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

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== Solution ==
 
== Solution ==
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{{solution}}
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=== Solution 1 ===
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At each point of reflection, we pretend instead that the light continues to travel straight.
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{{image}}
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<asy>
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pathpen = linewidth(0.7); size(250);
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real alpha = 28, beta = 36;
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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);
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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));
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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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}
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</asy>
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 19:34, 6 December 2008

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.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

[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, $\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{083}$.

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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