2002 AMC 12A Problems/Problem 22
Problem
Triangle is a right triangle with
as its right angle,
, and
. Let
be randomly chosen inside
, and extend
to meet
at
. What is the probability that
?
Solution
Clearly and
. Choose a
and get a corresponding
such that
and
. For
we need
, creating an isosceles right triangle with hyptenuse
. Thus the point
may only lie in the triangle
. The probability of it doing so is the ratio of areas of
to
, or equivalently, the ratio of
to
because the triangles have identical altitudes when taking
and
as bases. This ratio is equal to
. Thus the answer is
.
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.