Difference between revisions of "2017 AIME I Problems/Problem 8"
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Solution by Leesisi | Solution by Leesisi | ||
+ | |||
+ | ==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 19:16, 20 January 2018
Contents
[hide]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.