|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[2022 AMC 10A Problems/Problem 18]] |
| − | | |
| − | Let <math>T_k</math> be the transformation of the coordinate plane that first rotates the plane <math>k</math> degrees counter-clockwise around the origin and then reflects the plane across the <math>y</math>-axis. What is the least positive
| |
| − | integer <math>n</math> such that performing the sequence of transformations <math>T_1, T_2, T_3, \cdots, T_n</math> returns the point <math>(1,0)</math> back to itself?
| |
| − | | |
| − | <math>\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 </math>
| |
| − | | |
| − | ==Solution 1==
| |
| − | Note that since we're reflecting across the <math>y</math>-axis, if the point ever makes it to <math>(-1,0)</math> then it will flip back to the original point. Note that after <math>T_1</math> the point will be <math>1</math> degree clockwise from the negative <math>x</math>-axis. Applying <math>T_2</math> will rotate it to be <math>1</math> degree counterclockwise from the negative <math>x</math>-axis, and then flip it so that it is <math>1</math> degree clockwise from the positive <math>x</math>-axis. Therefore, after every <math>2</math> transformations, the point rotates <math>1</math> degree clockwise. To rotate it so that it will rotate <math>179</math> degrees clockwise will require <math>179 \cdot 2 = 358</math> transformations. Then finally on the last transformation, it will rotate on to <math>(-1,0)</math> and then flip back to it's original position. Therefore, the answer is <math>358+1 = 359 = \boxed{A}</math>
| |
| − | | |
| − | ~KingRavi
| |
| − | | |
| − | ==Solution 2==
| |
| − | Let <math>A_{n}</math> be the point <math>(\cos n^{\circ}, \sin n^{\circ})</math>.
| |
| − | | |
| − | Starting with <math>n=0</math>, the sequence goes <cmath>A_{0}\rightarrow A_{179}\rightarrow A_{359}\rightarrow A_{178}\rightarrow A_{358}\rightarrow A_{177}\rightarrow A_{357}\rightarrow\cdots</cmath>
| |
| − | | |
| − | We see that it takes <math>2</math> turns to downgrade the point by <math>1^{\circ}</math>. Since the fifth point in the sequence is <math>A_{177}</math>, the answer is <math>5+2(177)=\boxed{\textbf{(A)}~359}</math>
| |
| − | | |
| − | ==Video Solution==
| |
| − | | |
| − | https://youtu.be/QQrsKTErJn8
| |
| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
| |
| − | | |
| − | ==See also==
| |
| − | {{AMC10 box|year=2022|ab=A|num-b=17|num-a=19}}
| |
| − | {{AMC12 box|year=2022|ab=A|num-b=17|num-a=19}}
| |
| − | {{MAA Notice}}
| |
