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

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At each point of reflection, we pretend instead that the light continues to travel straight.  
 
At each point of reflection, we pretend instead that the light continues to travel straight.  
  
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<asy>
 
<asy>
 
pathpen = linewidth(0.7); size(250);  
 
pathpen = linewidth(0.7); size(250);  
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Then each intersection of the extended line with the rotated segments corresponds to a reflection in the original problem. We quickly see that the extended line will intersect each rotation of the angle by <math>k \beta</math> until <math>k\beta \ge 180 - \alpha \Longrightarrow k \ge \frac{180 - \alpha}{\beta}</math>. Thus, our answer is, including the first intersection, <math>\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{083}</math>.
 
Then each intersection of the extended line with the rotated segments corresponds to a reflection in the original problem. We quickly see that the extended line will intersect each rotation of the angle by <math>k \beta</math> until <math>k\beta \ge 180 - \alpha \Longrightarrow k \ge \frac{180 - \alpha}{\beta}</math>. Thus, our answer is, including the first intersection, <math>\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{083}</math>.
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{{alternate solutions}}
  
 
== See also ==
 
== See also ==

Revision as of 15:41, 3 January 2009

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 $AB=AC,\,$ 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

Solution 1

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 = D(MP("B",(0,0))), C = MP("C",D((1,0))), A = MP("A",D(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,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)); } [/asy]

Then each intersection of the extended line with the rotated segments corresponds to a reflection in the original problem. We quickly see that the extended line will intersect each rotation of the angle by $k \beta$ until $k\beta \ge 180 - \alpha \Longrightarrow k \ge \frac{180 - \alpha}{\beta}$. Thus, our answer is, including the first intersection, $\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{083}$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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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