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

(Solution 1)
(Solution)
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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>.
  
{{alternate solutions}}
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=== Solution 2 ===
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Same as above except, the last paragraph of the above solution needs correction.
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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 13:38, 21 October 2010

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, $\left\lceil \frac{180 - \alpha}{\beta} \right\rceil = \boxed{081}$.

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 $C$) the extended line will form an angle $k \beta$ at point $B$. For the $k$th reflection to be just inside or at the point $C$, 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 (ProblemsAnswer KeyResources)
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