几何学 - 作业 1

T1

Draw a planar picture with at least 3 different non trivial symmetry transformations. Describe all its symmetry.

中文翻译

绘制平面图描绘至少三个非平凡的对称变换，并描述其对称性.

The picture above illustrates 3 diverse non trivial symmetry transformations.

1. The first one shows the transformation of reflecting the figure $△ABC$ across the line $\ell$ . This is the reflective symmetry in textbook.
2. The second one shows the transformation of rotating the ellipse $\Gamma$ $360^{\circ }$ around point $O$ . This is the cyclic symmetry.
3. The third one depicts the transformation of translating figure $\Sigma$ to ${\Sigma }^{\prime }$ where a fixed point $P$ overlapping its center of gravity.

T2

Classify all simple quadrilaterals based on their number of symmetry transformations.

中文翻译

将所有简单四边形基于它们对称变换的数量分类.

First, let’s denote four vertices with $v_1,v_2,v_3,v_4$ , and ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ for their interior angle. And we denote four edges with

$$\{\begin{array}{l}{f}_1=v_1{v}_2\\ f_2=v_2{v}_3\\ f_3=v_3{v}_4\\ f_4=v_4{v}_1\end{array}$$

and their length with $l_1,l_2,l_3,l_4$ .

If none of transmation $\phi (\cdot )$ is symmetric, then we got a quadrilateral with only $1$ symmetry $\mathrm{id}$ .

If a transformation $\phi (\cdot )$ is symmetric, then $\phi (v_1),\phi (v_2),\phi (v_3),\phi (v_4)$ will still be cyclic. Then we need to firstly categorize the question into 3 means:

① $\phi (v_1)=v_2$ .

If $\phi (v_4)=v_1$ , then

$$\phi (v_2)=v_3,\phi (v_3)=v_4$$

It means we can rotate $v_1$ to $v_2$ and then get an overlapping shape, which implies ${\alpha }_1={\alpha }_2$ . And the same goes to ${\alpha }_2={\alpha }_3$ and ${\alpha }_3={\alpha }_4$ . Moreover,

$${\alpha }_1={\alpha }_2={\alpha }_3={\alpha }_4,l_1=l_2=l_3=l_4$$

Therefore we have a square. It has 8 symmetries.

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If $\phi (v_4)=v_3$ , then

$$\phi (v_2)=v_1,\phi (v_3)=v_4$$

This implies:

$$l_4=l_2,{\alpha }_1={\alpha }_2,{\alpha }_3={\alpha }_4$$

Thus we have a trapezium. If it is a rectangle, then we have $4$ symmetries. Otherwise, we only have $2$ symmetries.

② $\phi (v_1)=v_4$ .

The same as ① .

③ $\phi (v_1)=v_3$ .

If $\phi (v_2)=v_4$ , then

$$\phi (v_3)=\phi (v_1),\phi (v_4)=v_2$$

which implies:

$${\alpha }_1={\alpha }_3,{\alpha }_2={\alpha }_4,l_1=l_2,l_3=l_4$$

Then it is a diamond and it has $4$ symmetries.

In conclusion, we have totally 4 types that has $1,2,4,8$ symmetries. $\square$