Difference between revisions of "1994 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | A beam of light strikes <math>\overline{BC}\,</math> at point <math>C\,</math> with angle of incidence <math>\alpha=19.94^\circ\,</math> and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments <math>\overline{AB}\,</math> and <math>\overline{BC}\,</math> according to the rule: angle of incidence equals angle of reflection. Given that <math>\beta=\alpha/10=1.994^\circ\,</math> and <math>AB= | + | A beam of light strikes <math>\overline{BC}\,</math> at point <math>C\,</math> with angle of incidence <math>\alpha=19.94^\circ\,</math> and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments <math>\overline{AB}\,</math> and <math>\overline{BC}\,</math> according to the rule: angle of incidence equals angle of reflection. Given that <math>\beta=\alpha/10=1.994^\circ\,</math> and <math>AB=BC,\,</math> determine the number of times the light beam will bounce off the two line segments. Include the first reflection at <math>C\,</math> in your count. |
[[Image:AIME_1994_Problem_14.png]] | [[Image:AIME_1994_Problem_14.png]] | ||
== Solution == | == 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 = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",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,D(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)); | ||
+ | } D(B); | ||
+ | </asy> | ||
+ | Note that after <math>k</math> 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 point <math>B</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 == | ||
{{AIME box|year=1994|num-b=13|num-a=15}} | {{AIME box|year=1994|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:03, 28 December 2017
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.
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.