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2022 AMC 10A 第 18 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

18.

设 TkT_k 是平面坐标变换：先将平面绕原点逆时针旋转 kk 度，再关于 yy 轴反射。求最小的正整数 nn，使得依次执行变换 T1,T2,T3,⋯ ,TnT_1, T_2, T_3, \cdots, T_n 后，点 (1,0)(1,0) 回到自身。

Let TkT_k be the transformation of the coordinate plane that first rotates the plane kk degrees counterclockwise around the origin and then reflects the plane across the yy-axis. What is the least positive integer nn such that performing the sequence of transformations T1,T2,T3,⋯ ,TnT_1, T_2, T_3, \cdots, T_n returns the point (1,0)(1,0) back to itself?

359359

360360

719719

720720

721721

答案：A
知识点：变换找规律
难度评级：1950
解答：

因为题目涉及角度和反射，使用极坐标会更方便。

设极坐标为 (r,θ)(r, \theta)。逆时针旋转 kk 度后变为 (r,θ+k∘)(r, \theta + k^{\circ})，再反射后变为 (r,180∘−θ−k∘)(r, 180^{\circ} - \theta - k^{\circ})。

因此 Tk(r,θ)=(r,180∘−θ−k∘). T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}).

由此可见 Tk+1(Tk(r,θ))= T_{k + 1}(T_k(r, \theta)) = Tk+1(r,180∘−θ−k∘)=T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) = (r,θ−1∘).(r, \theta - 1^{\circ}).

现在分析点 (1,0∘)(1, 0^{\circ}) 的变化。

经过 T1T_1 后，得到 (1,179∘)(1, 179^{\circ})。

经过 T2T_2 后，得到 (1,−1∘)(1, -1^{\circ})。

经过 T3T_3 后，得到 (1,178∘)(1, 178^{\circ})。

经过 T4T_4 后，得到 (1,−2∘)(1, -2^{\circ})。

⋮\vdots

经过 T2m−1T_{2m - 1} 后，得到 (1,180∘−m∘)(1, 180^{\circ} - m^{\circ})。

经过 T2mT_{2m} 后，得到 (1,−m∘)(1, -m^{\circ})。

由此可见，角度第一次回到 0∘0^{\circ} 是在 T2(180)−1=T359T_{2(180)-1}=T_{359} 之后。因此 n=359n=359。

所以正确答案是 A。

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with.

Let (r,θ)(r, \theta) be a polar coordinate. Rotating this by kk degrees counterclockwise maps the point to (r,θ+k∘)(r, \theta + k^{\circ}) and then reflecting it maps it to (r,180∘−θ−k∘).(r, 180^{\circ} - \theta - k^{\circ}).

Therefore, we have that Tk(r,θ)=(r,180∘−θ−k∘). T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}).

From this, we can see that Tk+1(Tk(r,θ))= T_{k + 1}(T_k(r, \theta)) = Tk+1(r,180∘−θ−k∘)=T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) = (r,θ−1∘).(r, \theta - 1^{\circ}).

Now, let's analyze what happens to the point (1,0∘).(1, 0^{\circ}).

After T1,T_1, we get (1,179∘).(1, 179^{\circ}).

After T2,T_2, we get (1,−1∘).(1, -1^{\circ}).

After T3,T_3, we get (1,178∘).(1, 178^{\circ}).

After T4,T_4, we get (1,−2∘).(1, -2^{\circ}).

⋮\vdots

After T2m−1,T_{2m - 1}, we get (1,180∘−m∘).(1, 180^{\circ} - m^{\circ}).

After T2m,T_{2m}, we get (1,−m∘).(1, -m^{\circ}).

From this, we can see that the first time the angle is back to 0∘0^{\circ} is after T2(180)−1=T359.T_{2(180)-1}=T_{359}. Therefore n=359.n=359.

Thus, A is the correct answer.

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