Difference between revisions of "1994 AIME Problems/Problem 14"
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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{081}</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{081}</math>. | ||
| − | {{ | + | === Solution 2 === |
| + | Same as above except, the last paragraph of the above solution needs correction. | ||
| + | |||
| + | Note that after k 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 the point <math>C</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 == | ||
Revision as of 12:38, 21 October 2010
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
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]](http://latex.artofproblemsolving.com/5/d/7/5d7543d66ce2d6d2c9f40193f6f340ad12927ad1.png)
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 
until 
. Thus, our answer is, including the first intersection, 
.
Solution 2
Same as above except, the last paragraph of the above solution needs correction.
Note that after k 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 | ||
