as a uniform wire. The centroid of the side $s_i$ is its midpoint $M_i$, and the mass of that side is proportional to its length $\ell_i$. Hence the centroid $K$ of the whole boundary satisfies

$$\overrightarrow{OK} = \frac{\sum_i \ell_i,\overrightarrow{OM_i}} {\sum_i \ell_i}.$$

Comparing this with the preceding formula yields

$$\overrightarrow{OP} = \frac23,\overrightarrow{OK}.$$

Thus $\overrightarrow{OP}$ is a positive scalar multiple of $\overrightarrow{OK}$, so the points $P$, $O$, and $K$ are collinear.

Furthermore,

$$OP=\frac23,OK.$$

Since $P$ lies on the segment $OK$,

$$PK=OK-OP = OK-\frac23,OK = \frac13,OK.$$

Therefore

$$OP=2PK.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the area formula for the triangles $T_i$. The polygon is tangential, so every side is tangent to the same circle. The perpendicular distance from the center $O$ to every side is exactly the inradius $r$. Taking side $s_i$ as the base, the corresponding altitude in triangle $T_i$ is therefore $r$, giving

$$A_i=\frac12 r\ell_i.$$

Without the tangential condition, the altitudes would vary from side to side and the argument would fail.

The second delicate point is the vector relation

$$\overrightarrow{OG_i}=\frac23\overrightarrow{OM_i}.$$

In a triangle, the centroid divides every median in the ratio $2:1$ from the vertex. Since $M_i$ is the midpoint of the base, $OM_i$ is a median. Hence $G_i$ lies on the segment $OM_i$ at two thirds of its length from $O$, yielding the stated vector equation.

The third delicate point is the comparison of the two weighted averages. For the lamina centroid, the weights are the areas $A_i$. For the wire centroid, the weights are the lengths $\ell_i$. Lemma 1 shows that all areas satisfy

$$A_i=\frac12 r,\ell_i,$$

so the two systems of weights differ only by the same constant factor. After cancellation, the averages involve exactly the same coefficients, producing

$$\overrightarrow{OP}=\frac23\overrightarrow{OK}.$$

Alternative Approaches

One may formulate the proof entirely in terms of moments. Let the origin be at $O$. The moment of the polygonal lamina is the sum of the moments of the triangles $T_i$. The moment of a triangle about the origin equals its area multiplied by the position vector of its centroid. Since each triangle centroid is located at $\frac23 M_i$, its contribution is

$$A_i\cdot \frac23,\overrightarrow{OM_i}.$$

Using $A_i=\frac12 r\ell_i$, the total moment of the lamina becomes a constant multiple of

$$\sum_i \ell_i,\overrightarrow{OM_i},$$

which is precisely the moment determining the wire centroid.

Another approach uses continuous mass distributions. Regard the boundary measure and the area measure as two related measures generated by the family of triangles with common height $r$. Integrating the centroid position along each median shows that the area centroid is obtained from the boundary centroid by a homothety centered at $O$ with ratio $\frac23$. The discrete decomposition into triangles is preferable because it reduces the argument to elementary centroid facts and requires no integration.