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Difference between revisions of "1981 AHSME Problems/Problem 20"

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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.} </math>
 
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.} </math>
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==Solution==
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Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be <math>0</math>, but that would be impossible. Therefore we start by working backwards. Since angle <math>CDA</math> is <math>8</math>, the reflection would give us a triangle with angles <math>16, 90</math>, and <math>74</math>. Then, when we reflect again, we will have <math>180 - 74 - 74</math> = <math>32</math>. Since the other side of the reflection when we had the <math>82</math> degrees had carried over to the other side, we have a <math>32-82-66</math> triangle.
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Notice that we keep decreasing by increments of <math>8</math>. This is because the starting angle was <math>8</math> and since we always have to decrease <math>8</math> every time and that every triangle has every increasing angles of <math>8</math>, we must decrease by <math>8</math> every time. This is the most optimal path of the light beam.
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The pattern of light will be <math>82-74-66-58-50-42-34-26-18-10</math>. When we get to the angle of <math>2</math> degrees, we have reached angle <math>A</math>. Therefore, we don't count the <math>2</math>, so our total number of reflections between <math>CD</math> and <math>AD</math> is <math>\boxed {(B) 10}</math>
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~Arcticturn

Latest revision as of 18:35, 23 October 2021

Problem

A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$). If $\measuredangle CDA=8^\circ$, what is the largest value $n$ can have?

[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$

Solution

Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$, but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$, the reflection would give us a triangle with angles $16, 90$, and $74$. Then, when we reflect again, we will have $180 - 74 - 74$ = $32$. Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle.

Notice that we keep decreasing by increments of $8$. This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$, we must decrease by $8$ every time. This is the most optimal path of the light beam.

The pattern of light will be $82-74-66-58-50-42-34-26-18-10$. When we get to the angle of $2$ degrees, we have reached angle $A$. Therefore, we don't count the $2$, so our total number of reflections between $CD$ and $AD$ is $\boxed {(B) 10}$

~Arcticturn

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