Difference between revisions of "2017 AIME I Problems/Problem 8"
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<cmath>(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} </cmath> | <cmath>(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} </cmath> | ||
<cmath> \sin^{2} (b-a) \le \frac{1}{4} </cmath> | <cmath> \sin^{2} (b-a) \le \frac{1}{4} </cmath> | ||
− | So we want <math> -\frac{1}{2} \le \sin b-a \le \frac{1}{2} </math> or <math> -30 \le b-a \le 30</math>. | + | So we want <math> -\frac{1}{2} \le \sin b-a \le \frac{1}{2} </math> or <math> -30 \le b-a \le 30</math>. This cuts off two isosceles right triangles from opposite corners with side length 45 from the <math>75</math> by <math>75</math> box. Hence the probability is <math>1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}</math> and the answer is <math>16+25 = \boxed{41}</math> |
Solution by Leesisi | Solution by Leesisi |
Revision as of 16:32, 10 March 2017
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
or
. This cuts off two isosceles right triangles from opposite corners with side length 45 from the
by
box. Hence the probability is
and the answer is
Solution by Leesisi
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.