2024 AMC 12A Problems/Problem 18

Problem

Let $P_{1}$ and $P_{2}$ be distinct points in the plane, and for positive integers $n \geq 3$ , $P_{n}$ is constructed according to the following rules:

If $n$ is odd, then $P_{n}$ is obtained by rotating $P_{n - 2}$ about $P_{n - 1} ~ 60^{\circ}$ clockwise.
If $n$ is even, then $P_{n}$ is obtained by rotating $P_{n - 2}$ about $P_{n - 1} ~ 45^{\circ}$ clockwise.

What is the least positive integer $k > 1$ for which $P_{k} = P_{1}$ ?

$\textbf{(A)}~25 \qquad \textbf{(B)}~31\qquad \textbf{(C)}~37\qquad \textbf{(D)}~49 \qquad \textbf{(E)}~61$

Solution

Due to the repeating nature of the process, note that each of the segments $\overline{P_{2k-1}P_{2k}}$ for positve integers $k \geq 1$ is a rotation of another about a common center, which we shall call $O$ . In order to show that $O$ exists, note that the figure formed by $P_{1}P_{2}P_{3}P_{4}$ is congruent to the figure formed by $P_{3}P_{4}P_{5}P_{6}$ . Furthermore, there is a unique rotation that sends $\overline{P_{1}P_{2}}$ to $\overline{P_{3}P_{4}}$ , and it is the same as the rotation sending $\overline{P_{3}P_{4}}$ to $\overline{P_{5}P_{6}}$ , which will continue throughout the process. Therefore, all rotations have the same center $O$ .

In each such rotation, $P_{2k - 1}$ maps to $P_{2k + 1}$ , and $P_{2k}$ maps to $P_{2k + 2}$ . Thus all points with an odd index lie on a circle, all points with an even index lie on a circle, and both circles have center $O$ . Note that this also means if the process arrives at $P_{1}$ again, it must do so at an odd index. The location of $O$ is at the point where the perpendicular bisectors of segments $\overline{P_{2}P_{4}}$ and $\overline{P_{1}P_{3}}$ intersect.

Note that since $P_{2}P_{3} = P_{3}P_{4}$ (the distance between any two consecutive points is constant throughout the process), the perpendicular bisector of $\overline{P_{2}P_{4}}$ coincides with the angle bisector of $\angle P_{2}P_{3}P_{4}$ . Thus $\angle P_{2}P_{3}O = \tfrac{1}{2}\angle P_{2}P_{3}P_{4} = \tfrac{1}{2}(45^{\circ}) = 22.5^{\circ}$ . Also, $\triangle P_{1}P_{2}P_{3}$ is equilateral, so $\angle P_{1}P_{3}O = \angle P_{1}P_{3}P_{2} - \angle P_{2}P_{3}O = 60^{\circ} - 22.5^{\circ} = 37.5^{\circ}$ , and since $P_{1}O = P_{3}O$ , we have $\angle P_{1}OP_{3} = 180^{\circ} - 2 \cdot 37.5^{\circ} = 105^{\circ}$ .

Thus points with an odd index rotate $105^{\circ}$ about $O$ each cycle. Since $\tfrac{105^{\circ}}{360^{\circ}} = \tfrac{7}{24}$ , it will take $24$ cycles to reach the location of $P_{1}$ again, and that point will be at index $1 + 2 \cdot 24 = \boxed{\textbf{(D)}~49}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2024 AMC 12A Problems/Problem 18

Problem

Solution

See also