AMC10 2024 A

AMC10 2024 A · Q13

AMC10 2024 A · Q13. It mainly tests Basic counting (rules of product/sum), Transformations.

Two transformations are said to commute if applying the first followed by the second gives the same result as applying the second followed by the first. Consider these four transformations of the coordinate plane: Of the $6$ pairs of distinct transformations from this list, how many commute?

如果先应用第一个变换再应用第二个变换的结果，与先应用第二个再应用第一个相同，则称这两个变换交换。将以下四个坐标平面变换考虑在内：在这 6 对不同的变换中，有几对交换？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Label the given transformations $T_1, T_2, T_3,$ and $T_4,$ respectively. The rules of transformations are: - $T_1: \ (x,y)\to(x+2,y)$ - $T_2: \ (x,y)\to(-y,x)$ - $T_3: \ (x,y)\to(x,-y)$ - $T_4: \ (x,y)\to(2x,2y)$ Note that: - Applying $T_1$ and then $T_2$ gives $(x,y)\to(x+2,y)\to(-y,x+2).$ Applying $T_2$ and then $T_1$ gives $(x,y)\to(-y,x)\to(-y+2,x).$ Therefore, $T_1$ and $T_2$ do not commute. One counterexample is the preimage $(0,0).$ - Applying $T_1$ and then $T_3$ gives $(x,y)\to(x+2,y)\to(x+2,-y).$ Applying $T_3$ and then $T_1$ gives $(x,y)\to(x,-y)\to(x+2,-y).$ Therefore, $T_1$ and $T_3$ commute. They form a glide reflection. - Applying $T_1$ and then $T_4$ gives $(x,y)\to(x+2,y)\to(2x+4,2y).$ Applying $T_4$ and then $T_1$ gives $(x,y)\to(2x,2y)\to(2x+2,2y).$ Therefore, $T_1$ and $T_4$ do not commute. One counterexample is the preimage $(0,0).$ - Applying $T_2$ and then $T_3$ gives $(x,y)\to(-y,x)\to(-y,-x).$ Applying $T_3$ and then $T_2$ gives $(x,y)\to(x,-y)\to(y,x).$ Therefore, $T_2$ and $T_3$ do not commute. One counterexample is the preimage $(1,0).$ - Applying $T_2$ and then $T_4$ gives $(x,y)\to(-y,x)\to(-2y,2x).$ Applying $T_4$ and then $T_2$ gives $(x,y)\to(2x,2y)\to(-2y,2x).$ Therefore, $T_2$ and $T_4$ commute. - Applying $T_3$ and then $T_4$ gives $(x,y)\to(x,-y)\to(2x,-2y).$ Applying $T_4$ and then $T_3$ gives $(x,y)\to(2x,2y)\to(2x,-2y).$ Therefore, $T_3$ and $T_4$ commute. Together, $\boxed{\textbf{(C)}~3}$ pairs of transformations commute: $T_1$ and $T_3, T_2$ and $T_4,$ and $T_3$ and $T_4.$ Remark To show that two transformations do not commute, we only need one counterexample.

分别将给定的变换标记为 $T_1, T_2, T_3,$ 和 $T_4$。变换规则为： - $T_1: \ (x,y)\to(x+2,y)$ - $T_2: \ (x,y)\to(-y,x)$ - $T_3: \ (x,y)\to(x,-y)$ - $T_4: \ (x,y)\to(2x,2y)$ 注意： - 先 $T_1$ 后 $T_2$：$(x,y)\to(x+2,y)\to(-y,x+2)$。先 $T_2$ 后 $T_1$：$(x,y)\to(-y,x)\to(-y+2,x)$。因此 $T_1$ 和 $T_2$ 不交换。反例 $(0,0)$。 - 先 $T_1$ 后 $T_3$：$(x,y)\to(x+2,y)\to(x+2,-y)$。先 $T_3$ 后 $T_1$：$(x,y)\to(x,-y)\to(x+2,-y)$。因此 $T_1$ 和 $T_3$ 交换。 - 先 $T_1$ 后 $T_4$：$(x,y)\to(x+2,y)\to(2x+4,2y)$。先 $T_4$ 后 $T_1$：$(x,y)\to(2x,2y)\to(2x+2,2y)$。因此 $T_1$ 和 $T_4$ 不交换。 - 先 $T_2$ 后 $T_3$：$(x,y)\to(-y,x)\to(-y,-x)$。先 $T_3$ 后 $T_2$：$(x,y)\to(x,-y)\to(y,x)$。因此 $T_2$ 和 $T_3$ 不交换。 - 先 $T_2$ 后 $T_4$：$(x,y)\to(-y,x)\to(-2y,2x)$。先 $T_4$ 后 $T_2$：$(x,y)\to(2x,2y)\to(-2y,2x)$。因此 $T_2$ 和 $T_4$ 交换。 - 先 $T_3$ 后 $T_4$：$(x,y)\to(x,-y)\to(2x,-2y)$。先 $T_4$ 后 $T_3$：$(x,y)\to(2x,2y)\to(2x,-2y)$。因此 $T_3$ 和 $T_4$ 交换。总共有 $\boxed{\textbf{(C)}~3}$ 对交换：$T_1$ 和 $T_3$，$T_2$ 和 $T_4$，$T_3$ 和 $T_4$。备注：证明两个变换不交换只需一个反例。

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