Difference between revisions of "2017 AIME I Problems/Problem 8"
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Solution by Leesisi | Solution by Leesisi | ||
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+ | ==Solution 3 (Quicker Trig)== | ||
+ | <asy> | ||
+ | pair O, P, Q, R; | ||
+ | draw(circle(O, 10)); | ||
+ | O = (10, 0); | ||
+ | P = (-10, 0); | ||
+ | Q = (10*cos(pi/3), 10*sin(pi/3)); | ||
+ | R = (10*cos(5*pi/6), 10*sin(5*pi/6)); | ||
+ | dot(Q); | ||
+ | dot(O); | ||
+ | dot(P); | ||
+ | dot(R); | ||
+ | draw(P--O--Q--P--R--O); | ||
+ | draw(Q--R, red); | ||
+ | label("$O$", O, 2*E); | ||
+ | label("$P$", P, 2*W); | ||
+ | label("$Q$", Q, NE); | ||
+ | label("$R$", R, NW); | ||
+ | label("$200$", (0,0), 2*S); | ||
+ | label("$x$", (Q+R)/2, N); | ||
+ | draw(rightanglemark(O, Q, P, 38)); | ||
+ | draw(rightanglemark(O, R, P, 38)); | ||
+ | </asy> | ||
+ | Let <math>QR=x.</math> Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: <math>OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.</math> Now observe that quadrilateral <math>OQRP</math> is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it: | ||
+ | <cmath>200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),</cmath> | ||
+ | <cmath>x + 200 (\cos a \sin b) = 200 (\sin a \cos b),</cmath> | ||
+ | <cmath>x = 200(\sin a \cos b - \sin b \cos a),</cmath> | ||
+ | <cmath>x = 200 \sin(a-b).</cmath> | ||
+ | We want <math>|x| \le 100</math> (the absolute value comes from the fact that <math>a</math> is not necessarily greater than <math>b,</math> so we cannot assume that <math>Q</math> is to the right of <math>R</math> as in the diagram), so we substitute: | ||
+ | <cmath>|200 \sin(a-b)| \le 100,</cmath> | ||
+ | <cmath>|\sin(a-b)| \le \frac{1}{2},</cmath> | ||
+ | <cmath>|a-b| \le 30 ^\circ,</cmath> | ||
+ | <cmath>-30 \le a-b \le 30.</cmath> | ||
+ | By simple geometric probability (see Solution 2 for complete explanation), <math>\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},</math> so <math>m+n = \boxed{041}.</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=7|num-a=9}} | {{AIME box|year=2017|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:16, 20 January 2018
Problem 8
Two real numbers and are chosen independently and uniformly at random from the interval . Let and be two points on the plane with . Let and be on the same side of line such that the degree measures of and are and respectively, and and are both right angles. The probability that is equal to , where and are relatively prime positive integers. Find .
Solution 1
Noting that and are right angles, we realize that we can draw a semicircle with diameter and points and on the semicircle. Since the radius of the semicircle is , if , then must be less than or equal to .
This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:
Given such that , what is the probability that ? Through simple geometric probability, we get that .
The answer is
~IYN~
Solution 2 (Trig Bash)
Put and with on the origin and the triangles on the quadrant. The coordinates of and is , . So = , which we want to be less then . So So we want , which is equivalent to or . The second inequality is impossible so we only consider what the first inequality does to our by box in the plane. This cuts off two isosceles right triangles from opposite corners with side lengths from the by box. Hence the probability is and the answer is
Solution by Leesisi
Solution 3 (Quicker Trig)
Let Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: Now observe that quadrilateral is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it: We want (the absolute value comes from the fact that is not necessarily greater than so we cannot assume that is to the right of as in the diagram), so we substitute: By simple geometric probability (see Solution 2 for complete explanation), so
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.