Difference between revisions of "1994 AIME Problems/Problem 14"
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== Solution == | == Solution == | ||
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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. | ||
Revision as of 01:50, 20 August 2011
Problem
A beam of light strikes 
at point 
with angle of incidence 
and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments 
and according to the rule: angle of incidence equals angle of reflection. Given that 
and 
determine the number of times the light beam will bounce off the two line segments. Include the first reflection at in your count.
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 = 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]](http://latex.artofproblemsolving.com/5/d/7/5d7543d66ce2d6d2c9f40193f6f340ad12927ad1.png)
Note that after 
reflections (excluding the first one at 
) the extended line will form an angle 
at point 
. For the th reflection to be just inside or at the point , we must have 
. Thus, our answer is, including the first intersection, 
.
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 | ||
