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Difference between revisions of "2024 AMC 12A Problems/Problem 18"

(Removed solution 4, which was not relavent to math)
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==Problem==
 
==Problem==
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Let <math>P_{1}</math> and <math>P_{2}</math> be distinct points in the plane, and for positive integers <math>n \geq 3</math>, <math>P_{n}</math> is constructed according to the following rules:
  
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On top of a rectangular card with sides of length <math>1</math> and <math>2+\sqrt{3}</math>, an identical card is placed so that two of their diagonals line up, as shown (<math>\overline{AC}</math>, in this case).
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* If <math>n</math> is odd, then <math>P_{n}</math> is obtained by rotating <math>P_{n - 2}</math> about <math>P_{n - 1} ~ 60^{\circ}</math> clockwise.
 +
* If <math>n</math> is even, then <math>P_{n}</math> is obtained by rotating <math>P_{n - 2}</math> about <math>P_{n - 1} ~ 45^{\circ}</math> clockwise.
  
−
<asy> defaultpen(fontsize(12)+0.85); size(150); real h=2.25; pair C=origin,B=(0,h),A=(1,h),D=(1,0),Dp=reflect(A,C)*D,Bp=reflect(A,C)*B; pair L=extension(A,Dp,B,C),R=extension(Bp,C,A,D); draw(L--B--A--Dp--C--Bp--A); draw(C--D--R); draw(L--C^^R--A,dashed+0.6); draw(A--C,black+0.6); dot("$C$",C,2*dir(C-R)); dot("$A$",A,1.5*dir(A-L)); dot("$B$",B,dir(B-R)); </asy>
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What is the least positive integer <math>k > 1</math> for which <math>P_{k} = P_{1}</math>?
  
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Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled <math>B</math> in the figure?
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<math>\textbf{(A)}~25 \qquad \textbf{(B)}~31\qquad \textbf{(C)}~37\qquad \textbf{(D)}~49 \qquad \textbf{(E)}~61</math>
  
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<math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{No new vertex will land on }B.</math>
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==Solution==
 +
Due to the repeating nature of the process, note that each of the segments <math>\overline{P_{2k-1}P_{2k}}</math> for positve integers <math>k \geq 1</math> is a rotation of another about a common center, which we shall call <math>O</math>. In order to show that <math>O</math> exists, note that the figure formed by <math>P_{1}P_{2}P_{3}P_{4}</math> is congruent to the figure formed by <math>P_{3}P_{4}P_{5}P_{6}</math>. Furthermore, there is a unique rotation that sends <math>\overline{P_{1}P_{2}}</math> to <math>\overline{P_{3}P_{4}}</math>, and it is the same as the rotation sending <math>\overline{P_{3}P_{4}}</math> to <math>\overline{P_{5}P_{6}}</math>, which will continue throughout the process. Therefore, all rotations have the same center <math>O</math>.
  
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==Solution 1==
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In each such rotation, <math>P_{2k - 1}</math> maps to <math>P_{2k + 1}</math>, and <math>P_{2k}</math> maps to <math>P_{2k + 2}</math>. 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 <math>O</math>. Note that this also means if the process arrives at <math>P_{1}</math> again, it must do so at an odd index. The location of <math>O</math> is at the point where the perpendicular bisectors of segments <math>\overline{P_{2}P_{4}}</math> and <math>\overline{P_{1}P_{3}}</math> intersect.
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Let the midpoint of <math>AC</math> be <math>P</math>.
 
  
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We see that no matter how many moves we do, <math>P</math> stays where it is.
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Note that since <math>P_{2}P_{3} = P_{3}P_{4}</math> (the distance between any two consecutive points is constant throughout the process), the perpendicular bisector of <math>\overline{P_{2}P_{4}}</math> coincides with the angle bisector of <math>\angle P_{2}P_{3}P_{4}</math>. Thus <math>\angle P_{2}P_{3}O = \tfrac{1}{2}\angle P_{2}P_{3}P_{4} = \tfrac{1}{2}(45^{\circ}) = 22.5^{\circ}</math>. Also, <math>\triangle P_{1}P_{2}P_{3}</math> is equilateral, so <math>\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}</math>, and since <math>P_{1}O = P_{3}O</math>, we have <math>\angle P_{1}OP_{3} = 180^{\circ} - 2 \cdot 37.5^{\circ} = 105^{\circ}</math>.
  
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Now we can find the angle of rotation (<math>\angle APB</math>) per move with the following steps:
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Thus points with an odd index rotate <math>105^{\circ}</math> about <math>O</math> each cycle. Since <math>\tfrac{105^{\circ}}{360^{\circ}} = \tfrac{7}{24}</math>, it will take <math>24</math> cycles to reach the location of <math>P_{1}</math> again, and that point will be at index <math>1 + 2 \cdot 24 = \boxed{\textbf{(D)}~49}</math>.
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<cmath>AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}</cmath>
 
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<cmath>1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB</cmath>
 
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<cmath>1=2(2+\sqrt{3})(1-\cos\angle APB)</cmath>
 
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<cmath>\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}</cmath>
 
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<cmath>\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}</cmath>
 
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<cmath>\cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}</cmath>
 
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<cmath>\angle APB=30^\circ</cmath>
 
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Since Vertex <math>C</math> is the closest one and <cmath>\angle BPC=360-180-30=150</cmath>
 
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Vertex C will land on Vertex B when <math>\frac{150}{30}+1=\fbox{(A) 6}</math> cards are placed.
 
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(someone insert diagram maybe)
 
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~lptoggled, minor Latex edits by eevee9406
 
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==Solution 2==
 
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[[Image:2024_amc12A_p18.png|thumb|center|600px|]]
 
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Let AC intersect BD at O,
 
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We want to find <math>\angle AOB </math>
 
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Since <math>tan(75^\circ) = 2+ \sqrt{3} =\frac{AD}{AB} </math>,  <math>\angle CBD = \angle BCA = 15^\circ </math>
 
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<cmath> \angle AOB  = \angle CBD  + \angle BCA  =30^\circ  </cmath>
 
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So each time we rotate BD to AC for <math>30^\circ </math>,  and we need to rotate <math> 180^\circ / 30^\circ = 6 </math> times to overlap a point with B
 
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Therefore, the answer is <math>\fbox{\textbf{(A) } 6}</math>
 
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Note: If you don't remember  <math>tan(75^\circ)</math>
 
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<math> tan(75^\circ) = \frac{tan(45^\circ) + tan(30^\circ)}{ 1 - tan(45^\circ)\cdot tan(30^\circ)} </math>
 
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<math> = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} </math>
 
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<math> = \frac{(\sqrt{3}+1)^2  }{ (\sqrt{3})^2-1} = 2+ \sqrt{3} </math>
 
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 
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==Solution 3(In case you have no time and that's what I did) ==
 
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<math>\tan{15}=\frac{\sin{15}}{\cos{15}}=\frac{1}{2+\sqrt3}</math> and it eliminates all options except <math>6</math> and <math>12</math>. After one rotation it has turned <math>30^{\circ}</math>, so to satisfy the problem, divide <math>\frac{180}{30}</math> and get <math>\boxed{\textbf{A. }6}</math>.
 
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==Solution 4==
 
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Memorize that in a 15-75-90 right triangle, the ratios of the side lengths are <math>1</math>, <math>2+\sqrt3</math>, and <math>\sqrt2+\sqrt6</math>. So, we have that the diagonal of the rectangle forms a 15 degree angle. Drawing it out we see the answer is <math>\boxed{\textbf{(A) }6}</math>, and this makes sense because 15 times 6 is 90, thus rotating a vertex back to B.
 
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==Solution 5 (the simplest solution ever)==
 
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Look at the picture and draw the next one and continue draw down the line and then when you first hit B, count how many rectangles you’ve drawn (excluding the first which hasn’t been rotated), and you should get <math>\boxed{\textbf{(A) or 6}}</math> as the answer.
 
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~EaZ_Shadow
 
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==Solution 6==
 
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[[File:Rotation of rectangle.png|250px|right]]
 
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Process is the rotation around the center of the card point <math>O</math> at the angle <math>\alpha = \angle AOB.</math>
 
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<cmath>AO = BO = R, BD^2 = 4R^2 = AB^2 + AD^2 = 4 \cdot (2+\sqrt{3}).</cmath>
 
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By applying the Law of Cosines, we get
 
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<cmath>2R^2 (1-\cos \alpha) = AB^2 \implies \cos \alpha = \frac {\sqrt{3}}{2} \implies</cmath>
 
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<cmath>\alpha = 30^\circ \implies \boxed{\textbf{(A) or 6}}. </cmath>
 
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'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2024|ab=A|num-b=17|num-a=19}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:43, 20 March 2025

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 $1 + 2 \cdot 24 = \boxed{\textbf{(D)}~49}$.

See also

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

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