2003 AMC 12A Problems/Problem 16
Contents
[hide]Problem
A point P is chosen at random in the interior of equilateral triangle . What is the probability that
has a greater area than each of
and
?
Solution
Solution 1
After we pick point , we realize that
is symmetric for this purpose, and so the probability that
is the greatest area, or
or
, are all the same. Since they add to
, the probability that
has the greatest area is
Solution 2
We will use geometric probability. Let us take point , and draw the perpendiculars to
,
, and
, and call the feet of these perpendiculars
,
, and
respectively. The area of
is simply
. Similarly we can find the area of triangles
and
. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose
= the height of the triangle. Setting the area of triangle
greater than
and
, we want
to be the largest of
,
, and
. We then realize that
when
is the incenter of
. Let us call the incenter of the triangle
. If we want
to be the largest of the three, by testing points we realize that
must be in the interior of quadrilateral
. So our probability (using geometric probability) is the area of
divided by the area of
. We will now show that the three quadrilaterals,
,
, and
are congruent. As the definition of point
yields,
=
=
. Since
is equilateral,
is also the circumcenter of
, so
. By the Pythagorean Theorem,
. Also, angles
, and
are all equal to
. Angles
are all equal to
degrees, so it is now clear that quadrilaterals
are all congruent. Summing up these areas gives us the area of
.
contributes to a third of that area so
.
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |
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