Difference between revisions of "2013 AMC 10B Problems/Problem 16"
(→Solution) |
(→Solution) |
||
Line 37: | Line 37: | ||
pair A,B,C,D,E,P; | pair A,B,C,D,E,P; | ||
A=(0,0); | A=(0,0); | ||
− | B=( | + | B=(800,0); |
− | C=( | + | C=(200,400); |
− | D=( | + | D=(500,200); |
− | E=( | + | E=(400,0); |
− | P=( | + | P=(333,133); |
draw(A--B); | draw(A--B); | ||
draw(B--C); | draw(B--C); |
Revision as of 20:34, 22 February 2013
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
Solution
Let us use mass points:
Assign mass
. Thus, because
is the midpoint of
,
also has a mass of
. Similarly,
has a mass of
.
and
each have a mass of
because they are between
and
and
and
respectively. Note that the mass of
is twice the mass of
, so AP must be twice as long as
. PD has length
, so
has length
and
has length
. Similarly,
is twice
and
, so
and
. Now note that triangle
is a
right triangle with the right angle
. This means that the quadrilateral
is a kite. The area of a kite is half the product of the diagonals,
and
. Recall that they are
and
respectively, so the area of
is
Solution 2
Note that triangle is a right triangle, and that the four angles that have point
are all right angles. Using the fact that the centroid (
) divides each median in a
ratio,
and
. Quadrilateral
is now just four right triangles. The area is