2001 AMC 12 Problems/Problem 17
Contents
Problem
A point is selected at random from the interior of the pentagon with vertices
,
,
,
, and
. What is the probability that
is obtuse?
Solution
The angle is obtuse if and only if
lies inside the circle with diameter
. (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)
The area of is
, and the area of
is
.
From the Pythagorean theorem the length of is
, thus the radius of the circle is
, and the area of the half-circle that is inside
is
.
Therefore the probability that is obtuse is
.
Solution 1.5
Remember that an obtuse angle is any angle greater than a right angle (90 degrees). This means that if we can figure out all of the points at which creates a right angle with
and
, any points within these bounds will yield an obtuse angle. Notice that by the Pythagorean theorem, the length of
is
. Also notice that when there is a right angle at
,
becomes the hypotenuse. Therefore, the sum of the squares of lengths
and
(or the "legs") must equal
, or
. This can be written as an equation.
If Point is situated at
, then the distance from
to
is as follows:
. Accordingly, the distance from
to
is
. Since absolute value and squaring (because we will be squaring the lengths) both eliminate negatives, we can ignore the absolute value and solve. Setting the sum of the squares of these lengths equal to 20 yields
. Completing the square for both
and
, we get
, or the equation of a circle with center
and radius
. Graphing both the pentagon and the circle, we see that the area enclosed is a semicircle. This is
or
.
Calculating the area of the pentagon yields
. The probability of point P being within the bounds of the semicircle is then
or
.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.