From the regular pentagon $A B C D E$, we cut out the rhombus $A B F E$ determined by the edges $A B$ and $A E$. Determine the centroid of the remaining concave pentagon $B C D E F$.

 untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

Note: The untranslated portion is a note in Chinese and should not be included in the translated text. Here is the final translation without the note:

From the regular pentagon $A B C D E$, we cut out the rhombus $A B F E$ determined by the edges $A B$ and $A E$. Determine the centroid of the remaining concave pentagon $B C D E F$.
I. Solution. In both solutions, we will use the following known statement.

Lemma. If a plane figure is divided into two parts, with areas $T_{1}$ and $T_{2}$, and the centroids of these parts are $S_{1}$ and $S_{2}$, then the centroid $S$ of the figure is the point on the segment $S_{1} S_{2}$ such that $T_{1}: T_{2}=S_{2} S: S S_{1}$.

We know that the diagonal of a regular pentagon is parallel to the opposite side of the pentagon. Since the opposite sides of a rhombus are also parallel and only one parallel can be drawn through a given point to a given line, $F$ coincides with the intersection of the diagonals $E C$ and $B D$.

Let the centroid of the $B C D E F$ pentagonal plate be $S$. If we divide the plate by the perpendicular bisector $f$ of the segment $C D$, the resulting parts are symmetric with respect to $f$, so their centroids are also symmetric, from which it follows by the lemma that $S$ lies on $f$.

Now, divide the pentagonal plate by the line $E C$ into two parts. The resulting two triangles, $B C F$ and $D E C$, are isosceles because the equality of the sides of the rhombus implies $B F=A E$, so each triangle has two sides whose lengths are equal to the side length of the regular pentagon. Let the midpoints of the segments $C F$ and $C E$ be $M_{1}$ and $M_{2}$, respectively, and the centroids of the triangles $B C F$ and $D E C$ be $S_{1}$ and $S_{2}$ (Figure 1). Since the centroid of any triangle trisects the medians, $S_{1}$ is the point that trisects $B M_{1}$ closer to $E C$, and $S_{2}$ is the point that trisects $D M_{2}$ closer to $E C$. Since both triangles are isosceles, $B M_{1} \perp C F$ and $D M_{2} \perp C E$. Therefore, in the axial affinity whose axis is the line $E C$ and whose ratio is $1 / 3$, the image of point $B$ is $S_{1}$, the image of point $D$ is $S_{2}$, and the image of point $F$ on the axis is itself. Therefore, the image of the line $B F D$ under the affinity is the line $S_{1} F S_{2}$, which means $S_{1} S_{2}$ passes through $F$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c16f8be79ba5059e77b8g-1.jpg?height=452&width=466&top_left_y=953&top_left_x=819)

Figure 1

By the lemma, the centroid of the concave pentagonal plate lies on the segment $S_{1} S_{2}$, so it must be the intersection of $f$ and $S_{1} S_{2}$. Therefore, the centroid of the $B C D E F$ concave pentagonal plate is the point $F$.

II. Solution. Let the center of the circle circumscribed around the regular pentagon be $K$, and the intersection of the diagonals of the rhombus $A B F E$ be $M$. By symmetry, these points are also the centroids of the regular pentagon and the rhombus, respectively. If the centroid of the $B C D E F$ concave pentagonal plate is $N$, then by the lemma, $K$ is the point on the segment $M N$ such that $M K: K N=T_{B C D E F}: T_{A B F E}$.

The ratio of the areas can be easily determined. Let the side length of the regular pentagon be 1. It is known that the length of the diagonals is $(\sqrt{5}+1) / 2$. Every angle of the regular pentagon is $108^{\circ}$, so the inscribed angle corresponding to the sides in the circumscribed circle is $\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$ (Figure 2). The area of a triangle can be calculated from two sides and the sine of the included angle, so we have:

$$
\begin{aligned}
T_{A B C D E} & =T_{A E B}+T_{B E C}+T_{C E D}= \\
& =\frac{1}{2}\left(1 \cdot \frac{\sqrt{5}+1}{2} \cdot \sin 36^{\circ}+\left(\frac{\sqrt{5}+1}{2}\right)^{2} \cdot \sin 36^{\circ}+1 \cdot \frac{\sqrt{5}+1}{2} \cdot \sin 36^{\circ}\right)= \\
& =\frac{5+3 \sqrt{5}}{4} \cdot \sin 36^{\circ}, \\
T_{A B F E} & =2 \cdot T_{A E B}=1 \cdot \frac{\sqrt{5}+1}{2} \cdot \sin 36^{\circ}, \\
T_{B C D E F} & =T_{A B C D E}-T_{A B F E}=\frac{5+3 \sqrt{5}}{4} \cdot \sin 36^{\circ}-\frac{\sqrt{5}+1}{2} \cdot \sin 36^{\circ}= \\
& =\frac{3+\sqrt{5}}{4} \cdot \sin 36^{\circ} .
\end{aligned}
$$

Thus,

$$
T_{B C D E F}: T_{A B F E}=\frac{3+\sqrt{5}}{2+2 \sqrt{5}}=\frac{\sqrt{5}+1}{4}
$$

The triangle $A K E$ is isosceles with a vertex angle of $360^{\circ} / 5=72^{\circ}$, so the base angles are $54^{\circ}$. Therefore,

$$
M E K \varangle=A E K \varangle-A E M \varangle=54^{\circ}-36^{\circ}=18^{\circ}=\frac{M E F \varangle}{2},
$$

which means that $E K$ is the angle bisector in the triangle $M E F$. Thus, by the angle bisector theorem,

$$
\frac{M K}{K F}=\frac{M E}{E F}=\frac{\frac{\sqrt{5}+1}{4}}{1}=\frac{\sqrt{5}+1}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c16f8be79ba5059e77b8g-2.jpg?height=458&width=488&top_left_y=508&top_left_x=813)

Figure 2

Thus, for the point $F$, it holds that $M K: K F=T_{B C D E F}: T_{A B F E}$, and since there is only one such point on the ray $M K$, it follows from the lemma that the centroid of the $B C D E F$ concave pentagonal plate is $F$.

Remark. Generally, the centroid of a polygonal plate does not coincide with the centroid of the point system formed by the vertices of the polygon; this is only true for triangles. More details can be found in the article by Zoltán Bogdán related to problem B. 3295 and on Wikipedia [²]. Those who tried to determine the centroid of the concave pentagonal plate using the arithmetic mean of the vectors pointing to the vertices provided an incorrect solution.[^0]

[^0]:    ${ }^{1}$ Zoltán Bogdán: Remark on problem B. 3295 - on the centroids of figures, KöMaL, 50 (February 2000), 72-75. http://db.komal.hu/KomalHU/cikk.phtml?id=200059

    https://en.wikipedia.org/wiki/Center_of_mass.