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Difference between revisions of "1994 AIME Problems/Problem 14"

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== Problem ==
 
== Problem ==
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A beam of light strikes <math>\overline{BC}\,</math> at point <math>C\,</math> with angle of incidence <math>\alpha=19.94^\circ\,</math> and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments <math>\overline{AB}\,</math> and <math>\overline{BC}\,</math> according to the rule: angle of incidence equals angle of reflection.  Given that <math>\beta=\alpha/10=1.994^\circ\,</math> and <math>AB=AC,\,</math> determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at <math>C\,</math> in your count.
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A beam of light strikes <math>\overline{BC}\,</math> at point <math>C\,</math> with angle of incidence <math>\alpha=19.94^\circ\,</math> and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments <math>\overline{AB}\,</math> and <math>\overline{BC}\,</math> according to the rule: angle of incidence equals angle of reflection.  Given that <math>\beta=\alpha/10=1.994^\circ\,</math> and <math>AB=BC,\,</math> determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at <math>C\,</math> in your count.
  
 
[[Image:AIME_1994_Problem_14.png]]
 
[[Image:AIME_1994_Problem_14.png]]
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
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At each point of reflection, we pretend instead that the light continues to travel straight.
 +
<asy>
 +
pathpen = linewidth(0.7); size(250);
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real alpha = 28, beta = 36;
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pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180);
 +
D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9));
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for(int i = 0; i < 180/alpha; ++i){
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path l = B -- (1+i/2)*expi(-i * alpha * pi / 180);
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D(l, linetype("4 4"));
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D(IP(l,r));
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} D(B);
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</asy>
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Note that after <math>k</math> reflections (excluding the first one at <math>C</math>) the extended line will form an angle <math>k \beta</math> at point <math>B</math>. For the <math>k</math>th reflection to be just inside or at point <math>B</math>, we must have <math>k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27</math>. Thus, our answer is, including the first intersection, <math>\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}</math>.
  
 
== See also ==
 
== See also ==
Line 11: Line 23:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 23:03, 27 December 2017

Problem

A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection. Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C\,$ in your count.

AIME 1994 Problem 14.png

Solution

At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9)); for(int i = 0; i < 180/alpha; ++i){ path l = B -- (1+i/2)*expi(-i * alpha * pi / 180); D(l, linetype("4 4")); D(IP(l,r)); } D(B); [/asy] Note that after $k$ reflections (excluding the first one at $C$) the extended line will form an angle $k \beta$ at point $B$. For the $k$th reflection to be just inside or at point $B$, we must have $k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27$. Thus, our answer is, including the first intersection, $\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}$.

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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