Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
− | Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot | + | Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math> |
Revision as of 08:01, 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