中点五边形面积比介于1/2到3/4之间

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The sidesplitting story of the midpoint polygon
The Mathematics Teacher; Apr 1994; 87, 4; Social Science Premium Collection
pg. 249
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Area ratios greater than $1 / 2$. In figure $\mathbf{8 a}$, FGHIJ is the midpoint polygon for pentagon $A B C D E$. In $\triangle A B E$, if we choose a point $K$ (we have chosen the intersection of $\overline{B E}$ and $\overline{A D}$ ) on $\overline{B E}$ (which we know is parallel to $\overline{J F}$ from the sidesplitting theorem), then $\triangle J F K$ and $\triangle A F J$ have the same area. Similarly, in figure $\mathbf{8 b}$, $\triangle F G L$ and $\triangle F G B, \triangle G H M$ and $\triangle G H C, \triangle H I N$ and $\triangle H I D$, and $\triangle I J O$ and $\triangle I J E$, by pairs, have the same area. This process is like moving the triangles as we did with quadrilaterals. In this case the triangles, though not congruent, still have the same area. Since pentagon FGHIJ is not fully covered when the outside triangles are moved inside, the area of the midpoint pentagon is greater than half the area of the original pentagon $A B C D E .$

Area ratios less than $3 / 4$. The proof that the ares ratios for pentagons are less than $3 / 4$ is best understood in terms of what has been called flexing (Anderson and Arcidiacono 1989). Suppose that $A,B, C, D$, and $E$ are five consecutive vertices of any polygon. To flex a polygon at the vertex $C$ means that the polygon is deformed in such a way that $C$ moves along the line that passes through $C$ and is parallel to $\overline{B D}$ (see fig. 9). Notice that the area of the polygon remains unchanged because the area the moving part, $\triangle B C D$, does not change. We can also flex the polygon at $C$ even if $B, C$, and $D$ are collinear, and again, the area of the polygon does not change under flexing. We shall consider flexing that make the polygon more degenerate, that is, the polygon will approach becoming a polygon with fewer sides. For example, in figure $\mathbf{9}$, if $A B C D E$ are the vertices of a pentagon, then it becomes a quadrilateral when the vertex $C$ is translated along to the intersection of $\overleftrightarrow{C C^{\prime}}$ and $\overleftrightarrow{A B}$.

What happens to the midpoint polygon when a polygon is flexed? Let $F, G, H$, and $I$ be the midpoints of the consecutive sides $A B, B C, C D$, and $D E$ of a polygon (fig. 10). Suppose $E$ is closer to $\overleftrightarrow{B D}$ than $A$ is; then $I$ is closer to $\overleftrightarrow{G H}$ than $F$ is. Let us flex $C$ to $C^{\prime}$, Since $G H=G^{\prime} H^{\prime}=(1 / 2) B D, G G^{\prime}=H H^{\prime}$ and area $\left(\triangle G G^{\prime} F\right)$ is greater than area $\left(\triangle H H^{\prime} I\right)$ (because the height of $\triangle G G^{\prime} F$ is greater than the height of $\left.\triangle H H^{\prime} I\right)$, making area $\left(F G^{\prime} H^{\prime} I\right)$ greater than area $(F G H I)$. Since the area of the original polygon does not change, we have increased the area ratio by flexing. (If we had flexed $C$ to $C^{\prime \prime}$, a similar argument shows that the area ratio decreases. If we flex $C$ past $C^{\prime}$ or $C^{\prime \prime}$, then the pentagon becomes nonconvex.) The limit for the area ratio by flexing at any vertex occurs when the vertex becomes collinear with an adjacent side.

We can then show that the area ratio of a convex pentagon is less than $3 / 4$. Figure $\bf11$ shows the process of flexing pentagon $A B C D E$ four times into a limiting triangle. Figure $\bf11 a$ shows pentagon $A B C D E$ and its associated midpoint pentagon $F G H I J$. First we flex $B$ to $B^{\prime}$ on $\overleftrightarrow{D C}$ ) because $E$ is closer to $\overline{A C}$ than $D$ is [fig. $\bf11b$]). Then we flex vertex $A$ to $A^{\prime}$ on $\overleftrightarrow{D E}$ (because $C$ is closer to $\overrightarrow{B^{\prime} E}$ than $D$ is [fig. $\bf11c$]). Next we flex $C$ to $C^{\prime}$ on $\overline{B^{\prime} A^{\prime}}$ (because $E$ is closer to $\overline{B^{\prime} D}$ than $A^{\prime}$ is [fig. $\bf11d$]). (Note that $C^{\prime}$ is also $B^{\prime}$ and consequently also $G^{\prime \prime}$.) Finally we flex vertex $E$ to $A^{\prime}$ (which is also $J^{\prime \prime}$ ) (fig. 11e). In figure $\bf11e$ we see that area $\left(F^{\prime} G^{\prime \prime} H^{\prime} T^{\prime} J^{\prime \prime}\right)=3 / 4$ area $\left(A^{\prime} B^{\prime} D\right)$ as a consequence of the sidesplitting theorem for triangles. Because each step in this specific flexing process increases the area ratio, and because the resulting triangle obviously is not really a pentagon, we know that the area ratio for the original pentagon must be strictly less than $3 / 4$.

Convex pentagons exist whose area ratios are arbitrarily close to $1 / 2$ or $3 / 4$ (see pentagon₁ and pentagon₂ in fig. 12). By continually deforming pentagon₁ into pentagon₂, we obtain a family of convex pentagons whose area ratios range between the area ratios of pentagon₁ and pentagon₂. It follows that any number between $1 / 2$ and $3 / 4$ is the area ratio of some convex pentagon.

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What about hexagons and other $n$-gons? For the upper bound, the arguments presented previously still apply, but the degenerate polygon (limiting triangle) one obtains at the end of the flexing process will coincide with its own midpoint polygon, making the area ratio 1 (see fig. 13 for an example). For the lower bound, the argument presented earlier for pentagons can be applied to general $n$-gons to show that if $n>4$, the area ratios of $n$-gons are greater than $1 / 2$. So in general we have the following theorem:

AREA RATIO FOR GENERAL POLYGONS THEOREM. The range of the area ratios for $n$-gons$(n>5)$ is between $1 / 2$ and $1 .$
Students can demonstrate that the area ratios of polygons are greater than $1 / 2$ with graph paper using the midpoint formula and other ideas from coordinate geometry. Suggest that they use lattice points as vertices of the original polygon to make the identification of the midpoints easier. Introduce an origin and axis on the drawing and have students find the midpoints of each side of their $n$-gon (see fig. 14a). After constructing their midpoint polygon, students can use the process for the pentagons shown in figure 8 with their polygons. Students will find it much easier to see if they use one color for the original polygon, a second color for their midpoint polygon, and a third color to draw the diagonals needed. The intersections of diagonals that connect every other point on the original polygon can be used to identify the triangles inside their midpoint polygon that have the same areas as the triangles formed outside the midpoint polygon (see fig. 14b). Students will be able to see that the triangles do not cover the entire area of the midpoint polygon. This activity makes continual use of the sidesplitting theorem for triangles and will strengthen students' understanding of the area of a triangle as well. The dot-paper activities involving Pick's theorem described earlier could be used to demonstrate the area relationships for general polygons as well.


REGULAR POLYGONS
What is the pattern for regular polygons? In this situation, the midpoint polygon is also regular and, therefore, similar to the original polygon. So the area ratio is the square of the ratio between the lengths of the sides of the midpoint polygon and those of the original polygon. From figure 15 we can see that the latter ratio is $\cos (180 \%$ ). So the area ratio for a regular $n$-gon is $\cos ^{2}\left(180^{\circ} / n\right) .$ For $n=5,6,7$, the area ratios are approximately $.65, .75$, and $. 81$, respectively. Note that the area ratio approaches 1 as $n$ gets larger, which means that as the number of sides $(n)$ increases, the area outside the midpoint polygon gets smaller.

Archimedes used area ratios for regular polygons and their midpoint polygons in a very different context to arrive at an approximation of pi (Dijksterhuis 1987). In figure 16, the inscribed hexagon is the midpoint polygon of the circumscribed hexagon. Archimedes reasoned that the area of the circle of radius one unit $(\pi)$ was larger than the inscribed polygon and smaller than the circumscribed polygon. So he calculated the area of each hexagon and was able to limit the range of the value of pi. He then doubled the number of sides and calculated the area of each polygon again. Through this process of doubling sides, he eventually arrived at a polygon with 96 sides and estimated pi correctly to two decimal places. At the same time he was also demonstrating that as the number of sides of the regular polygon increased, the area of the associated midpoint polygon approached the area of the original polygon.

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对于n边形,中点多边形的面积比的最大值是多少呢?文中只说了小于1
能否用五边形类似的flexing方法